IONIC EQUILLIBRIUM FOR IIT-JEE ENTRANCE TEST by S.K.sinha See Chemistry Animations at http://openchemistry.in

MASTERING

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IONIC EQUILLIBRIA By :

S.K.Sinha SINHA'S

IIT CHEMISTRY

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ACIDS AND BASES

By-- S.K.SINHA

1

The Nature of Acids and Bases Arrhenius concept: Acid

a substance that produces hydrogen ion in aqueous solution.

Base a substance that produces hydroxide ion in aqueous solution. Examples of Acids: 1. 2. 3.

HCl(aq)  H+(aq) + Cl-(aq);

(a strong acid)

HNO3(aq)  H+(aq) + NO3-(aq); CH3COOH(aq)

⇋

H+(aq)

(a strong acid)

+ CH3COO (aq); (a weak acid) -

Examples of bases: 1.

NaOH(aq)  Na+(aq) + OH-(aq);

8.

NH3(aq) + H2O

2.

(a strong base)

Ba(OH)2(aq)  Ba2+(aq) + 2OH-(aq); (a moderately strong base)

⇋ NH4+(aq) + OH-(aq); (a weak base)

* All hydroxides and oxides of metal are bases. (Oxides of metals in veryhigh O.N. is acidic. Ex CrO3 , Mn2O7

.

The Brønsted-Lowry concept: Acid

a substance that acts as a proton donor in chemical reaction;

Base a substance that acts as a proton acceptor in chemical reaction; * Reactions that involve the transfer of protons (H+) are acid-base reactions. The Bronsted-Lowry acid-base reaction can be represented as follows: HA + B ⇋ acid1 base2

For examples: 1. 2. 3.

BH+ + conjugate acid2

HCl + H2O  H3O+(aq) + acid1 base2 conjugate acid2 HC2H3O2 acid1

+

H2O base2

NH3 + H2O base1 acid2

⇋

Aconjugate base1

Cl-(aq) conjugate base1

⇋ H3O+(aq)

conjugate acid2

+

C2H3O2-(aq); conjugate base1

NH4+(aq) + OH-(aq); conjugate conjugate acid1 base2

In each reaction, a conjugate base is what remains of the acid molecule after it loses a proton (H+), and a conjugate acid is what becomes of the base after it gains a proton. acid1 - conjugate base1 and acid2 - conjugate base2 are called conjugate acid-base pairs; they are a pair of species that are related to each other by the loss or gain of a single

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proton (H ). Thus, H2O and H3O ion. +

+

are conjugate acid-base pair; H3O

By-- S.K.SINHA

+

is called hydronium

A BrØnsted-Lowry acid-base reaction involves a competition between two bases for a proton, in which the stronger base ends up being the most protonated at equilibrium. In the reaction: HCl + H2O 

H3O+(aq) + Cl-(aq),

H2O is a stronger base than Cl-; at equilibrium, a solution of hydrochloric acid contains mostly H3O+ and Cl- ions.

In reaction: HC2H3O2 



+

H2O

⇋ H3O+(aq) + C2H3O2-(aq);

C2H3O2- is the stronger base; at equilibrium, a solution of acetic acid contains mostly HC2H3O2 and a small amount of H3O+ and C2H3O2- ions.

In the reaction: NH3 + H2O

⇋ NH4+(aq) + OH-(aq),

water is an acid. The competition for protons is between NH3 and OH-, in which OHis the strong base; the above equilibrium shifts to the left. That is, an aqueous ammonia solution contains mostly NH3 molecule and a small amount of NH4OH, NH4+, and OH-. Exercises: 

1.

Write the conjugate base for each of the following acids:

2.

Write the conjugate acid for each of the following bases:

(a) H2PO4-;

(a) NH3

(b) H2C2O4;

(b) [Al(H2O)2(OH-)4]-

(c) [Al(H2O)6]3+;

(c) SO32-

(d) NH3;

(d) (CH3)3N:

The fects of Structure of Acid-Base Properties

The Relative Strength of Acids

The strength of acids is determined by a combination of various factors, such as the strength and polarity of XH bond in the molecule, and the hydration energy of the ionic species in aqueous solution. For binary acids such as HF, HCl, HBr, and HI, HX bond strength decreases down the group. The weaker the bond, the easier the molecule ionizes in aqueous solution. Hence, the stronger will be the acid. Thus, for this group of acids, their strength increases down the group: HF < HCl < HBr < HI; H2O < H2S < H2Se < H2Te Among the hydrohalic acids, HF is the only weak acid; the others are strong acids. The relative strength of HCl, HBr, and HI cannot be differentiated in aqueous solution, because each of them dissociates almost completely. Less polar solvents are used to determine their relative strength. For example, HCl, HBr and HI ionize only partially in acetone or methanol, which have a weaker ionizing strength than water. The ionization of HCl in acetone can be represented by the following equilibrium: (CH3)2CO(l) + HCl(acetone)

⇋ (CH3)2COH+(acetone) + Cl-(acetone)

The degree of ionization in acetone increases in the order of HCl < HBr < HI. Based on this observation, it was determined that the acidity of hydrogen halides increases down the

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group: HF < HCl < HBr < HI. A similar trend of relative acidity is also observed for the hydrides of Group VIA elements, such that, H2O < H2S < H2Se < H2Te. For the same period hydrides, the relative acidity increases from left to right, such that: CH4 < NH3 < H2O < HF;

PH3 < H2S H2SO4 > H3PO4. For oxo-acids with central atoms from the same group, their relative strength decreases from top to bottom as the electronegativity of the central atom decreases: HOCl > HOBr > HOI;

HOClO > HOBrO > HOIO;

HOClO2 > HOBrO2 > HOIO2

For oxo-acids containing identical central atom, acidity increases as more oxygen atoms are bonded to it. For example, acidity increases in the following order: HOCl < HOClO < HOClO2 < HOClO3;

H2SO4 > H2SO3;

HNO3 > HNO2

Oxygen is a very electronegative atom; when more oxygen atoms are bonded to the central atom, the OH group in the molecule becomes more polarized (due to inductive effect), and the more readily it ionizes to release H+ ion. Acetic acid is an organic acid, which contains the acidic carboxyl group, -COOH. When acetic acid (CH3COOH) ionizes, only the OH bond of the carboxyl group breaks, but not the CH bonds. However, if one or more of the hydrogen atoms in the methyl group (CH3) is substituted with a more electronegative atom, the OH of -COOH group becomes more polarized and ionizes more readily, and increasing the acidity. The following Ka values illustrate the effect on the acidity of COOH when one or more of the methyl hydrogen is substituted with more electronegative atoms: CH3COOH(aq)

⇋ CH3COO-(aq) + H+(aq);

Ka = 1.8 x 10-5

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ClCH2COOH(aq)

⇋ ClCH2COO-(aq) + H+(aq);

Ka = 1.4 x 10-3

FCH2COOH(aq)

⇋ FCH2COO-(aq) + H+(aq);

Ka = 2.6 x 10-3

chloroacetic acid fluoroacetic acid

CCl3COOH(aq)

⇋ CCl3COO-(aq) + H+(aq);

trichloroacetic acid

Ka = 3.0 x 10-1

Exercise: 1. Rank the following acids in order of increasing strength: (a) CH3COOH, CH3CH2COOH, HCOOH;

2.

