1.Ionic Equilibrium(1-26) (1)
Short Description
Relative Strength of Inorganic Acids * Hydracids of the elements of the same period: As we know, if the charge is spre...
Description
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IONIC EQUILIBRIUM CH4 (10-58) < NH3 (10-35) < H2O (10-14)< HF (10-4) · Basicity constant is decreasing therefore, the acidic character is increasing
SYNOPSIS Relative Strength of Inorganic Acids Hydracids of the elements of the same period: As we know, if the charge is spread over larger volume then, charge density is small and basic character of the ion is also small because of the ability of ion to attract a proton becomes less and hence acidic strength of conjugate acid increases. · Now, consider the hydracids of 2nd period CH4, NH3, H2O, HF. We find that, from left to right · Size of central atom over which the negative charge is present is decreasing · The volume of central atom overlapped by hydrogen is 3/4th, 2/3rd, ½ and 1 respectively. H
C– H
N– H
CH3–
H
O– H
NH–2
F–
(CH 4 NH 3 H 2O HF) .
*
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* *
H OH–
F–
Increasing volume available to electron Increasing electron delocalisation Decreasing electron density Decreasing bascity
· Volume available for negative charge is increasing in the conjugate bases (CH 3 NH 3 OH F ) , therefore, charge
density of the conjugate bases is decreasing · Basicity of the conjugate bases is decreasing and acidity of the acids is increasing · Stability of conjugate bases is increasing (CH 3 NH 3 OH F )
· Acid dissociation constant is increasing Narayana Junior Colleges
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Hydracids of the elements of the same group: In each Group : from top to bottom Atomic size of the central atom is increasing. Volume available for the negative charge is increasing Charge density is decreasing Basicity of the ions is decreasing Acidity of conjugate acids is increasing Acidity constant is increasing Basicity constant is decreasing Therefore; order of acidic character can be also explained as following along with the above reasons a) VII A group (Halogens) - HF < HCl < HBr < HI (Due to decreasing bond energy of H - X bond). b) VI A group - H2O < H2S < H2Se < H2Te (Due to decreasing trend in electron donor ability of OH-, HS-, HSe-, HTe- ions). c) V A group - NH3 < PH3 < AsH3 < BiH3 (Due to decreasing order of electron density). d) IV A group - CH4 < SiH4 < GeH4 < SnH4 < PbH4 (Due to decreasing order of electronegativity.) e) III A group - BH3 < AlH3 < GaH3 < InH3 < TlH3 (Due to decreasing order of electronegativity.) Among hydrides of elements with same electronegativity, acidic strength increases with size of the central atom e.g. CH 4 H 2S HI .
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The ability of Borontrihalides to act as Lewis acid increases in the order of BF3 < BCl3 < BBr3 < BI3 due to decreasing order of strength of bond.
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X
X
B
X
X
X
X
Acids that require only one electron pair to complete an outer shell are stronger than those requiring two. Thus GaCl3 > ZnCl2. Oxyacids Oxyacids with same central atom at different oxidation state: Among oxyacids of the same central atom having general formula. (HO)m ZOn Where Z = central atom m = no of oxygen atoms attached to z and attached to H. n = No of oxygen atoms attached to z but not attached to H. Acidic strength increases with increasing oxidation no e.g.-
* *
1
3
5
7
Cl HO Cl O HO Cl O 2 HO Cl O 3 a) HO very weak weak strong
very strong
They give OCl, OClO, OClO2 , OClO3 anions after removal of a proton. The species which have more than one oxygen atom show resonance. We see that from left to right, acidic strength increases Due to : · More the no. of oxygen atoms, more the resonance. · Volume available for the negative charge is increasing · Charge density is decreasing · Basicity of the anion is decreasing · Acidity of the conjugate acid is increasing. 4
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c) HPO3 < HNO3 d) H3AsO4 < H3PO4. Among above; acidity increases from left to right Due to: · Decreasing size of the central atom · Increasing order of the electro negativity · Increasing order of pull of electron of O-X bond towards X. · Resulting into increasing positive character on oxygen. · Ultimately, increasing order of release of H+. The strength of oxyacids increases from left to right across a period e.g. H2SiO4 < H3PO4 < H2SO4 < HClO4, Which is attributable to increase in electronegativity of the central atom, Hydrated metal ions: Under favourable conditions, one or more protons may dissociate to form the coordinated aqua groups. [M(H 2 O)6 ]n H 2 O
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B
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IONIC EQUILIBRIUM
( Hydrated metal ion )
M(H 2O)5 OH
n 1
H 3O
For hydrated metal ion or aqua acids, acidity increases with the increase of positive charge of the central metal ion and with decreasing ionic radius. [Fe(H2O)6]3+ > [Fe(H2O)6]2+. Exceptions are some times seen due to the effect of covalent bonding. [Fe(OH 2 )6 ]2 [Fe(OH 2 )6 ]3 [Al(OH 2 )6 ]3 [Hg(OH 2 )6 ]2
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Oxyacids of Phosphorous
H 3 PO2 H 3 PO3 H 3 PO4 Relative Strength of Inorganic Bases
6
b) H 2 SO3 H 2 SO 4 3
5
c) H NO 2 H NO3 *
Oxyacids with different central atom of same oxidation state and same configuration. a) HOI < HOBr < HOCl b) HIO4 < HBrO4 < HClO4 2
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In Period: The basicity of the compound decreases from left to right along a period e.g.: NH3 H 2O HF
Due to: a) Increase in the electronegativity of the atom holding the electron pair(s) Narayana Junior Colleges
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b) Decrease in the availability of electron pair(s) for sharing with the proton. In group: The basicity of a compound decreases from top to bottom in a group e.g. : a) F > Cl > Br > I b) O2 > S2 c) NH3 > PH3 > AsH3 > SbH3 > BiH3. Due to: a) Increase in the size of atom holding the unshared electron pair(s). b) Decrease in the availability of electrons. Presence of negative charge on the atom holding the electron pair(s) increases the basicity while presence of positive charge decreases basicity
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H O+ HA + H2O(l) 3
C1(1 - 1) C2(1 - 2)
(B)
pH calculation of solution of a mixture of two weak Monobasic Acids in water Let two weak acids be HA and HB and their conc. are C1 and C2, 1 is the degree of dissociation of HA in presence of HB (due to common ion effect) and 2 be degree of dissociation of HB in presence of HA. In aqueous solution of HA and HB following equilibrium exists.
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C1 1 + B-
(C1 1 + C2 2)
C2 2
K a[HA]
[H 3O ][A ] [C11 C2 2 ][C11 ] [HA] C1 (1 1 )
K a[ HB]
[H 3O ][B ] [C11 C2 2 ][C2 2 ] [HB] [C2 (1 2 )]
pH – log[H ] – log[C11 C 2 2 ] = -
log K a C1 K a C2 1
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(A)
Alkali and Alkaline Hydroxides: Basic nature increases on going down the group e.g. a) LiOH < NaOH < KOH < RbOH < CsOH b) Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2. pH OF THE MIXTURES : pH for the mixture of Weak Acid and Strong Acid Let strong acid be HB whose conc. is C1 + HB H + B 0 C1 C 1 and weak acid whose concentration is C2 and degree of dissociation of HA = H+ + A C2(1- ) C2 C2 Total (H+) conc. = C1 + C2 pH = - log [C1 + C2 ]
C 1 1 + C2 2
H O+ HB + H2O(l) 3
e.g. H O H — OH H3 O *
+ A-
2
where K a = Ka(HA) and K a = Ka(HA) 1
(C)
2
pH of a dibasic Acids and Polyprotic Acid Let's take the e.g. of a dibasic acid H 2A. Assuming both dissociation is weak. Let the initial conc. of H2A is C and 1 and 2 be degree of dissociation for first and second dissociation. HAH2A
H+
+
C(1 - 1) C 1 (1 - 2) HA-
H+
+
C 1 + C 1 2 A2
C 1 (1 - 2) C 1 + C 1 2 C 1 2. Ka1
[HA ][H ] [H 2 A]
Ka1
[C1 (1 2 )][C1 C1 2 ] C(1 1 )
Ka 2
[H ][A ] [C1 C1 2 ][C1 2 ] [HA ] [C1 (1 2 )]
After solving for 1 and 2. We can calculate the H+ conc. [H+] = C 1 + C 1 2 pH = - log [C 1 + C 1 2] 3
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(D) pH of mixture of acids Let one litre of an acidic solution of pH = 2 be mixed with two litre of other acidic solution of pH = 3. The resultant pH of the mixture can be evaluated in the following way. Sample 1 Sample 2 pH = 2 pH = 3 [H+] = 10-2 M [H+] = 10-3 M V = 1 litre V = 2 litre M1V1 + M2V2 = MR(V1 + V2) 10-2 1 + 10-3 2 = MR(1 + 2) 12 103 MR 3
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pH R – log[4 10 –3 ] 2.3980
pKa and pKb for a conjugate acid-base pair For an acid HX H X – HX
Ka
–
[H ][X ] ............... (A) [HX]
For conjugate base X – of acid HX
HX OH – X – H 2O –
Kb
[HX][OH ] [X – ]
K a K b [H ][OH – ] K w
or pK a pK b pK w 14 Note : 1. Stronger is acid, weaker is its conjugate base. 2. Higher is the value of pK a of an acid, lower is acid strength and higher is basic strength of its conjugate base. Hydrolysis of Salts Salts are strong electrolytes when dissolved in 4
NH 4 OH HCl NH 4 Cl H 2 O NH 4OH H Cl NH 4 Cl H 2 O
NH 4 OH H NH 4 H 2O
(acidic)
Kh
*
[NH 4OH][H ] [NH 4 ]
Relation between K h , K b and K w : NH 4 OH NH 4OH
..................... (B)
By eqs. (A) and (B),
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4 103 M R (Here, MR = Resultant molarity)
water, they dissociated almost completely into cation or anions. If anion interacts with water it is called as anionic hydrolysis. A + H2O HA + OH Akaline solution (pH increases). If cation intereacts with water it is called as cationic hydrolysis. + B+ + 2H2O B(OH) + H3O A c i d i c solution (pH lowers down). "The phenomenon of the interaction of anions and cations of the salt with H+ and OH- ions furnished by water yielding acidic or alkaline solution is known as salt hydrolysis. For the study of hydrolysis, salts are divided into four groups. 1) Hydrolysis of salt of strong Acid and weak base: NH4Cl is a salt of weak base (NH4OH) and strong acid (HCl). After hydrolysis resultant solution will be acidic due to presence of strong acid HCl.