(b) CH3COOH, ClCH2COOH, CHCl2COOH, CCl3COOH, CF3COOH Rank the following bases in order of increasing strength:

(a) NH3, CH3NH2, (CH3)2NH2, (CH3)3N, (CH3CH2)3N ________________________________________________________________________ Acid-Base Properties of Oxides

Metal oxides are either basic or have amphoteric properties. The oxides the Group IA metals are strongly basic because they are generally very soluble in water. Other metal oxides are less soluble, but they react with strong acids: Na2O(s) + H2O(l)  2NaOH(aq);

BaO(s) + H2O(l)  Ba(OH)2(aq)

MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O(l) The oxide ion, O2-, has a very strong affinity for protons and reacts with water to produce hydroxide ions: O2-(aq) + H2O(l)  2OH-(aq) Some metal oxides are amphoteric; for example, Al2O3, Cr2O3, PbO, SnO, and ZnO. They react with both strong acids and strong bases: Al2O3(s) + 6HCl(aq)  2AlCl3(aq) + 3H2O(l);

Al2O3(s) + 2NaOH(aq) + 3H2O(l)  2Na[Al(OH)4](aq); Oxides of nonmetals are acidic. They form acidic solution when dissolved in water. CO2(g) + H2O(l) SO2(g) + H2O(l)

⇋ H2CO3(aq) ⇋ H+(aq) + HCO3-(aq);

⇋ H2SO3(aq) ⇋ H+(aq) + HSO3-(aq);

P4O10(g) + 6H2O(l)

⇋ 4H3PO4(aq); H3PO4(aq)  H+(aq) + H2PO4-(aq);

Metal Hydroxides and Hydrides Hydroxides of Group IA metals are strongly basic. For example, LiOH, NaOH, and KOH, are strong bases. The basicity of other metal hydroxides is limited by their low solubility. The hydroxides of some metals exhibit amphoteric properties. For example, aluminum hydroxide, Al(OH)3 is an amphoteric hydroxides. Al(OH)3(s) + OH-(aq)

⇋ Al(OH)4-(aq);

Al(OH)3(s) + 3H3O+(aq)

⇋ [Al(H2O)6]3+(aq)

Cr(OH)3, Zn(OH)2, Sn(OH)2, and Pb(OH)2 are also amphoteric hydroxides.

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The hydrides of reactive metals, such as NaH and CaH2, are strong bases. The hydride ion has a strong affinity for protons and reacts with water to produce hydroxide ions and hydrogen gas: H-(aq) + H2O(l)  H2(g) + OH-(aq) The Lewis Acids and Bases According to G.N. Lewis, an acid is the reactant that is capable of sharing a pair of electrons from another reactant to form a covalent bond; a base is the reactant that provides the pair of electrons to be shared to form a covalent bond. Based on the Lewis’s definition, hydrogen ion (H+) is considered a Lewis acid and water and ammonia are the Lewis bases in the following reactions:

H+ + H2O  H3O+; H+ + NH3  NH4+; Lewis Lewis Lewis Lewis acid base acid base Species with incomplete octet (or electron deficient species) may act as Lewis acid and those with lone pair of electrons may act as Lewis bases. For example, in the following reactions, BF3, AlCl3, and FeBr3 are Lewis acids; while NH3, Cl-, and Br- are Lewis bases. BF3 + NH3  F3B:NH3; AlCl3 + Cl-  AlCl4-; FeBr3 + Br-  FeBr4-; In the formation of complex ions, the positive ions act as Lewis acids and the ligands (anions or small molecules) are Lewis bases: Cu2+(aq) + 4NH3(aq) ⇋ Cu(NH3)42+(aq); Lewis acid Lewis base

Al3+(aq) + 6H2O(l) ⇋ [Al(H2O)6]3+(aq); Exercise 1. Determine the Lewis acids and Lewis bases in the following reactions: (a) CO2(g) + H2O(l) (b) SO3(g) + H2O(l)

⇋ H2CO3(aq)

⇋ H2SO4(aq)

(c) AlCl3 + (CH3)3N:

⇋ (CH3)3N:AlCl3

(d) Zn(OH)2(s) + 2OH-(aq) (e) AuCl3(s) + Cl-(aq)

⇋ Zn(OH)42-(aq);

⇋ AuCl4-(aq)

Acids Strength

The strength of an acid is defined by the extent of its dissociation (ionization) in aqueous solution. HA(aq) + H2O

H3O+(aq) + A-(aq)

Strong acids, such as HClO4, HCl, HNO3, and H2SO4, are those that are considered to completely ionize at 1 M concentration. The above equilibrium would lie far to the right. Acids that are only partially ionized at this concentration, such as HC2H3O2, HF, HNO2, H2SO3, H3PO4, and HClO, are considered weak acids. Their dissociation equilibria would lie far to the left. The equilibrium constant, Ka, for the acid ionization is given by the following expression: Ka

[H3O+][A-] [H+][A-] =  =  [HA] [HA]

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1.

HF(aq)

2.

HNO2(aq)

3.

HC2H3O2(aq)

4.

HOCl(aq)

5.

HCN(aq)

CHEMISTRY

By-- S.K.SINHA

For strong acids, Ka would have a very large value; for weak acids, Ka < 10-1. The following are the Ka values of some weak acids:

⇋ H+(aq) + F-(aq);

Ka = [H+][F-] = 7.2 x 10-4 [HF]

⇋ H+(aq) + NO2-(aq); ⇋ H+(aq) + C2H3O2-(aq);

⇋ H+(aq) + ClO-(aq);

Ka = [H+][NO2-] = 4.0 x 10-4 [HNO2]

Ka = [H+][ C2H3O2-] = 1.8 x 10-5 [HC2H3O2] Ka = [H+][ClO-] = 3.5 x 10-8 [HOCl]

⇋ H+(aq) + CN-(aq);

Ka = [H+][CN-] = 6.2 x 10-10 [HCN] The value of Ka is a measure of the extent of acid dissociation; hence, the strength of the acid. A stronger acids have larger Ka values. For strong acids such as HClO4, HCl, H2SO4, and HNO3, their Ka’s are very large (not given in any tables of Ka values). The Conjugate Acid-Base Relationships

Strong acids have weak conjugate bases, and weak acids have strong conjugate bases; the weaker the acid, the stronger will be its conjugate base. Likewise, weak bases have strong conjugate acids; the weaker the base, the stronger will be its conjugate acid. Examples:  HCl is a strong acid and Cl- is a very weak base; 

HF is a weak acid, and F- is a stronger conjugate base than Cl-.

According to BrØnsted-Lowry, acid-base reactions favor the direction from strong acidstrong base combinations to weak acid-weak base combinations: The following acid-base reactions go completely in the forward direction: 1. 2.

HClO4(aq) + H2O(l)  H3O+(aq) + ClO4-(aq); HCl(aq) +

NH3(aq)  NH4+(aq) + Cl-(aq);

(HClO4 is a very strong acid)

(HCl is a strong acid)

3. + CN (aq)  HCN(aq) + (HSO4- is a stronger than HCN) Many acid-base reactions reach a state of equilibrium. The following acid-base reactions do not go to completion; their equilibrium positions may shift to the right or to the left, depending on the relative strength of the acid: 1.

HSO4-(aq)

SO42-(aq);

-

H2PO4-(aq) + C2H3O2-(aq)

⇋ HC2H3O2(aq) + HPO42-(aq);

Equilibrium shifts left; HC2H3O2 is the stronger acid and HPO42- is the stronger base. 2.

HNO2(aq) + C2H3O2-(aq)

⇋ HC2H3O2(aq) + NO2-(aq);

Equilibrium shifts right; HNO2 is the stronger acid, C2H3O2- is the stronger base. Water as an Acid and a Base

Water is amphoteric because it can behave as an acid and a base. The following equilibrium occurs in pure water: H2O + H2O Acid1 base1

⇋ H3O+(aq) + OH-(aq); conjugate acid2

conjugate base1

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The above equilibrium can be simplified as follows, which is also called the auto-ionization of water: H2O

⇋ H+(aq) + OH-(aq);

Kw = [H+][OH-] = 1.0 x 10-14 at 25 oC.

Kw is called the ion product constant for water. An aqueous solution contains both H+ (in the form of H3O+) and OH-, such that the product of their concentrations is constant equal to 1.0 x 10-14 at 25 oC. In pure water (or in a neutral solution, [H+] = [OH-] = 1.0 x 10-7 M. + If [H ] increases (> 1.0 x 10-7 M), [OH-] will decreases (< 1.0 x 10-7 M), and vice versa. Thus, a solution with:  [H+] > 1.0 x 10-7 M, or [H+] > [OH-],  the solution is acidic;  

[H+] < 1.0 x 10-7 M, or [H+] < [OH-],  the solution is basic; [H+] = [OH-] = 1.0 x 10-7 M,  the solution is neutral;

Exercise-2: 1. What is [OH-] if [H+] = 0.0050 M? Is the solution acidic, basic or neutral? 2. What is the [H+] if [OH-] = 6.0 x 10-4 M? Is the solution acidic, basic or neutral? ________________________________________________________________________ The pH Scale pH is a scale that measures the acidity of an aqueous solution where [H+] is very small. pH = -log[H+],

If [H+] = 1.0 x 10-7 M, pH = -log(1.0 x 10-7) = -(-7.00) = 7.00 We can also express basicity using the log scale for [OH-], such that pOH = -log[OH-]

That is, if a solution has [OH-] = 1.0 x 10-7 M, pOH = -log(1.0 x 10-7) = -(7.00) = 7.00 Since, at 25oC, Kw = [H+][OH-] = 1.0 x 10-14

pKw = -log Kw = -log[H+] + (-log[OH-]) = -log(1.0 x 10-14) = -(-14.00) = 14.00 pKw = pH + pOH = 14.00;

and pOH = 14.00 – pH

Thus, in aqueous solutions, if pH > 7.00  pOH < 7.00, and vice versa;

In neutral solution, [H+] = [OH-] = 1.0 x 10-7 M, and pH = pOH = 7.00; 

[H+] > 1.0 x 10-7 M,  pH < 7.00, the solution is acidic.

 [H+] < 1.0 x 10-7 M,  pH > 7.00, the solution is basic. _________________________________________________________________________. Calculating the pH of Strong Acid and Strong Base Solutions Strong acids are assumed to ionize completely in aqueous solution. For monoprotic acids (that is, those acids with a single ionizable hydrogen atom), the concentration of hydrogen ion in solution is the same as the molar concentration of the acid. That is [H+] = [HA] For example, in 0.10 M HCl(aq), [H+] = 0.10 M, and pH = -log(0.10) = 1.00. A strong base such as NaOH, contains OH- which concentration is equal to that of dissolved NaOH. That is, a solution of 0.10 M NaOH(aq) contains [OH-] = 0.10 M pOH = -log[OH-] = -log(0.10) = 1.00;

pH = 14.00 – 1.00 = 13.00

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A strong base such as Ba(OH)2 produces twice the concentration of OH- as the concentration of Ba(OH)2 in solution. Ba(OH)2 dissociates as follows: Ba(OH)2(aq)  Ba2+(aq) + 2OH-(aq); [OH-] = 2 x [Ba(OH)2]

In a solution of 0.010 M Ba(OH)2, [OH-] = 0.020 M, pOH = 1.70, and pH = 12.30 Exercise-3: 1. Calculate the pH of the following solutions: 2. 3.

(a) 0.0050 M HCl; (b) 0.00060 M NaOH. What is [H+] and [OH-], respectively, in a solution where,

(a) pH = 4.50; (b) pOH = 5.40 The pH of an HCl solution is found to be 3.00. To what final volume must a 100.-mL sample of this acid be diluted so that the pH of the solution becomes 3.50? (316 mL)

4.