Kb
[NH 4 ][OH ] [NH 4OH]
… (A)
H OH H 2O K w [H ][OH ] Dividing (B) by (A)
… (B)
Kw [H ][OH ] [NH 4OH] K b [NH 4 ][OH ] Kw Kh Kb Narayana Junior Colleges
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H2O H OH
Degree of hydrolysis NH 4 OH NH 4 H 2 O C(1 h)
Ch . Ch Kh Ch 2 C(1 h)
h
h
K w [H ][OH ]
Ch
Ch
Dividing equation (B) ÷ (A) (1 - h 1)
Kw [H ][OH ][CH3COOH] [OH ][CH3COOH] Kh Ka [CH3COO ][H ] [CH3COO ]
Kh C
*
t=0 t=t
1 pH [pK w pK b log C] 2 1 1 pH 7 pK b log C 2 2 2) Hydrolysis of salt of weak acid and strong base: CH 3COONa is a salt of weak acid (CH3COOH) and strong base (NaOH). After hydrolysis resultant solution will be basic due to presence of strong base (NaOH).
CH 3COOH NaOH CH 3COONa H 2 O()
CHCOO Na H2O() Na OH CHCOOH 3 3
Kh
[CH 3COOH][OH ] [CH 3COO ]
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Ch . Ch Ch 2 C(1 h)
h
Kh C
h
Kw KaC
[ [OH ] Ch [OH ] C
[H3O ]
Kw Kw C Ka C Ka
Kw Kw [OH ]
Ka Ka K w Kw C C
Taking -log both sides log[H ]
1 1 1 log K w log K a log C 2 2 2
1 1 pK a log C 2 2 pH will be more than 7, hence resultant solution will be basic in nature. 3) Hydrolysis of salt of Weak Acid and Weak Base: Let's take the salt CH3COONH4 made of salt of weak acid (CH3COOH) and Weak base (NH4OH).
pH = 7
CH 3COO H CH 3COOH
[CH 3COO ][H ] [CH 3COOH]
0 Ch
1 [pK w pK a log C] 2
Relation between, K h , Kw and Ka
Ka
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Kw C Kb
0 Ch
if h is smaller than 0.1 then, (1 - h) = 1.
Taking -log10 both side
C C(1 - h) Kh
Kw [H ] Ch C KbC
*
Degree of Hydrolysis CH 3COOH OH CH 3COO H 2 O()
Kw K bC
[H ]
… (B)
… (A)
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CH 3COOH NH 4 OH CH 3COONH 4 H 2 O
neglect the acetate ion coming from CH 3COOH – Therefore CH3COOH CH3COO H
CH 3COOH NH 4 OH CH 3COO NH 4 H 2O
Ka
C(1 – )[H ] C
Ka
[H ]
Relation B/w, K h , K w , K a & K b CH 3COO H CH 3COOH
Ka
[CH 3COO ][H ] [CH 3COOH]
[H ] K a K a
… (A)
[NH 4 ][OH ] [NH 4OH]
… (B)
K w [H ][OH ] … (C)
Kw [H ][OH ] [CH 3COOH][NH 4 OH] Ka K b [CH 3COO ][H ][NH 4 ][OH ] Kw Kh Ka Kb
at the first glance that CN ions hydrolysed to a much greater extent than NH 4 ions. However,, the hydrolysis of CN- ions produced OH- ions according to the equation. HCN OH CN H 2 O
Degree of Hydrolysis CH3COOH NH4OH CH3COO NH4 H2O
t=0 C t = t C(1 - h)
Kh
C C(1 - h)
Kw K w Ka Ka Kb Kb
ions are hydrolysed, if we assume K a K b , then the hydrolysis of the cation and anion of the salt occur approximately to equal extent for a salt which has K a K b , it would be expected
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H OH H 2 O
*
(1- 1)
This expression is independent of conc.of the salt. i) if Ka = Kb, pH = 7 solution will be neutral ii) if Ka > Kb, pH < 7, acidic solution iii) if Ka < Kb then pH > 7, alkaline solution In the hydrolysis of salt of weak acid and a weak base such as NH4CN, CH3COONH4. Both the
NH 4 OH NH 4 OH
Kb
C(1 – )
C
[CH 3COOH][NH 4OH] Kh [CH 3COO ][NH 4 ]
0 Ch
which can react with NH 4 ions as
0 Ch
NH 4OH OH NH 4
This latter reaction causes equilibrium in the form reaction to be displaced to the right. Because OH- ions are removed from the solution. Also the production of OH- by the former reaction displaces the latter reaction to the right. Therefore the hydrolysis of one ion drags the hydrolysis of the other ion along so that both the hydrolysis are fairly extensive not too far in extent from each other so it is fairly safe to assume that [HCN] = [NH4OH], even in the case of the salt where
Ch . Ch h2 C(1 h)C(1 h) (1 h)2
if h 0.1, 1 h 1 Kw h 1 h Ka K h The acetic acid formed would partially decompose to give CH 3COO – and H . But because of common ion effect (that is, due to the
*
Ka K b . Buffer capacity or Buffer Index
unhydrolysed CH 3COO – ) it is possible to 6
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Buffer capacity of a solution is defined in terms of buffer index which is the change in the concentration of Buffer acid (or base) required for change of it's pH value by one, keeping (Csalt + Cacid) or (Cbase + Csalt) constant. Let there be a buffer solution of volume 1 litre with `x' mole of acid and `S' moles of `salt'. pH = pKa + log10 pH pK a
pK a log 0.1 pH pK a log10 10 (pKa 1) pH (pKa 1)
*
S x S
1 S log e 2.303 x S
(pH) 1 1 1 S 2.303 S x S
(pH) 1 x S S S 2.303 S(x S)
Maximum value of Buffer Index B.I = 2.303
S(x S) x
d 1 (B.I) = 2.303 (x 2S) x dS
for maximum value of Buffer index d (B.I) = 0 dS After solving S = x/2
Thus
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[Salt] S x/2 1 [Acid] x S x x / 2
H Ph HPh (Pink)
Colourless
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S S(x S) 2.303 (pH) x
Outside this range the Buffer capacity is too small to be of any practical application. Acid-base Indicators An acid & base indicator is a substance which changes it's colour within limits with variation in pH of the solution to which it is added. Indicators, in general are either organic weak acids or weak bases with a characteristics of having different colours in the ionized and unionized form e.g. phenolphthalein is a weak acid (ionized form is pink and unionized form is colourless). Acidic Indicator Action (e.g. HPh) HPh (Phenolpthalein) is a colourless weak acid
K In
[H ][Ph ] [HPh]
If the solution is acidic, the H+ by the acid increases and since Kin is constant and it does not depend upon the concentration so HPh also increases means equilibrium will shift towards left means solution remain colourless. By addition of alkali, OH- will be furnished and that OH- will combines with H+ of HPh to form water and equilibrium will moves towards right and therefore solution becomes pink. Thus HPh appears colourless in acidic and pink in alkaline solution pH range of HPh is (8.3 -10).
[In ] [HIn]
Hence max. value of Buffer index occurs when,
pH pK In log
[Salt] 1 [Acid]
The colour of the indicator changes from colour A to colour B at a particular point known as end
Buffer Range It is difficult to give an exact limit upto which a buffer can be used it is generally accepted that a solution has useful buffer capacity provided that the value of [Salt]/[Acid] lie within the range of 10 to 0.1.Hence from Henderson equation
point of indicator. At this point [HIn] [In ]
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means pH pK In (at this point half of indicator is in the acid form and half in the form of its conjugate form. 7
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Indicator (Basic) action of MeOH (Methyl Orange) When MeOH is dissolved in water and undergoes dissociation to a small extent. The undissociated molecules are yellow while dissociated Me+ are red in colour.
ii)
Percentage of ionization of indicator is [In ] 10 [HIn] 100% 100% [I n ] [HIn] 10 [HIn] [HIn] 1000 91% 11 Thus most of the indicator is present in the ionized
Me OH MeOH red
colourless
If the solution is acidic, the H+ furnished by the acid combines with OH- ions furnished by the indicators to form undissociated water. This shifts the equilibrium towards right giving red coloured solution. Therefore in acid solution, this indicator gives red colour. In the presence of alkali, OHincreases and due to common ion effect the dissociation of MeOH supress means equilibrium will shifts towards left. Hence the solution in alkaline medium remains yellow in colour. Colour of solution depends upon relative amount of ionized form to unionized form (ratio of Me+/ MeOH). In general pH range of indicator lies between pK in 1 to pK in 1 i)
pH = pK in 1 I n 0.1 10% Means [HIn]
Percentage ionization of indicator would be In 0.10 HI n 100% 100% [I n ] [HI n ] 0.10 HIn (HI n ) 1 100 9.1% 11 Infact, pH = pKin - 1 is the maximum pH upto which the solution has a distinct colour characteristic of HIn. At pH smaller than this value, more of the indicator is present in the unionized form. Thus at pH pKin -1, the solution has a colour characteristic of HIn.
=
8
form I n and solution gets the characteristic colour. In fact pH = pKin + 1 is the minimum pH upto which the solution has a distinct characteristic colour of I n . At pH greater than this value, more of the indicator is present in the ionized form. Thus at pH pK in 1 , the solution has a colour Narayana Junior Colleges
yellow
at pH = pK in 1
[In ] Mean [HIn] 10
characteristics of I n . Ostwald's Theory According to this theory: a) The colour change is due to ionization of the acidbase indicator. The unionized form has different colour than the ionized form. b) The ionization of the indicator is largely affected in acids and bases as it is either a weak acid or a weak base. In case, the indicator is a weak acid, its ionization is very much low in acids due to common H+ ions while it is fairly ionized in alkalies. Similarly if the indicator is a weak base, its ionization is large in acids and low in alkalies due to common OH- ions. Considering two important indicators phenolphthalein (a weak acid) and methyl orange (a weak base), Ostwald theory can be illustrated as follows: Phenolphthalein: It can be represented as HPh. It ionizes in solution to a small extent as: H Ph HPh Pink
Colourless
[H ][Ph ] [HPh] The undissociated molecules of phenolphthalein are colourless while Ph- ions are pink in colour. In presence Applying law of mass action, K
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This theory also explains the reason why phenolphthalein is not a suitable indicator for titrating a weak base against strong acid. The OH- ions furnished by a weak base are not sufficient to shift the equilibrium towards right hand side considerably, i.e., pH is not reached to 8.3. Thus, the solution does not attain pink colour. Similarly, it can be explained why methyl orange is not a suitable indicator for the titration of weak acid with strong base. Quinonoid Theory: According to this theory: a) The acid-base indicators exist in two tautomeric forms having different structures. Two forms are in equilibrium. One form is termed benzenoid form and the other quinonoid form.
of an acid, the ionization of HPh is practically negligible as the equilibrium shifts to left hand side due to high concentration of H+ ions. Thus, the solution would remain colourless. On addition of alkali, hydrogen ions are removed by OH- ions in the form of water molecules and the equilibrium shifts to right hand side. Thus, the concentration of Ph- ions increases in solution and they impart pink colour to the solution. H 3O In HIn H2O
`Acid form'
`Baseform'
Conjuage acid-base pair
K In
[In ][H 3 O] ; [HIn]
K In Ionization constant of indicator,, [HIn] [In ]
pH = - log10 [H 3 O] = - log10 [Kin] - log10
[HIn] [In ]
[In ] pH = pKIn + log10 [HIn] (Handerson equation for indicator) At equivalence point; [In ] [HIn] and pH = pKIn Methyl orange: It is a weak base and can be represented as MeOH. It is ionized in solution to give Me+ and OH- ions.
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[H 3 O] K In
b) The two forms have different colours. The colour change is due to the interconversion of one tautomeric form into other. c) One form mainly exists in acidic medium and the other in alkaline medium. Thus, during titration the medium changes from acidic to alkaline or vice-versa. The change in pH converts one tautomeric form into other and thus, the colour change occurs. Phenolphthalein has benzenoid form in acidic medium and thus, it is colourless while it has quinonoid form in alkaline medium which has pink colour. OH OH H
C O
Me OH MeOH Yellow
O
C
OH
C
COO
OH
Re d
O
Applying law of mass action [Me ][OH ] [MeOH] In presence of an acid, OH- ions are removed in the form of water molecules and the above equilibrium shifts to right hand side. Thus, sufficient Me+ ions are produced which impart red colour to the solution. On addition of alkali, the concentration of OH- ions increases in the solution and the equilibrium shifts to left hand side (due to common ion effect), i.e., the ionization of MeOH is practically negligible. Thus, the solution acquires the colour of unionized methyl orange molecules, i.e., yellow. K
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Methyl orange has quinonoid form in acidic solution and benzenoid form in alkaline solution. The colour of benzenoid form is yellow while that of quinonoid form is red. CH 3
O 3S
NH
N
N
CH 3 Quinonoid form — Acidic solution (red) -
OH H
O 3S
N
N
CH 3 N CH 3
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Solubility and solubility Product A solution which remain in contact with excess of the solute is said to be saturated. The amount of a solute, dissolved in a given volume of a solvent (in 1 litre) to form a saturated solution at a given temperature, it termed as the solubility of the solute in the solvent at that temperature. Molar Solubility: No. of moles of solute dissolved in per litre of solution Solubility Product: In a saturated solution of a salt, there exists a dynamic equilibrium b/w the excess of the solute and ions furnished by that parts of the solute which has gone in solution. The solubility product of a sparingly soluble salt is given as product of the conc. of the ions raised to the power equal to the no. of times the ion occur in the equation after the dissociation of the electrolyte. xAy+ + yBx AxBy
K sp [A y ]x [Bx ]y Let the solubility of AxBy is S then
K sp [xS]x [yS]y x
y
xy
K sp x .y [S
]
The principle of solubility product is applicable for sparingly soluble salt. Calculation of Solubilities of Salts We shall now discuss the solubilities types of salts under various conditins. Solubilities of AgCl(salt of a strong acid and strong base) in Water : AgCl would disolve in water as, AgCI (S) Ag+ (aq) +Cl- (aq)’ At saturation point,
i)
AgCI(s)
Ag+ (aq) +Cl- (aq)
If the solubility of the salt is x moles/ lit. [ Ag+] = xM, [Cl ] = xM x2 = KSP x = K SP 10
ii)
Solubility of AgCl in a solution that is having 0.1M in AgNO3: AgCI would dissolve and finally reach saturation. AgCl(s) Ag+(aq) + Cl-(aq) The KSP of AgCl is approximately 10-10 . If AgCl were to be dissolved in water (pure), its solubility would have been 10-5M (previous section).In the presence of 0.1 M AgNO3 its solubility will decrease due to common ion effect. This means that [Ag+] from AgCl would be less than 10-5 M . Hence, we can ignore the contribution of Ag+ from AgCl. If the solubility of AgCl is x’ moles /L in the presence of 0.1 M AgNO3, then [Ag+] = 0.1 M, [Cl-] = x’ M x’ =
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K SP = 10-9 moles / l 0.1