What is the pH of a saturated aqueous solution of Ca(OH)2 in which 0.165 g of Ca(OH)2 is dissolved in 100. mL solution at 25 oC? (Answer: pH = 11.65) _________________________________________________________________________. Calculating the pH of Weak Acid Solutions Weak acids do not ionize completely. At equilibrium, [H+] is much less than the concentration of the acid. The concentration of H+ in a weak acid solution depends on the initial acid concentration and the Ka of the acid. To determine [H+] of a weak acid we can set up the “ICE” table as follows: Consider a solution of 0.10 M acetic acid and its ionization products: Concentration: HC3H3O2(aq) ⇋ H+(aq) + C2H3O2-(aq)  Initial [ ], M: 0.10 0.00 0.00 Change, [ ], M:

-x

+x

+x

Equilibrium [ ], M : (0.10 - x) x x  The acid ionization constant, Ka, is given by the expression: Ka = [H+][C2H3O2-] = x2 [HC2H3O2] (0.10 – x)

= 1.8 x 10-5

Since Ka HPO42-; Sulfuric acid is a strong acid because the first hydrogen ionizes completely: H2SO4(aq)  H+(aq) + HSO4-(aq);

Ka1 = very large

However, the second hydrogen does not dissociate complete. Thus, HSO4- is a weak acid, which dissociates as follows: HSO4-(aq) ⇋ H+(aq) + SO42-(aq); Ka2 = 1.2 x 10-2 Exercise: 1. What is [H+], [HSO4-], and [SO42-] in 0.10 M H2SO4(aq)? What is the pH of the solution? (Ka1 is very large; Ka2 = 1.2 x 10-2) (Answer: pH = 0.96) 2.

What is the pH of 0.10 M oxalic acid, H2C2O4, and what is [C2O42-] in this solution? (Ka1 = 6.5 x 10-2; Ka2 = 6.1 x 10-5 ) (Answer: [H+] = 0.054 M; pH = 1.26; [C2O42-] = 6.1 x 10-5 M) _________________________________________________________________________ Acid-Base Properties of Salts

When salts (ionic compounds) dissolve in water, we assume that they completely dissociate into separate ions. Some of these ions can react with water and behave as acids or bases. Salts are also products of acid-base reactions. The acidic or basic nature of a salt solution depends on whether it is a product of:  a strong acid-strong base reaction;  a weak acid-strong base reaction;  a strong acid-weak base reaction, or  a weak acid-weak base reaction.

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Salts of Strong Acid-Strong Base Reactions: such as NaCl, NaNO3, KBr, etc. 

This type of salt forms a solution that is neutral, because neither the cation ion nor the anion reacts with water and offset the equilibrium between the H3O+ and OHconcentrations in the aqueous solution.

Salts of Weak Acid-Strong Base Reactions: such as NaF, NaNO2, NaC2H3O2, etc. 

A salt that is the product of a reaction between a weak acid and a strong base will form an aqueous solution that is basic in nature.



the anion of such salt reacts with water to increase [OH-].

For example, Sodium acetate (NaC2H3O2) is a product of reaction between a weak acid (HC2H3O2) and a strong base (NaOH). HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l)

When sodium acetate is dissolved in water, it dissociates into sodium and acetate ions: NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)

The acetate ion undergoes hydrolysis in aqueous solution as follows: C2H3O2-(aq) + H2O(l)

⇋ HC2H3O2(aq) + OH-(aq);

Note the for the dissociation of acetic acid: HC2H3O2(aq) + H2O(l)

Kb = [HC2H3O2][OH-] [C2H3O2-]

⇋ H3O+(aq) + C2H3O2-(aq); Ka = [H3O+][C2H3O2-] [HC2H3O2]

Ka x Kb = [H3O+][C2H3O2-] x [HC2H3O2][OH-] = [H3O+][OH-] = Kw = 1.0 x 10-14 [HC2H3O2] [C2H3O2-] Thus, for C2H3O2-, Kb (for =

Ka

Kw

(for HC2H3O2)

=

1.0 x 10-14 = 5.6 x 10-10 1.8 x 10-5

Thus, the hydrolysis of acetate ion has Kb = 5.6 x 10-10 (> Kw), which implies that the solution of sodium acetate will be basic (pH > 7). Salts of Strong Acid-Weak Base Reactions: NH4Cl, NH4NO3, (CH3)2NH2Cl, C5H5NHCl.  

Aqueous solutions of this type of salts are acidic, because the cation (such as NH4+ ion) is the conjugate acid of a weak base (NH3) the cations react with water, which increases [H3O+] in solutions.

For example, ammonium chloride, NH4Cl, is produced by the reaction of a strong acid HCl and a weak base NH3. HCl(aq) + NH3(aq)  NH4Cl(aq)  NH4+(aq) + Cl-(aq)

In an aqueous solution, ammonium ion (NH4+) establishes the following equilibrium that increases [H3O+] and makes the solution acidic: NH4+(aq) + H2O(l)

⇋ H3O+(aq) + NH3(aq);

Ka = [H3O+][NH3] [NH4+]

For the reaction of NH3 with water, the following equilibrium occurs: NH3(aq) + H2O(l)

⇋ NH4+(aq) + OH-(aq);

Kb = [NH4+][OH-] [NH3]

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Ka x Kb = [H3O ][NH3] x [NH4 ][OH ] = [H3O ][OH ] = Kw = 1.0 x 10 [NH4+] [NH3] +

Thus, for NH4+, Ka =

+

Kw = Kb (for NH3)

-

+

-

13

-14

1.0 x 10-14 = 5.6 x 10-10 1.8 x 10-5

Thus, the hydrolysis of ammonium ion has Ka = 5.6 x 10-10 (> Kw), which implies that aqueous solution of ammonium chloride will be acidic (pH < 7). Salts of Weak Acid-Weak Base Reactions: such as NH4C2H3O2, NH4CN, NH4NO2, etc..    

Aqueous solutions of this type of salts can either be neutral, acidic, or basic, depending on the relative magnitude of the Ka of the weak acid and the Kb of the weak base. If Ka ~ Kb, the salt solution will be approximately neutral; If Ka > Kb, the salt solution will be acidic, and If Ka < Kb, the salt solution will be basic.