Solubility of CH3COOAg(salt of weak acid and strong base) in water: CH 3COOAg dissolves and reaches saturation.Since it is a salt of weak acid and strong base, it would hydrolyse. If the solubility of the salt is x moles /then
CH 3COO aq Ag aq CH 3COOAg s At eq : x-y x CH3COO aq H2O CH3COOH aq OH
x-y y y where y is the amount of CH3COO ion that is hydrolysed. (x - y) x = KSP y2 K w (x y) K a
iv)
Knowing the values of KSP and Ka, solubility of the salt can be calculated. Solubility of CH3COOAg (salt of a weak acid and strong base) in an acid buffer of pH = 4 (assuming that the buffer does not have any common ion by CH3COOAg): CH 3 COOAg would dissolve and reach equilibrium. It would then be hydrolysed.If the solubility of the salt is x’ M in this solution ,then At eq.; Narayana Junior Colleges
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Knowing Ksp and Kf , the solubility can be calculated. The common ion presence in the solution decrease the solubility of a given compound e.g. The solubility of BaSO4 in Na2 SO4 solution is smaller than that in an aqueous solution . Common-ion Effect on Solubility The common ion presence in the solution decrease the solubility of a given compound e.g. The solubility of BaSO4 in Na2SO4 solution is smaller than that in an aqueous solution. Consider saturated solution of AgCl. If a salt having either of the ion common to AgCl say KCl is added to it, then
CH 3COO aq Ag aq CH 3COOAg s
x’ - y’ x’ CH3COO (aq) + H2O CH3COOH (aq) + OHx’- y’ y’ 10-10 Since the solution is a buffer, the pH will be maintained. (x ’ - y’) x’ = KSP -
y2 K w (x y) K a
knowing KSP and K a, the solubility can be calculated. Solubilty of CH3COOAg in an buffer solution of pH =9: Following the same logic as give in the earlier section, CH3COOAg(s) CH3COO- (aq) + Ag+ (aq) At eq : x’’- y’’ x’’ CH3COO (aq) + H2O CH3COOH + OHAt eq : x’’- y’’ y’’ 10-5 where x’’ M is the solubility of the salt and y’’ the extent to which it is hydrolysed. ( x’’ - y’’) x’’ = Ksp y '' 105 K w (x '' y '') K a
Ag Cl – AgCl(s) aq. KCl(s) aq. K Cl –
For AgCl
KCl and thus to have K SP constant, [Ag ] will decrease or AgCl will precipitate out from solution, i.e., solubility of AgCl will decrease with increasing concentration of KCl in solution. Let 0.1 M KCl(aq.) solution with AgCl(aq.) . If solubility of AgCl is s mol litre-1, then,
Knowing K sp and K a, the solubility can be calculated. vi)
For AgCl
AgCI (s)
y Kf (x y)(a 2y)2
( formation constant of [Ag(NH3)2]+ Narayana Junior Colleges
K SP s 0.1 or s AgCl
-
Ag (aq) + Cl (aq) x-y x + Ag (aq) + 2NH3 (aq) [Ag(NH3)2]+(aq) x-y a - 2y y Where y is the amount of Ag+ which has reacted with NH3. (x -y ) x = Ksp
K SP [Ag ][Cl – ]
KSP s(s 0.1) is being small in comparison to 0.1 and thus may be neglected therefore,
Solubility of AgCl in an aqueous solution containing NH3. Let the amount of NH3 initially be ‘a’ M. If the solubility of the salt is x moles /L, then +
K SP [Ag ][Cl – ]
[Cl – ] Increases in solution due to presence of
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v)
K SP 0.1
where s is
solubility of AgCl in presence of 0.1 M KClaq. *
Ionic Product For a solution of a salt at a specified concentration, the product of the concentration of the ions, each raised to the proper power, is called as the ionic product for a saturated solution in equilibrium with excess of solid, the ionic product is equal to solubility product. At equilibrium, ionic product = solubility product 11
JEE ADVANCED-VOL - III
and CrO 4–2 both of these ions form ppt with Ag+ though the Ksp (AgCl) > Ksp (Ag2CrO4). Yet it is AgCl (less soluble) which ppted first when Ag+ ions added to the solution. In order to predict which ion (Cl- or CrO 42 ) ppt first. We have to calculate the conc. of Ag+ ion needed to start ppt through the Ksp and given conc. of Cl- and CrO 42 , if the conc. of Ag+ ions needed to start
i) Solubility of a salt of weak acid and strong base in Basic Buffer suppresses than pure water due to common ion effect. But in acidic buffer solution soubility increase than pure water. ii) Solubility of salt of weak acid and weak base in pure water: Let the solubility of salt be S, and y mol/litre is the amount of salt getting hydrolysed. CH3COONH4 CH3COO- + NH 4 … (A)
S-y
NH4OH S-y S-y
as AgNO3 is added to the solution, the minimum of the two conc. of Ag+ to start the ppt will be reached first and thus the corresponding ion (Clin this case) will be ppted in preference to the other. During the course of ppt conc. of Cldecreases and conc. of Ag+ increases when its's conc. become equals to the conc. required (of Ag+) for . At this stages the whole of Cl- ions have been ppted the addition of more of AgNO3 causes the ppt of both the ions together. 12
y
y ... (B)
K sp (S y)(S y) (S y) 2 Due to hydrolysis of salt from equation (B) Kh
[CH 3COOH][NH 4OH] y.y [CH 3COO ][NH 4 ] (S y)(S y) 2
y Kh S y and we also know that Kh
Kw Ka Kb
Solubility of a salt of weak acid and weak base in acidic buffer Let the solubility of salt be S and y be the amount of weak acid being formed. CH COO- + NH 4 CH3COONH4 3
S-y y CH3COO + H CH3COOH … (B) S - y (from Acidic Buffer) y -
the ppt of CrO 42 is larger than that of Cl-. Hence
S-y
CH3COO - + NH 4 + H 2O CH 3COOH +
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If ionic product is less than solubility product it means solution is unsaturated and more salt can be dissolve in it. If ionic product is greater than solubility it means solution is holding more salt than it can dissolve in therefore precipitation starts and continues till, ionic product becomes equal to Ksp. Preferential Precipitation of Salts Frequently, a solution contains more than one ion capable of forming a ppt. with another ion which is added to the solution. e.g., in a solution containing Cl , Br and I , if Ag+ ions are added then out of the three, the least soluble silver salt is ppt first. If the addition of Ag+ ions is continued, eventually a stage is reached when the next lesser soluble salt starts ppt along with the least soluble salt and so on. If the stoichiometry of the ppted salts is the same, then the salt with the minimum Ksp or minimum solubility will ppted first followed by higher Ksp. If the stoichiometry of the ppted salts is not the same, then with Ksp alone, we can't predict which ion will ppted first. e.g. a solution containing Cl-
IONIC EQUILIBRIUM
+
K sp [CH3COO ][NH 4 ] = [S - y] [y] = y [S - y] for equation (B) 1 Ka(CH3COOH)
Ka
[CH3COOH] y [CH3COO ][H ] (S y)(H )
Solubility of CH3COONH4 in acidic buffer would be higher than in pure water Solubility of a salt of weak acid and weak base in basic buffer Narayana Junior Colleges
JEE ADVANCED-VOL - III
IONIC EQUILIBRIUM
CH COO- + NH 4 Similarly CH3COONH4 3
y NH 4 OH
S-y
S-y
NH4OH
(from buffer)
y
K sp (CH3COONH4 ) [CH3COO ][NH 4 ] K sp y(S – y)
(ii) Weak acid and strong base: p H curve
1 [NH 4 OH] y Kb Kb [NH 4 ][OH ] (S y)[OH ]
of weak acid (say CH 3COOH or oxalic acid) and strong base (say NaOH) is
The solubility of CH 3COONH 4 in basic buffer
*
Acid- Base Titrations: In acid-base titrations, pH of solution changes as the reaction prooceeds and if pH of solution is plotted against volume of tirant, the curves obtained are known as titration curves or neuttralization curves or pH curve. Generally, they are S-shaped if base is being used as titrant. Titration curves are useful in detection of end point and around the equivalence point, titration curve remains almost vertical. It indicates that initially pH changes very slowly but change is rapid near equivalence point. After equivalence point again pH change becomes slow. The extension of vertical zone gives idea about pH range and correct choices of the indicators. For a particular titration, the indicator should be so selected that it changes its changes its colour within vertical distance of the curve. (i) Strong acid and strong base: p H curve
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would be higher than pure water.
vertical over the approximate pH range 7 to11. So, phenolphthalein is the suitable indicator for such a titration. (iii) Strong acid and weak base: pH curve of strong acid (say Hcl or H 2 SO4 or HNO3 ) with a weak base say NH 4OH is vertical over the pH range of 4 to 7. So, the indica tors methyl red and methyl orange are suitable for such a titration.
of strong acid (say HCl)and strong base base (say NaOH) is vertical over almost the p H range 4-10. So, the indicators phenolphtthalein ( p H range 8.3 to 10.5), methyl red ( p H range 4.4 - 6.5) and methyl orange ( p H range 3.2 - 4.5) are suitable for such a titration. Narayana Junior Colleges
13
JEE ADVANCED-VOL - III
IONIC EQUILIBRIUM
(iv) Weak aicd and weak base: pH curve of weak acid and weak base indiactes that there is no vertical part and hence, no suitable indicator can be used for such a titration.