For example, Ka of HC2H3O2 = 1.8 x 10-5, Kb of NH3 = 1.8 x 10-5, and aqueous solution of ammonium acetate, NH4C2H3O2, is approximately neutral. NH4C2H3O2(aq)  NH4+(aq) + NH4+(aq) + H2O(l)

C2H3O2-(aq) + H2O(l)

C2H3CO2-(aq)

⇋ H3O+(aq) + NH3(aq);

⇋ HC2H3O2(aq) + OH-(aq);

Ka = 5.6 x 10-10;

Kb = 5.6 x 10-10;

Ka=Kb  a neutral solution, because the hydrolyses result in a solution with [H3O+] = [OH-]. For NH4CN, Ka(HCN) = 6.2 x 10-10, Kb(NH3) = 1.8 x 10-5, and aqueous solution of NH4CN will be basic. NH4CN(aq)  NH4+(aq) + CN-(aq) NH4+(aq) + H2O(l)

⇋ H3O+(aq) + NH3(aq);

Ka = 5.6 x 10-10;

CN-(aq) + H2O(l) ⇋ HCN(aq) + OH-(aq); Kb = 1.6 x 10-5; Ka < Kb  basic solution, because the hydrolyses result in a solution in which [H3O+] < [OH-]. For NH4NO2, Ka(HNO2) = 4.0 x 10-4, Kb(NH3) = 1.8 x 10-5, and aqueous solution of NH4NO2 will be acidic. NH4NO2(aq)  NH4+(aq) + NO2-(aq) NH4+(aq) + H2O(l)

NO2-(aq) + H2O(l)

⇋ H3O+(aq) + NH3(aq);

⇋ HNO2(aq) + OH-(aq);

Ka = 5.6 x 10-10;

Kb = 2.5 x 10-11;

Ka > Kb  acidic solution, because the hydrolyses result in a solution with [H3O+] > [OH-]. Calculating the pH of a Basic or an Acidic salt Solution

1. Consider a solution of 0.050 M sodium acetate, which dissociates completely as follows: NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)

The acetate ion undergoes hydrolysis in aqueous solution as follows: C2H3O2-(aq) + H2O(l)

By approximation, [OH-] =

⇋ HC2H3O2(aq) + OH-(aq);

Kb = [HC2H3O2][OH-] = 5.6 x 10-10 [C2H3O2-]

(Kb[C2H3O2-]) = {(5.6 x 10-10)(0.050)} = 5.3 x 10-6 M

pOH = -log(5.3 x 10-6) = 5.28, and pH = 8.72, ( a basic solution)

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2. Consider a solution of 0.050 M NH4Cl, which dissociates as follows: NH4Cl(aq)  NH4+(aq) + Cl-(aq)

In an aqueous solution, the following equilibrium occurs: NH4+(aq) + H2O(l)

⇋ H3O+(aq) + NH3(aq);

By approximation, [H3O+] =

Ka = [H3O+][NH3] = 5.6 x 10-10 [NH4+]

(Ka[NH4+]) = {(5.6 x 10-10)(0.050)} = 5.3 x 10-6 M

pH = -log(5.3 x 10-6) = 5.28, ( an acidic solution) Exercise: 1. Predict whether aqueous solution of each of the following salts is acidic, basic, or neutral?

(a) KNO3

(b) CH3NH3Cl

(c) Na2SO3

(d) (NH4)2HPO4

Ka (for HSO3-) = 1.0 x 10-7; Ka (for H2PO4-) = 6.2 x 10-8; Kb (for CH3NH2) = 4.4 x 10-4 2.

Calculate the pH of a solution that is 0.10 M in sodium cyanide, NaCN(aq). (Ka for HCN = 6.2 x 10-10) (Answer: pH = 11.10)

3.

What is the pH of a solution containing 1.5 g of pyridinium chloride, C5H5NHCl, in 100.0 mL solution? (Kb of C5H5N = 1.7 x 10-9) In aqueous solution C5H5NHCl dissociates as follows: C5H5NHCl(aq)  C5H5NH+(aq) + Cl-(aq) (Answer: pH = 3.56)

________________________________________________________________________ The Common Ion Effect in Acid-Base Equilibria

Common ions are ions produced by more than one solute in the same solution. For example, in a solution containing sodium acetate and acetic acid, the acetate ion (C2H3O2-) is the common ion. According to the Le Chatelier's principle, the following equilibrium for acetic acid: HC2H3O2(aq) + H2O(l)

⇌ H3O+(aq) + C2H3O2-(aq)

will shift to the left if C2H3O2- (from another source) is added to the solution. This will reduce the acid dissociation, which lowers [H3O+] and increases the pH of the solution. Addition of ammonium chloride, NH4Cl, to ammonia solution will cause the following equilibrium to shift to the left: NH3(aq) + H2O(l)

⇌ NH4+(aq) + OH-(aq)

NH4Cl dissociates to produce NH4+ and Cl- ions, where NH4+ is a common ion in the equilibrium of ammonia solution. The introduction of NH4+ ion reduces the ionization of NH3, which decreases [OH-] and lowers the pH of the solution. Calculating [H+] and pH of a solution containing weak acid and the salt of its conjugate base. Consider a solution containing 0.10 M HNO2 and 0.050 M NaNO2, where the latter dissociates completely as follows: NaNO2(aq)  Na+(aq) + NO2-(aq) The concentration of H+ can be calculated using the following “ICE” table:

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Concentration: HNO2(aq) ⇌ H+(aq) + NO2-(aq)  Initial [ ], M 0.10 0.00 0.050 (from salt) Change, [ ], M

-x

+x

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15

+x

Equilibrium [ ], M (0.10 – x) x (0.050 + x)  Ka = Since Ka 7 at the equivalent point. In the case of acetic-NaOH titration, the pH at the equivalent point is about 8.72. The larger the Ka of the weak acid, the closer the pH to 7 at equivalent point. The smaller Ka, the higher the pH than 7 at equivalent point. -

Exercise 1. In a titration of 20.00 mL of 0.100 M formic acid, HCHO2, with 0.100 M NaOH, calculate the pH: (a) before NaOH was added; (b) after 15.00 mL of NaOH is added; (b) after 20.00 mL of NaOH is added. (Ka = 1.8 x 10-4 for HCHO2) 2.

In a titration of 20.00 mL of 0.100 M NH3(aq) with 0.100 M HCl(aq), calculate the pH: (a) before titration begins; (b) after 10.0 mL of HCl(aq) is added; (c) at equivalent point, that is after 20.0 mL of HCl(aq) is added. (Kb = 1.8 x 10-5 for NH3)

_________________________________________________________________________ The Use of Indicators in Acid-Base Titration

    

An indicator - a substance that changes color to mark the end-point of a titration. Most indicators used in acid-base titration are weak organic acids. Indicators exhibit one color in the acid or protonated form (HIn) and another in the base or deprotonated form (In-). Each indicator has a range of pH = pKa  1, where the change of colors occurs; A suitable indicator gives the end-point that corresponds to the equivalent point of the titration; this would be one with a range of pH that falls within the sharp increase (or decrease) of pH in the titration curves.

Like a weak acid, an indicator exhibits the following equilibrium in aqueous solution: HIn(aq)

⇌

Ka =

H+(aq) + In-(aq);

[H  ][In - ] ; [HIn]

Re-arranging, we obtain [H+] = Ka x ([HIn]/[In-]);  pH = pKa + log(

[In - ] ) [HIn]

When [HIn] = 10 x [In-], pH = pKa + log(

[In - ] ) = pKa - 1; indicator assumes the color of acid form [HIn]

When [In-] = 10 x [HIn],

[In - ] pH = pKa + log( ) = pKa + 1; indicator assumes the color of base. [HIn] Phenolphthalein, which is the most common acid-base indicator, has Ka ~ 10-9. Its acid form (HIn) is colorless and the conjugate base form (In-) is pink. It is colorless when the solution pH < 8, that is when 90% or more of the species are in the acid form (HIn), and pink at pH > 10, when 90% or more of the species are in the conjugate base form (In-). The pH range at which an indicator changes depends on the Ka. For phenolphthalein with Ka ~ 10-9, its color changes at pH = 8 – 10. It is suitable for strong acid-strong base titration and weak

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acid-strong base titration. The range of pH for other common acid-base indicators are listed below:

Indicators: Methyl orange

Acid Color orange

Base Color yellow

pH Range 3.2 – 4.5

Type of Titration strong acid-strong base strong acid-weak base

Bromocresol green

yellow

blue

3.8 – 5.4

Methyl red

Red

Yellow

4.5 - 6.0

strong acid-strong base strong acid-weak base

Bromothymol blue

Yellow

blue

6.0 - 7.6

strong acid-strong base

Phenol Red

orange

red

6.8 – 8.2

strong acid-strong base strong acid-weak base

strong acid-strong base weak acid-strong base

Solubility Equilibria Solubility Equilibria and the Solubility Product Constants When a slightly soluble salt such as silver chloride, AgCl, is dissolved in water, a saturated is quickly obtained, because only a very small amount of the solid dissolves, while the rest remains undissolved. The following equilibrium between solid AgCl and the free ions occurs in solution: AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = [Ag+][Cl-]

Ksp is called the solubility product constant or just solubility product. General Expressions of Solubility Product Constant, Ksp

⇌

For solubility equilibrium: MaXb(s)

aMb+(aq) + bXa-(aq),

Ksp = [Mb+]a[Xa-]b A. For ionic equilibria of the type: MX(s)

⇌

Mn+(aq) + Xn-(aq);

Ksp = [Mn+][Xn-]

If the solubility of compounds is S mol/L, Ksp = S2;  S = (Ksp) For example, the solubility equilibrium for BaSO4 is BaSO4(s)

⇌

Ba2+(aq) + SO42-(aq);

Ksp = [Ba2+][SO42-] = 1.5 x 10-9

If the solubility of BaSO4 is S mol/L, a saturated solution of BaSO4 has

[Ba2+] = [SO42-] = S mol/L, Ksp = S2; and S = (Ksp) = (1.5 x 10-9) = 3.9 x 10-5 mol/L B. For ionic equilibria of the type: MX2(s) and for the type:

M2X(s)

⇌

⇌

M2+(aq) + 2X-(aq),

2 M+(aq) + X2-(aq);

Ksp = [M2+][X-]2 ;

Ksp = [M+]2[X2-]

For both types, if the solubility is S mol/L, Ksp = 4S3;  S = (Ksp/4)1/3 For example, CaF2(s)

⇌

Ca2+(aq) + 2F-(aq);

Ksp = [Ca2+][F-]2 = 4.0 x 10-11

The solubility of calcium fluoride is S = (Ksp/4)1/3 = (4.0 x 10-11/4)1/3 = 2.2 x 10-4 mol/L

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C. For solubility equilibria of the type: MX3(s) Or one of the type: M3X(s)

⇌

⇌

M3+(aq) + 3X-(aq),

3 M+(aq) + X3-(aq),

Ksp = [M3+][X-]3

Ksp = [M+]3[X3-]

If the solubility of the compound (MX3 or M3X) is S mol/L, Ksp = 27S4;  S = For example, Ag3PO4(s)

⇌

3Ag+(aq) + PO43-(aq);

4

( K sp / 27)

Ksp = [Ag+]3[PO43-] = 1.8 x 10-18

The solubility of silver phosphate is

S=

4

( K sp / 27) =

4

(1.8 x 10 -18 )/27 = 1.6 x 10-5 mol/L

D. For solubility equilibria: M2X3(s) OR, one of the type: M3X2(s)

⇌

⇌

2M3+(aq) + 3X2-(aq),

Ksp = [M3+]2[X2-]3

3M2+(aq) + 2X3-(aq),

Ksp = [M2+]3[X3-]2

If the solubility of the compound M3X2 is S mol/L, then [M2+] = 3S, and [X3-] = 2S; Ksp = (3S)3(2S)2 = 108S5;  S = For example, Ca3(PO4)2(s)

⇌

5

K sp / 108

3Ca2+(aq) + 2PO43-(aq),

Ksp = [Ca2+]3[PO43-]2 = 1.3 x 10-32;  the solubility of Ca3(PO4)2 is S =

5

(1.3 x 10 -32 )/108 = 1.6 x 10-7 mol/L

Calculating Ksp from Solubility

1. Suppose the solubility of PbSO4 in water is 4.3 x 10-3 g/100 mL solution at 25 oC. What is the Ksp of PbSO4 at 25 oC? Solution: Solubility of PbSO4 in mol/L = 4.3 x 10-3 g x 1000 mL/L = 1.40 x 10-4 mol/L 100. mL 303.26 g/mol

That is, a saturated solution of PbSO4, contains [Pb2+] = [SO42-] = 1.4 x 10-4 mol/L For the equilibrium: PbSO4(s)

⇌

Pb2+(aq) + SO42-(aq);

Ksp = [Pb2+][SO42-] = S2 = (1.4 x 10-4 mol/L)2 = 2.0 x 10-8 Exercise 1. If the solubility of MgF2 is 7.3 x 10-3 g/100 mL solution at 25oC, what is the Ksp of MgF2? 2.

A saturated solution of calcium hydroxide has a pH = 12.17. Calculate the solubility of Ca(OH)2 and its Ksp at 25oC.