LEVEL-IV SINGLE ANSWER QUESTIONS 1.
A solution contains 0.1 M H2S and 0.3 M HCl. Calculate the conc. of S2- and HSions in solution. Given K a1 and K a 2 for H2S are 10-7 and 1.3 10-13 respectively.. (A) 1.44 10 19 ,3.3 10 7 (B) 1.44 10 20 ,3.3 10 8 (C) 1.44 10 22 ,3.3 10 3
pH 8.5 indicates conversion of carbonate into bicarbonate. Na2CO3 HCl NaHCO3 NaCl As the inflection point lies in the pH range 8 to 10, phenolphthalein can be used to indicate the above conversion. The second inflection
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2. (v) Titration of diacidic base (sodium car bonate) with strong acid. pH curve of sodium carbonate with HCl shown two inflection points. First inflection point
3.
CrO 24 . If solid AgNO3 is gradually added
to this solution, what will be the concentration of Cl when Ag2CrO4 begins to precipitate?
point pH 4.3 indicates the following reac tion:
(Ksp (AgCl) = 10-10M2, Ksp (Ag2CrO4) = 10-12 M3)
NaHCO3 HCl NaCl CO2 H 2O As the point lies between 3 to 5, methyl or ange can be used. 4.
5.
14
(D) 1.44 10 18 ,3.3 10 5 An aqueous solution of metal bromide MBr2 (0.05M) is saturated with H2S. The minimum pH at which MS will ppt. will be Ksp for (MS) = 6 10-21 Concentration of saturated H2S = 0.1 Ka1(H2S) = 1 10-7 Ka2(H2S) = 1.3 10-13 (A) 0.982 (B) 0.0983 (C) 1.96 (D) 2.96 A solution of 0.1 M is Cl and 10-4 M
A) 10-6 M B) 10-4 M C) 10-5 M D) 10-9 M One litre of a buffer solution containing 0.02 mol of propanoic acid and some sodium propanoate has pH = 4.75. What will be the pH if 0.01 mol of hydrogen chloride is dissolved in the above buffer solution ? [Dissociation constant of propanoic acid at 250C is [1.34 10-5. ] (A) 4.11 (B) 2.11 (C) 0.11 (D) 3.11 A weak base (BOH) with Kb = 10-5 is titrated with a strong acid, HCl. At 3/4th of the equivalence point, pH of the solution is: (A) 5 + log3 (B) 14 - 5 - log3 (C) 14 - 5 + log3 (D) 9.523. Narayana Junior Colleges
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IONIC EQUILIBRIUM
7.
8.
9.
If the pH of solution is adjusted to 9.0. How much Mg2+ ion will be precipitated as Mg(OH)2 from a 0.1M MgCl2 solution at 25°C? Assume that MgCl2 is completely dissociated. (A) 0.011 (B) 0.089 (C) 0.11 (D) 0.89 Two buffers, (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ? (KHA = 1.0 10-5). (A) 4.7033 (B) 5.7033 (C) 6.7033 (D) 8.7033 0.1 mole of CH3NH2 (Kb = 5 10-4) is mixed with 0.08 mole of HCl and diluted to one litre.
JEE 2005) (A) 8 10-2 M (C) 8 10-15 M
(B) 8 10-11 M (D) 8 10-5 M
K
sp
12.
c) NaCN NH 4CI NaCI HCI 13.
D) 2 102 M BaCl and 2 102 M NaF 2 Narayana Junior Colleges
d) HCI NaCI NaCN NH 4CI The following pH range where the indicator shows change in colour are given i. 4 – 9.7 ii. 7.46 – 10.0 iii. 6.5 – 4 which of the above pH range represent titration of I. Strong acid/strong base (SA/SB) II. Weak acid/ strong base (WA/SB) III. weak base/strong acid (WA/SA) a. i I , ii II , iii III b) iii I , ii II , i III c. ii I , iii II , i III
14.
15.
A) 103 BaCl and 2 102 M NaF 2
C) 1.5 102 M BaCl and 10 2 M NaF 2
a) 2.0 b) 2.4 c) 2.8 d) 1.6 The pH of 0.1M solution of the following salts increases in the order. a) NaCl NH 4Cl NaCN HCI b) HCI NH 4CI NaCI NaCN
of BaF2 1.7 107
B) 103 M BaCl and 1.5 102 M NaF 2
2
aq H 3O aq
is 6.5 103 , what is the maximum pH value which could be used so that at least 80% of the total iron (III) in a dilute solution exists as Fe3 ?
If equal volumes of BaCl2 and NaF solutions are mixed which of these combination will not give a precipitate?
K a for the reaction
F e 3 aq H 2 O l F e O H
The K sp of Mg(OH)2 is 8.9 10-12 at 25°C.
What will be the H in the solution. (IIT
10.
11.
pKa (CH3COOH) is 4.74. x mol of lead acetate and 0.1 mol of acetic acid in one L solution make a solution of pH = 5.04. Hence, x is : (A) 0.2 (B) 0.05 (C) 0.1 (D) 0.02.
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6.
16.
d) i I , iii II , ii III What will be the pH of an aqueous solution of 1.0 M ammonium formate? Given :- pK a 3.8 and pK b 4.8 A) 7.5 B) 3.4 C) 6.5 D) 10.2 The percentage of hydrolysis of a salt of weak acid (HA) and weak base (BOH) in its 0.1 M solution is found to be 10 %. If the molarity of the solution is 0.05 M. the percentage hydrolysis of the salt should be A) 5% B) 10% ‘ C) 20% D) None of these 0.1 M formic acid solution is titrated against 0.1 M NaOH solution . What would be the difference in pH between 1/5 and 4/5 stages of neutralization of acid? A) 2 log ¾ B) 2 log 1/5 C) log 1/3 D) 2 log 4 15
JEE ADVANCED-VOL - III
17.
IONIC EQUILIBRIUM
The pH of a solution containing 0.4 M HCO32 is: Ka H 2CO3 4 107 ; K a HCO3 4 1011 2 1
18.
A) 10.4 B) 10.1 C) 6.1 D) 10.7 The pH of the resultant solution of 20 ml of
C)
0.1 M H3 PO4 and 20 mL of 0.1 M Na3 PO4 is
19.
A) pK a1 log 2
B) pK a1
C) pK a2
D)
pK a1 pKa2
2 In a saturated solution of AgCl, NaCl is added gradually, The concentration of Ag+ is plotted against the concentration of Cl . The graph appears as:
D)
A)
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20.
During the titration of a weak diprotic acid
H 2 A against a strong base NaOH , the pH of the solution at half –way to the first equivalent point and at the first equivalent point are given respectively by : A) pK a1 and pK a1 pK a2 B)
K a1 c and
C) pKa1 and
pK a1 pK a2 2
pK a1 pK a2 2
D) pK a1 and pK a2 B)
21.
Find the no.of moles of NH 4Cl required to prevent Mg OH
2 from precipitating in a
litre of solution which contains 0.02 mole of NH 3 and 0.001 mole of Mg2+ ions Given
:
K b NH 3 10 5 ; K SP Mg OH 2 10 11
A) 104 C) 0.02 16
B) 2 103 D) 0.1 Narayana Junior Colleges
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IONIC EQUILIBRIUM
22.
A 1.025 g sample containing a weak acid HX
28.
mol.wt. 82 is dissolved in 60 ml . water
where as second dissociation is 50% and third dissociation is negligible then the pH
and titrated with 0.25 M NaOH. When half of the acid was neutralized the pH was found to be 5.0 and at the equivalence point the pH is 9.0 . Calculate weight percentage of HX in sample A) 50% B) 75% C) 80% D) 60 % 23.
24.
of 4 103 M X OH 3 is: A) 11.78 B) 10.78 C) 2.5 29.
weak
triprotic
105 m K a2 109 , K a3 1013
acid
What is the value of pX of 0.1 M H 3 A (aq.)
H 2CO3 : pK a1 6.35, pK a2 10.33 log 3 0.477, log 2 0.30 A) 6.35 B) 6.526 C) 8.34 D) 6.173 Equilibrium constants of
A3 X HA2
solution? where pX= - log X and
A) 7 30.
5
20 ml of 0.1 M weak acid HA ka 10
A solution is 0.10 M in Ba NO3 2 and 0.10
K sp BaC rO 4 1.2 10 10 ; K sp SrCrO 4 3.5 10 5
31.
A) 7.4 107 B) 2.0 10 7 C) 6.110 7 D) 3.4 10 7 A solution is 0.01 M in KI and 0.1 M in KCl. If solid KI is added to the solution, what is the I when AgCl begins to precipitate?
K sp AgI 1.5 10 16 ; K sp AgCl 1.8 10 10
A / HA A in the resulting solution: A) 2 10 4 B) 2 105 C) 2 103 D) 0.05 Calculate the ratio of HCOO and F in a
D) 10
SrCrO4 beigns to precipitate?
is
mixed with solution of 10 ml of 0.3 M HCl and 10 ml of 0.1 M NaOH. Find the value of
C) 9
to the solution, what is Ba 2 , when
be diluted in
5
B) 8
M in Sr NO3 2 . If solid Na2CrO4 is added
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and H 2O are different at 298 K. Let at 298 K pure T2O has pT (like pH) is 7.62.The pT of a solution prepared by adding 10 ml, of 0.2 M TCl to 15 ml of 0.25 M NaOT is: A) 2- log 7 B) 14+ log7 C) 13.24 - log 7 D) 13.24 + log 7 To what volume, a soution of 10 litre of 0.5 order to double the hydroxide ion concentration: A) 20 L B) 30 L C) 40 L D) 50 L
27.
a1
is a
D) 2.22
0.1 M HCl . On adding 40 ml of HCl , pH of the solution will be
M CH 3COOH k a 10
26.
H3 A
K
50 m L of 0.05 M Na2CO3 is titrated against
T2 O T or 1 H 3 is an isotope of 1H 1
25.
If first dissociation of X OH 3 is 100%
32.
A) 3.5 107 B) 6.1 10 8 C) 2.2 10 7 D) 8.3 10 8 What is the minimum pH required to prevent the precipitation of ZnS in a solution that is 0.01 M ZnCl2 and saturated with 0.10 M
mixture of 0.2 M HCOOH ka 2 104
H2 S ?
and 0.1 M HF k a 6.6 10 4 :
[Given : K sp 1021 , K a1 ka2 1020 ]
A) 1 : 6.6 C) 2 : 3.3
A) 0
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B) 1 : 3.3 D) 3.3 : 2
B) 1
C) 2
D) 4 17
JEE ADVANCED-VOL - III
IONIC EQUILIBRIUM
38.
MULTIPLE ANSWER QUESTIONS
34.
35.
36.
Which of the following statement are true for a solution saturated with AgCl and AgBr. If their solubilities in mol lit-1 in separate solutions are x and y respectively. (A) [Ag+] = [Br-] + [Cl-] (B) [Cl-] > [Br-] (C) [Br-] > y (D) [Ag+] = x + y 10 mL of N/20 NaOH solution is mixed with 20 ml N/20 HC1 solution. The resulting solution will: (A) turn phenolphthalein solution pink (B) turn blue litmus red (C) turn methyl orange red (D) [H+]>[OH-] Choose the correct statement/s : a) pH of acidic buffer solution decreases if more salt is added b) pH of acidic solution increases if more salt is added c) pH of basic buffer decreases if more salt is added d) pH of basic buffer increases if more salt is added If you have a saturated solution of CaF2 , then 1/3
A) Ca 2 k sp / 4
B) 2 Ca 2 F C) Ca 2 2 F 37.
D) Ca 2 K sp If concentration of two weak acides are C1 and C2 mol / L and degree of ionization are 1 and 2 respectively then their relative strength can be compared by: H 1 A) H
2
C11 C) C 2 2 18
1 B) 2 K a1 C1
D) k C a2 2
third ionization constnat of H 3 PO4 respectively and K a1 K a2 K a3 which is / are correct: A) H ka1 H 3 PO4 2 B) H HPO4
39.
C) K a2 HPO42 D) K a2 HPO42 Which of the following mixtures can act as a buffer? A) NaOH +HCOONa (1: 1 molar ratio) B) HCOOH +NaOH(2 : 1 molar ratio) C) NH 4Cl NaOH (2 : 1 molar ratio) D) HCOOH + NaOH (1:1 molar ratio) COMPREHENSION TYPE QUESTIONS
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33.
If K a1 , K a2 and K a3 be the first , second and
Paragraph -I The solution which consumes [H+] or [OH-] or both simultaneously from externally added base in order to give negligible change in pH, is known as buffer solution. In general, the solution resists the change in pH. Buffer solution does not mean that there does not occur a pH change at all. It implies the pH change occurs but in negligible amount. There are two types of buffer(i) Acidic buffer : it is a mixture of weak acid and its slat with strong base. (ii) Basic buffer : it is a mixture of weak base and its salt with strong acid. 40. Which of the following mixture will be a buffer solution when dissolved in 500 mL of water ? (A) 0.200 mol of aniline and 0.200 mol of HCl (B) 0.200 mol of aniline and 0.400 mol of NaOH (C) 0.200 mol of NaCl and 0.100 mol of HCl (D) 0.200 mol of aniline and 0.100 mol of HCl 41. pH of 0.01 M (NH 4) 2SO 4 and 0.02 M NH 4OH buffer (pKa of NH4+ = 9.26) is (A) 4.74 + log2 (B) 4.74 - log 2 (C) 4.74 + log 1 (D) 9.26 + log 1 Narayana Junior Colleges
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IONIC EQUILIBRIUM
42.
To prepare a buffer of pH 8.26, amount of (NH4)2SO4 to be added into 500 mL of 0.01 M NH4OH solution [pKa (NH4+) = 9.26} (A) 0.05 mol (B) 0.025 mol (C) 0.01 mol (D) 0.005 mol Paragraph -II If a sparingly soluble salt is placed in water, after some time an equilibrium is established when the rate of dissolution of ions from the solid equals the rate of precipitation of ions from the saturated solution at a particular temperature. Thus, a dynamic equilibrium exists between the undissociated solid species and the dissolved ionic species in a saturated solution at a particular temperature. For example, in AgCl, we have the following equilibrium : AgCl (aq) Ag+ (aq) + Cl (aq.) The equilibrium constant
43.
44.
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For a sparingly soluble salt Ap Bq the relationship of its solubility product LS with its solubility S is (A) LS S p q . p p .q q (B) LS S p q . p q .q p (C) LS S pq . p p .q q (D) LS S pq . pq
46.
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Ag Cl K eq AgCl Keq [AgCl] = [Ag+] [Cl-] Ksp(AgCl) = [Ag+] [Cl-] [AgCl] is constant If there would not have been a saturated solution, then from equation (A), Keq. [AgCl] Ksp, but Keq. [AgCl] = QAgCl, where Q is ionic product. It implies that for a saturated solution, Q = Ksp Ksp is temperature dependent. When Q < Ksp, then the solution is unsaturated and there will be no precipitate formation. When Q = Ksp, then solution will be saturated, no ppt. will be formed when Q > Ksp, the solution will be supersaturated and there will be formation of precipitate. pH of a saturated solution of Ba(OH)2 is 12. Hence Ksp of Ba(OH)2 is: (A) 5 10-7 M3 (B) 5 10-4 M2 -6 3 (C) 1 10 M (D) 4 10-6 M3 A solution is a mixture of 0.05 M NaCl and 0.05 M NaI. The concentration of iodide ion in the solution when AgCl just starts precipitating is equal to: (Ksp AgCl = 1 10-10 M2; Ksp AgI = 4 10-16 M2) (A) 4 10-6 M (B) 2 10-8 M -7 (C) 2 10 M (D) 8 10-15 M
45.
pq
Slaked lime, Ca(OH)2 is used extensively in sewage treatment. What is the maximum pH that can be established in Ca(OH)2(aq.)