_________________________________________________________________________ Determining Solubility from Ksp

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If the Ksp of Mg(OH)2 is 6.3 x 10 at 25 C, what is its solubility at 25 C (a) in mol/L, and (b) in g/100 mL solution at 25 oC? -10

o

o

(a) Solubility equilibrium for Mg(OH)2 is: Mg(OH)2(s)

⇌ Mg2+(aq) + 2 OH-(aq)

Ksp = [Mg2+][OH-]2 = 4S3 = 6.3 x 10-10 Solubility of Mg(OH)2: S =

3

K sp / 4 =

3

(6.3 x 10 -10 )/4 = 5.4 x 10-4 mol/L

(b) Solubility in g/100 mL solution = (5.4 x 10-4 mol/L)(58.32 g/mol)(0.1L/100 mL) = 3.1 x 10-3 g/100 mL solution.

Exercise: 1. The Ksp of PbI2 is 1.4 x 10-8. What is its solubility in grams per 100. mL of solution at 25oC? 2. Calcium hydroxide, Ca(OH)2, has Ksp = 1.3 x 10-6. How many grams of Ca(OH)2 can be dissolved in 250-mL solution to make a saturated solution at 25oC? _________________________________________________________________________ Relative Solubility from Ksp For compounds whose formula yields the same number of ions, their Ksp value can be used to determine relative solubility. That is, larger Ksp implies greater solubility. Common Ion Effect of Solubility The presence of common ion decreases the solubility of a slightly soluble ionic compound. For example, in the following equilibrium PbCl2(s)

⇌

Pb2+(aq) + 2 Cl-(aq)

If some NaCl is added to a saturated solution of PbCl2, the [Cl-] will increase, and according to Le Chateliler’s principle, the equilibrium will shift in the direction that tends to reduce [Cl]. In this case, the equilibrium will shift left, to form more PbCl2 solid, hence decreasing the amount of PbCl2 that dissolves into solution. Sample Problem The Ksp of BaSO4 is 1.5 x 10-9 at 25 oC. (a) What is its solubility in water at 25oC? (b) What is the solubility in 0.10 M Na2SO4 at 25 oC? Solubility equilibrium for BaSO4: BaSO4(s)

⇌

Ba2+(aq) + SO42-(aq);

Ksp = [Ba2+][SO42-] = S2 = 1.5 x 10-9

Solubility of BaSO4 in water is S = (Ksp) =  (1.1 x 10-10) = 3.9 x 10-5 mol/L Let the solubility of BaSO4 in 0.10 M Na2SO4 be Then, the solution contains [Ba2+] =

x mol/L

x mol/L and [SO42-] = (0.10 + x) mol/L

Ksp = [Ba2+][SO42-] = (x)(0.10 + x) = 1.5 x 10-9 By approximation, since

x Ksp;  solution is saturated and a precipitate is formed.

Sample problem: If 20.0 mL of 0.050 M Pb(NO3)2 is mixed with 30.0 mL of 0.10 M NaCl, will PbCl2 precipitate form? Ksp of PbCl2 = 1.6 x 10-5. Pb2+(aq) + 2Cl-(aq)

[ ] after mixing

0.020 M

⇌ PbCl2(s)

0.060 M

Qsp = [Pb ][Cl-]2 = (0.020)(0.060)2 = 7.2 x 10-5 > Ksp 2+

Qsp > Ksp  PbCl2 ppt will form.

Exercise: 1. At what pH a solution containg 0.10 M Ca2+ ions will form a precipitate of Ca(OH)2? (Ksp = 1.3 x 10-6 at 25 oC for Ca(OH)2) 2. Will Ag2SO4 precipitate if 30.0 mL of 0.050 M AgNO3 is added to 20.0 mL of 0.10 M of Na2SO4 solution? Ksp = 1.2 x 10-5 for Ag2SO4 at 25oC. _________________________________________________________________________ Practical Application of Ionic Equilibria Qualitative Analysis

The presence of certain ions, such as Cl-, I-, SO42-, Ag+, etc. in water can be determined qualitatively and quantitatively by precipitation method. Cl- and I- can be precipitated as AgCl and AgI, respectively, by adding AgNO3 solution. Synthesis Certain industrial chemicals, such as AgCl, AgBr, AgI, that are needed in photographic industries are prepared by precipitation.

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AgNO3(aq) + NaBr(aq)  AgBr(s) + NaNO3(aq)

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Since the Ksp of these compounds are very small, the precipitation reaction practically goes to completion. The separation and purification processes are simple. Selective Precipitation

Because of the different solubility of compounds, the separation of ions from a mixture of solution can be accomplished by selectively precipitation. This selective precipitation is carried out by adding the precipitating ion until the ion product, Q, of the more soluble compound is almost equal to its Ksp, but the Ksp of the less soluble compound is exceeded as much as possible. For example, because of the different solubility of CaSO4 (Ksp = 6.1 x 10-5), and BaSO4 (Ksp = 1.5 x 10-9), Ca2+ and Ba2+ can be separated from a solution by selectively precipitating BaSO4 using Na2SO4 or H2SO4 solution. Sample problem-3: An aqueous solution contains Ca2+ and Ba2+ ions at concentrations 0.10 M and 0.010 M, respectively. What concentration of SO42- must be present so that the maximum amount of Ba2+ ion precipitates as BaSO4, but Ca2+ remains in solution ? Ksp of CaSO4 = 6.1 x 10-5; Ksp of BaSO4 = 1.5 x 10-9. Exercise-9: 1. How many grams of AgCl will precipitate out when 50.0 mL of 0.050 M AgNO3 is added to 50.0 mL of 0.10 M NaCl? What is the concentration of Ag+ that remains in solution after the precipitation? Ksp = 1.6 x 10-10 at 25 oC for AgCl. 2.

A solution contains 0.10 M in Cl- and 0.010 M in I-. At what concentration of Ag+ would (a) AgI begin to precipitate; (b) AgCl begins to precipitate? What is the concentration of I- when AgCl begins to precipitate? (Ksp[AgCl] = 1.6 x 10-10 and Ksp[AgI] = 1.5 x 10-16)

3.

Sea water contains Mg2+ at a concentration of about 0.015 M. Would precipitate of Mg(OH)2 form if a sample of sea water is added to a saturated solution of Ca(OH)2 ? What concentrion of Mg2+ is present when the solution is just saturated with Ca(OH)2 ? (Ksp = 1.3 x 10-6 for Ca(OH)2; Ksp = 8.9 x 10-12.

 Qualitative Analysis: Identifying Ions in Mixtures of Cations The separation and identification of ions in a mixture is called qualitative analysis. The technique employs the different solubility of ionic compounds in aqueous solution as well as the ability of certain cations to form complex ion with ligands. For many transition metals, their complex ions are often colored, which can be used in their identification. Separation into Groups The general approach in the qualitative analysis of cations is to separate them into various ion groups. Ion group 1: Insoluble chlorides. Treating the mixture with 6 M HCl will precipitate Ag+, Hg22+, and Pb2+ ions as chlorides, leaving other cations in solution. The formation of a white precipitate indicates the presence of at least one of these cations in the mixture. Ion group 2: Acid-insoluble sulfides. The supernatant from the above treatment with HCl is adjusted to pH ~ 0.5 and then treated with aqueous H2S. The high [H3O+] in solution keeps

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[HS ] very low, which precipitates only the following group of cations: Cu2+, Cd2+, Hg2+, Sn2+, and Bi3+. Centrifuging and decanting gives the next solution. -

Ion group 3: Base-insoluble sulfides. The supernatant from acidic sulfide treatment is treated with NH3/NH4+ buffer to make the solution slightly basic (pH~8). The excess OH- in solution increases [HS-], which causes the precipitation of the more soluble sulfides and some hydroxides. The cations that precipitate under this condition are: Zn2+, Mn2+, Ni2+, Fe2+, Co2+, as sulfides, and Al3+, Cr3+, and Fe3+ as hydroxides. The precipitate is centrifuged and the supernatant decanted to give the next solution. Ion group 4: Insoluble phosphates. The slightly basic supernatant separated from the group 3 ions is treated with (NH4)2HPO4, which precipitates Mg3(PO4)2, Ca3(PO4)2, and Ba3(PO4)2. Ion group 5: Alkali metal and ammonium ions. The final solution contains any of the following ions: Na+, K+, and NH4+. Sample qualitative analysis. A solution contains a mixture of Ag+, Al3+, Cu2+, and Fe3+ ions. Devise a scheme to separate and identify each of these cations. Mixture: Ag+, Al3+, Cu2+, Fe3+ (colored solution) add 3 M HCl, centrifuge

Precipitate (white) AgCl

Supernatant (colored) (Al3+, Cu2+, Fe3+) add 6 M NaOH, centrifuge

Precipitates (Cu(OH)2 & Fe(OH)3)

add 6 M NH3

Precipitate Fe(OH)3 (dark brown)

Supernatant (colorless) Al(OH)4-

Supernatant Cu(NH3)42+ (dark blue)

______________________________________________________________________________ Complex Ion Equilibria 15.8 Equilibria Involving Complex Ions

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A complex ion consists of a central metal ion that is covalently bonded to two or more ligands, which can be anions such as OH-, Cl-, F-, CN-, etc. or neutral molecules such as H2O, CO, and NH3. For example, in the complex ion [Cu(NH3)4]2+, Cu2+ is the central metal ion with four NH3 molecules covalently bonded to it. All complex ions are Lewis adducts; the metal ions act as Lewis acids (electron-pair acceptors) and the ligands are Lewis bases (electron-pair donors). Only species with at least one lone-pair electrons can be a ligand. The Formation of Complex Ion

In aqueous solutions, metal ions form complex ions with water molecules as ligands. When another ligand is introduced into the solution, ligand exchanges occur and equilibrium is established. For example, when NH3 is added to aqueous solution containing Cu2+ ion, the following equilibrium occurs: Cu(H2O)62+(aq) + 4 NH3(aq)

Kf =

⇌

[Cu(NH3)4]2+(aq) + 6H2O

[Cu(NH 3 ) 24  ] [Cu(H 2 O) 62  ][NH 3 ] 4

At molecular level, the ligand exchange process occurs in stepwise manner; water molecule is replace with NH3 molecule one at a time to give a series of intermediate species, each with its own formation constant. For convenience, the water molecules can be omitted from the equation. 1.

Cu2+(aq) + NH3(aq)

Kf1 = 2.

Cu(NH3)2+(aq);

[Cu(NH 3 ) 2 ] [Cu 2  ][NH 3 ]

Cu(NH3)2+(aq) + NH3(aq)

Kf2 = 3.

⇌

Cu(NH3)22+(aq);

[Cu(NH 3 ) 22  ] [Cu(NH 3 ) 2 ][NH 3 ]

Cu(NH3)32+(aq) + NH3(aq)

Kf4 =

⇌

⇌

Cu(NH3)42+(aq);

[Cu(NH 3 ) 24  ] [Cu(NH 3 ) 32 ][NH 3 ]

The overall formation constant is the product of all intermediate formation constants:

[Cu(NH 3 ) 24  ] Kf = Kf1 x Kf2 x Kf3 x Kf4 = [Cu 2  ][NH 3 ]4 Kf is called the formation constant for the complex equilibrium. Sample problem-4: If a 30.0-mL solution containing 0.020 M Cu2+ is mixed with 20.0 mL of 0.20 M NH3 solution, what is the concentration of Cu2+ in solution? Kf = 5.0 x 1012

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28 MASTERING PHYSICAL CHEMISTRY

By-- S.K.SINHA

Solution: First calculate the concentration of each species in the solution after mixing, but before formation of complex ion Cu(NH3)42+: [Cu2+] = 0.020 M x (30.0 mL/50.0 mL) = 0.012 M [NH3] = 0.20 M x (20.0 mL/50.0 mL) = 0.080 M

Write the equilibrium expression: Cu2+(aq) + 4 NH3(aq)

⇌

Cu(NH3)42+(aq) + 4 H2O

Since Kf is very large, we can assume that all of Cu2+ is converted to Cu(NH3)42+. Then, [Cu(NH3)42+] = 0.012 M and [NH3] = 0.080 M – (4 x 0.012 M) = 0.032 M Next, consider the following (reverse) equilibrium: Cu(NH3)42+(aq) Kc =

⇌

Cu2+(aq) + 4NH3(aq)

[Cu 2  ][NH 3 ]4 [Cu(NH 3 ) 24 ]

= 1/Kf = 1/(5.0 x 1012) = 2.0 x 10-13

Set up the following equilibrium table:

Concentration (M) Cu(NH3)42+(aq) ⇌ Cu2+(aq) + 4 NH3(aq)  Initial: 0.012 M 0.000 0.032 M Change: -x +x +x Equilibrium: (0.012 – x) x (0.032 + x)  Since Kc is very small, we assume that x