2+ Ca(OH)2(s) Ca (aq) + 2OH (aq); Ksp = 5.5 10-6 (A) 1.66 (B) 12.34 (C) 7 (D) 14 Paragraph -III Acidity or alkalinity of a solution depend upon the concentration of hydrogen ion relative to that of hydroxyl ions. The product of hydrogen ion & hydroxyl ion concentration is given by Kw = [H+] [OH-] the value of which depends only on the temperature & not on the individual ionic concentration. If the concentration of hydrogen ions exceeds that of the hydroxyl ions, the solution is said to be acidic; whereas, if concentration of hydroxyl ion exceeds that of the hydrogen ions, the solution is said to be alkaline. The pH corresponding to the acidic and alkaline solutions at 25ºC will be less than and greater than seven, espectively. To confirm the above facts 0.5 M CH3COOH is taken for the experiments. [Given : Ka of acetic acid = 1.8 ×10-5] 47. pH of the solution will be (A) 2.52 (B) 2.22 (C) 5 (D) 3.92 48. If pH of the solution is doubled, what will be the concentration of acetic acid (A) 1.8 × 10-5 M (B) 1.0 M 6 (C) 4.6 10 M (D) 1.25 × 10-3 M 49. To what volume at 25º C must 1 dm 3 of this solution be diluted in order to double the pH (A) 3.37 × 104 dm3 (B) 2.34 × 102 dm3 (C) 1.68 × 104 dm3 (D) 3.18 × 103 dm3
19
JEE ADVANCED-VOL - III
IONIC EQUILIBRIUM
pH pK a log
Conjugate base Acid
for basic buffer pOH pK b log
52.
Conjugate base Base
It is generally accepted that a solution has useful buffer capacity (pH change resisting power) provided that the value of [ salt or conjugate base]/ [acid] for acidic buffer lies within the range of 1:10 toi 10: 1. Buffer capacity is max. When[conjugate base ]= [acid] One litre of an aqueous solution contain 0.15
mole of CH 3COONa .After the addition of 0.05 mole of solid NaOH to this solution, the pH will be: A) 4.5 B) 4.8 C) 5.1 D) 5.4
Calculate the pH of a solution made by adding 0.01 mole of HCl in100 ml.of a solution which is 0.2 M in
NH 3 pKb 4.74 and 0.3 M in NH 4 :
54.
A) 5.34 B) 8.66 C) 7.46 D) None of these Use ful buffer range of weak acid HA ka 105 is :
55.
A) 5 to 7 B) 4 to 6 C) 3 to 5 D) None of these Select corect statement : A) When we add small amount of NaOH in acidic buffer solution,pOH of solution is increases B) When we add small amount of NaOH in basic buffer solution, pH of solution is increases C) When we add small amount of water in acidic buffer solution ,pH of solution is decreases D) When 100 ml of 0.2 M CH 3COOH react with 200 ml of 0.1 M NaOH buffer solution is formed MATRIX MATCHING QUESTIONS
:
mole of CH 3COOH pK a 4.8 and 0.15
20
53.
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Paragraph -IV In qualitative analysis, cations of group II as well as group IV precipitated in the form of sulphides. Due to low value of Ksp of group II sulphides, group reagent is H2S in presence of dil. HCl and due to high value of Ksp of group IV sulphides, group reagent is H2S in presence of NH4OH and NH4Cl. In a 0.1M H2S solution, Sn2+, Cd2+ and Ni2+ ions are present in equimolar concentration (0.1 M). Given: Ka1(H2S) = 10-7, Ka2(H2S) = 10-14, K sp (SnS) = 8 10 -29 K sp (CdS) = 10 -28 , Ksp(NiS) =3 × 10-21 50. At what pH precipitate of NiS will form (A) 12.76 (B) 7 (C) 1.24 (D) 4 51. Which of the following sulphide is more soluble in pure water(A) CdS (B) NiS (C) SnS (D) all have equal solubility Paragraph -V Solution of an acid and it’s anion (that is.it’s conjugate base ) or of a base and it’s common cation are buffered.When we add a small amount of acid or base to any one of them, the pH of solution changes very little.pH of butter solution can be computed as for acidic buffer :
56.
Consider a buffer of CH3COOH and CH3 COONa of maximum buffer-capacity and match the following : Column I (A) For maximum buffer capacity (B) Adding equal number of moles of CH3COOH and CH 3COONa (C) Diluting buffer 10 times (D) Adding some NaOH to buffer Column II (p) No change in pH (q) pH > pKa (r) pH = pKa (s) pOH = pKb Narayana Junior Colleges
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IONIC EQUILIBRIUM
57.
Match the following . Column- I Sparingly soluble salt of weak acid Ratio of solubilities of salt in buffer and in A) Ratio of solubility of a salt MCN in a B) Ratio of solubility of a salt MCN in a C) Ratio of solubility of a salt MCN in buffer
LEVEL - IV KEY 01. B 02.A 03.A 04.A 05.B 06.C 07.A 08.B 09.B 10.C 11.D 12.B 13.A 14.C 15.B 16.D 17.B 18.C 19.C 20.C 21.B 22.C 23.D 24.D 25.C 26.A 27.C 28.A 29.D 30.D 31.D 32.B 33.A,B,C 34.B,C,D 35.B,C 36. A,B 37.A,C 38.A,C 39.B,C 40.D 41.D 42.B 43.A 44.C 45.A 46.B 47.A 48.C 49.A 50.C 51.B 52.C 53.B 54.B 55.B 56. A r , s; B r; C p; D q,
D) Ratio of solubility of AgCI in 0.1 M HCI E) Ratio of solubility of Al OH 3 in 0.1 M HCI and 0.2M NaOH. Column-II p) 4 buffer of pH = 3. K a 10 3 / 3 and solubility in H 2O is q) 2 buffer of pH = 4, K a 10 4 / 7 and
r) 8 of pH = 5 K a 10 5 / 8 and solubility in H 2O is s) 3 and 0.2 M INTEGER TYPE QUESTIONS 58.
59. 60.
61.
What will be pH of the solution of the salt of weak acid and weak base? -6 -4 (Kb = 1 × 10 and Ka = 1× 10 ) Conjugate base of [Al(H2O)6]+3is Al (H2O)x(OH)y, the value of x/y is, The mixture of 10 ml 0.5 N CH3COOH and 10 ml of 0.25 M NaOH , having pH = 5, then find the value of pKa? Among these mixture (i) 50 ml of N/10 HCl + 50 ml of N/10 NaOH (ii) 55 ml of N/10 HCl + 45 ml of N/10 NaOH (iii) 75 ml of N/10 HCl + 25 ml of N/10 NaOH (iv) 30 ml of N/5 ml HCl + 70 ml of N/5 NaOH For how many of the above mixutres pH lies between 1 to 7
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solubility in H 2O is
57. Aq; B p;C s; D q; E r 58.6 59.5 60.5 61.2 HINTS - LEVEL - IV 1.
H HS ; ka 10 7 H 2 S 1
H S 2 ; ka 1.3 1013 HS 2 HCl H Cl
due to common ion effect the dissociation of H 2 S is suppressed and the H in solution is due H HS k to HCl. a1 H2S
0.3 HS H 0.1
10 7
HS
107 0.1 3.3 108 M 0.3
H S2 ka2 HS 1.31013
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from HCl 0.3
0.3 S2
3.3108 S 1.31013 3.3108 /0.03 1.431020 M 2
21
JEE ADVANCED-VOL - III
2.
IONIC EQUILIBRIUM
salt We get 4.75 = -log(1.34 10-5) + log acid
given MBr2 aq M 2 2 Br MBr2 g
MBr2 H 2 S MS 2 HBr
which gives 4.75 = 4.87 + log
K sp of MS M 2 s 2 6 10
or
0.05 S
21
2
S 2 1.2 10 19 M
thus MS will be precipitated if H2S provides 1.2 x 10-19M ions of S-2 5.
2 H S 2 now for H2S H 2 S
H S 2 ka1 ka2 H2S
[Ag+] required for precipitation of AgCl Ag
ksp AgCl
9
10 Cl [Ag+] required for precipitation of Ag2CrO4 Ag
ksp Ag 2 CrO 4 CrO 24
10 12 10 4 10 14
At this point, the concentration of Cl- ion in the solution can be calculated from Ksp (AgCl)
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H 1.2 10 19 0.1
H 1.04 101 p H 0.9826
0.76 or [salt] = 1.52 10-2 M
Hence, Amount of sodium propanoate to be added = 1.52 10-2 mol The addition of 0.01 mol of hydrogen chloride convert the equivalent amount of sodium propanoate into propanoic acid . Let the initial no. of equivalents of
x
At 3/4th eqv. pt.
2
3.
0.02 M
0.02M
BOH be x BOH + HCl BCl + H2O Initial equivalent x 3/4x 0 0
2
10 7 1.3 10 13
Salt
salt
3 x 4
3x x 4 4
0
3 x 4
pOH = pK b log
[salt] 3x 4 5 log [Base] 4 x
pH = 14 - 5 - log 3 = 8.523 6.
salt
pH = pKa + log acid 5.04 = 4.74 + log n CH3COO
2
Pb
CH 3 COO 2 Pb CH 3COOH
x
n CH COO 2x
3
n CH3COOH 0.1
10
Cl
4.
ksp(AgCl) 10 4 106 M Ag 10
5.04 = 4.74 + log
Using the expression
salt pH = pKa + log acid
0.3 = log 2=
2x 0.1
2x 0.1
2x 0.1
x = 0.1 22
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7.
9.
Ksp [Mg(OH)2] = 8.9 10-12
OH 10 5
2
2
14.
Mg 2 8.9 10 2 0.089
15.
Conc. of Mg2+ precipitated = 0.1 - 0.089 = 0.011 mol/Lt. pH of buffer is given by :
17.
CO32 pH pK a log 2 CO3
[Salt] = 0.1 × 0.5 = 0.05 M
Salt 0.5
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OH
Salt Case I : 4 = - log 1.0 10-5 + log 0.5
Case II : 6 = - log 1.0 10-5 + log
1 pK w pK a pKb 2 is not dependent on concentration in this case pH
21.
Salt pH = - log Ka + log Acid
or
0.08 log5 104 log 4 0.02
= 3.30 + 0.602 = 3.902 pH = 14 - p(OH) = 14 - 3.902 = 10.09 [H+] = 7.99 10-11 M = 8 10-11
8.9 10-12 = Mg 2 105
CH3NH3+ Cl0
HCl 0.08
p(OH ) pK b log
Ksp = Mg 2 OH
8.
+
0.1 Intialmoles 0.02 0 0.08 moles after mixing As it is a basic buffer, so
pH = 9 [H+] = 10-9
CH 3NH2
K SP Mg 2
pOH pK b log
1011 104 103
NH 4Cl NH 3
22.
When half of acid is neutralized pH pK a
23.
CO32
Salt
H
HCO3
log 0.5 = 1
Initial milli moles 50 0.05 Final milli moles __
[Salt] = 10 0.5 = 5 M Now the two buffer [(I. NaA = 0.05M and HA = 0.5 M) and (II. NaA = 5M and HA = 0.5M)] are mixed in equal proportion. Thus, new conc. of NaA is mixed buffer =
HCO3 H H 2CO3 Initial milli moles 2.5 1.5 Final milli moles 1 __
0.5 V 0.5 V 0.5M 2V
Thus, pH = - log (1.0 × 10-5) + log pH = 5 + 0.7033 = 5.7033. Narayana Junior Colleges
5.05/ 2 0.5
___ 1.5
HCO3 pH pK a1 log 6.173 H 2CO3
0.05 V 5 V 5.05 = 2V 2
New conc. of HA in mixed buffer =
40 0.1 1.5 2.5
24.
TCl NaOT NaCl T2O Initial 2 3.75 milli-moles milli -moles of remaining NaOT = 1.75 1.75 7 10 2 moles 25 pOT = 2- log 7 pT = 15.24 =2 +log 7 = 13.24 + log 7 OT
23
JEE ADVANCED-VOL - III
25.
IONIC EQUILIBRIUM
3 is very less so H K a C 3 10
Now H
1.5 10 3
3 103 1.5 103 2
HCOO x F y
ka .C2
C2 0.125 milli-moles acid remians constant 28.
10 0.5 0.125 V2
X OH 3 X OH OH
V2 40 litre
Second dissociaztion:
H
2
10 0.3 10 0.1 0.05 10 10 20
Total OH 4 10 3 2 103 6 10 3 pOH = 3 log 6 2.22
Ka
0.05 x .x x 0.05 x
x 105
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HA H A 0.05 - x 0.05+x x Due to common ion effect neglect x w.r.t. 0.05
29.
H K a C 105 0.1 103 H A3 103 A3 13 ka 3 10 HA2 HA2 A3 X 1010 ; HA2
x 105 2 104 0.05 5 102
HCOOH(aq.)+ H 2O l HCOO aq.
C1 x
x
pH 11.78 is negligible w.r.t. 1
H mainly from first step
A x HA A x 0.05
30.
CrO42
y x+y C2 y is very les for both acid so K a HF
24
y x y C2
pX = 10
K sp SrCrO4 Sr 2 CrO42
H 3O aq.
HF aq. H 2O l F (aq.) H 3O aq.
2
X OH 2 X OH OH
20 0.1 0.05 HA 20 20
27.
2 104 0.2 2 4 6.6 10 0.1 3.3 First dissociation
C1V1 C2V2
26.
x K a HCOOH C1 y K a HF C2
3.5 105 3.5 104 0.1
K sp BaCrO4 Ba 2 CrO42
CrO42 CrO42 from SrCrO4 total 1.2 1010 Ba 2 3.4 107 4 3.5 10 Narayana Junior Colleges
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31.
After mixing Ag 0.2 M ; NH 3 1M
43.
Due to very high value of K f ; Ag mainly converted into complex
S k sp 4s
102 5 103 2 k sp 4s 3 5 107 m3 s
Ag NH 3 2 aq. Ag NH 3 aq. NH 3 aq. At equilibrium 0.2 -y y 0.6 + y 0.2 0.6
44.
1 y 0 .6 0 .1 ; y A g N H 3 3 .3 3 1 0 5 M K f2 0 .2 104
K sp Zn2 S 2
1021 S 2 1019 0.01 2
H S 2 K . K for a1 a2 H2S 2
34.
H 1019 20 10 H 0.1 0.1 or pH= 1 10 ml + 20 ml N/20 NaOH N/20 ml HCl
milli milli
eq. equ.
= =
10
1 1 20 2
1 20 1 20
41.
solution also turns methyl orange rad [H +] > [OH - ] then (B) (C) (D) are correct For the formation of Buffer solution meq of weak base > meq of strong acid. Use Henderson’s equation.
42.
5.74 = 4.74 + log
40.
x , calculating x = 0.02, 5 10 3 hence [(NH4)2SO4] required = 0.01
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32.
2S
3
p H of Ba OH 2 12 OH 102 2s 102
Ag aq. 2 NH 3 aq. Ag NH 3 2 aq. Initial concentration 0.2 1 At equilibrium x 0.6 0.2
Ba 2 2OH 1 . Ba OH 2
Ag Cl 1010 m 2 . 1010 Ag 0.05 Ag I 4 1016 m 2 I
45.
4 1016 m 2 2 107 M Ag
Ap Bq pA qB pS
qS
Let solubility be s mol / lt . P
Thus K sp A B p
q
q
pS qS p p .q q S
46.
p q
.
Ca 2 2OH Ca OH 2 S
2S
k sp 4s 3 ; ksp 5.5 106 S
3
ksp 4
3
5.5 106 4
3 1.375 106 1.119 102 OH 2 s 2.2238 102 p OH log 2.2238 102 1.6529 p H 12.34 25
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47.
IONIC EQUILIBRIUM
59.
H ka c
3
ka 1.8105
60.
c 5101M H 1.8105 5101 9106 3103 pH 2.52
48.
P H 5.04
It is doubled
Al H 2O 6 Al H 2O 5 OH H 10 ml of 0.5 N CH3COOH mill equ. = 10 0.5 = 5 10 ml of 0.25 N NaOH Milli eq. = 0.25 10 = 2.5 On adding 2.5 CH3COOH 2.5 CH3COONa pH = pKa + log
H 9.120 106
2.5 2.5
pH = pka + log 1
pka = 5
H K a .C
C 4.6 106 . 49.
M1V1 M 2V2
50.
k sp Nis 3 1021 S 2
3 1021 0.1
H HS 1 ka 107 H 2 S 1 H S 2 ka 10 14 HS 1 2
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Amount of water to be added V2 V1 .
H S 2 k a2 HS 1 H k a 1 .k a 2 1 0 2 1
1 H 300 51.
2
20 3 1 0 0 .1
pH 1.2388
since ksp of NiS is greater then the remaining then NiS is more soluble
58.
1 2
pH= (14 pK a pK b ) 1 (14 4 6) = 6 2
26
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