DPP 04 Ionic Equilibrium JH Sir-4295

TARGET JEE (ADVANCED) : 2014 COURSE NAME : UMANG (UP)

PHYSICAL INORGANIC CHEMISTRY DAILY PRACTICE PROBLEMS

DPP

DATE ATE : 23.09.2013 to 28.09.2013

DPP NO. 55 & 57

DPP No.# 55  1.

2.

Solubility product  Write the equation for dissociation of following salts and determine their solubility in terms of Ksp. (i) PbBr2 (ii) Hg2CrO4 (iii) BaC2O4 (iv) Fe(OH)3 (v) Ag2CO3 (vi) Sb2S3 (vii) AgCNS (viii) Ag3PO4 (ix) Li3Na3(AlF6)2 (x) Hg2I2 (xi) Ba3(PO4)2 (xii) Ca5(PO4)3F (xiii) A3B4 (xiv) CaF2 (xv) Ag2CrO4 (i) Ksp of AgCl is 1.8 × 10 –10. The solubility of AgCl in pure water in moles/litre m oles/litre :  –5  –5  –5 (A) 1.34 × 10 (B) 4.24 × 10 (C) 1.8 × 10 (D) 3.6 × 10 –5 (ii) In the above question the solubility in gram per litre will be – (A) 180.26 × 10 –5 (B) 192.29 × 10 –5 (C) 210 × 10 10 –4

(D) 176.85 × 10 –4

3.

Calculate solubility of AB2 in pure water. Ksp of AB2 = 25.6 × 10 –8,

4.

Solubility of a A2B salt in pure water is 2 × 10 –5 moles in 100 ml. Calculate Ksp of salt.

5.

Ksp of AgBr is 4 × 10 –13 and [Al+] is a solution is 1 × 10 –6 m/l what is [Br –] in that solution.

6.

If Ksp of AgI is 8.5 × 10 –17. The maximum amount am ount of AgI which can be dissolved in 2500 ml of water will be – (A) 9.22 × 10 –9 gm  gm (B) 2.3 × 108 gm  gm (C) 5.42 × 10 –5 gm  gm (D) 2.17 × 10 –5 gm

7.

The volume of water needed to dissolve 1 g of BaSO4(Ksp = 1.1 × 10 –10) at 25ºC is  – (A) 820 litre (B) 410 litre (C) 205 litre (D) none of these

8.

How many grams of CaC2O4 will dissolve in distilled water to make one litre of saturated solution of it ? (Ksp of CaC2O4 = 2.5 × 10 –9 mol2 lit –2) (A) 0.0064 gm (B) 0.1028 gm (C) 0.1280gm (D) 0.2056 gm

9.

At 20ºC, the Ag+ ion concentration in a saturated solution Ag2CrO4  is 1.5 × 10 –4 mole/lit. At 20ºC, the solubility product of Ag2CrO4 could be  – (A) 3.37 × 10 –12 (B) 1.68 × 10 –10 (C) 1.68 × 10 –12 (D) 1.12 × 10 –10

10.

In the system CaF2(s) Ca2+(ag) + 2F – increasing the concentration of Ca2+ ions 4 times will cause the eq. concentration of F – ions to change to .............. times the initial value. (A) 4 (B) 1/2 (C) 2 (D) 1/4

11.

Concentration of Ag+ ion in a saturated solution of Ag2CrO4 is 5.4 × 10 –6 g/litre when the salt is 50% dissociated. Then solubility product of Ag2CrO4 is  is –  –17 (A) 7.03 × 10 (B) 6.25 × 10 –23 (C) 1.72 × 10 –23 (D) 2.15 × 10 –24

12.

(i) For an experiment Pb(OH)2 is taken , if salt is 80% dissociated & Ksp of Pb(OH)2 is 8 × 10 –6. Then solubility of salt in moles/litre is  – (A) 1.57 × 10 –2 (B) 2 × 10 –2 (C) 1.26 × 10 –5 (D) 2.3 × 10 –2 (ii) Considering the above question, what will be the solubility in gms/litre – (A) 3.32 (B) 3.65 (C) 3.05 (D) 3.79

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13.

pH of a saturated solution of A(OH)3 is 10. Calculate Ksp of A(OH)3 at 25 ºC.

14.

Solubility of As2S3 is 7.38 mg in 10ml of water. Calculate Ksp of As2S3.

15.

Ksp of A3B4 is 6.912 × 10 –18. Calculate solubility of A3B4.

DPP No.# 56  1.(a)

(b)

!3

The solubility of A2X is y mol dm  . Its solubility product is : (A) 6 y4 (B) 64 y4 (C) 36 y5

(D) 4 y3

The solubility of sparingly soluble electrolyte MmAa in water is given by the expression :

(  K sp  % (A) s = && m a ## ' m a  $

m"a

1 / m " a

(  K sp  % (B) s = && m a ## ' m a  $

(  K sp  % (C) s = && a m ## ' m a  $

m"a

1 / m " a

(  K sp  % (D) s = && a m ## ' m a  $

2.(a)

Three sparingly soluble salts M2X, MX and MX3 have the solubility product are in the ratio of 4: 1 : 27. Their solubilities will be in the order (A) MX3 > MX > M2 X (B) MX3 > M2X > MX (C) MX > MX3 > M2X (D) MX > M2X > MX3

(b)

A particular saturated solution of silver chromate, Ag2CrO4, has [Ag+]= 5×10 –5 and [CrO42 –] = 4.4×10 –4 M. What is value of Ksp for Ag2CrO4 ? (A) 1.1 × 10 –12 . (B) 1.5 × 10 –12 (C) 2 × 10 –6 (D) 1 × 1012.

3.(a)

If the solubility product of silver oxalate is 5 × 10 –10, what will be the weight of Ag2C2O4 in 2.5 litres of a saturated solution ? (Ag = 108, C = 12, O = 16). (A) 0.50 gm (B) 0.38 gm (C) 0.30 gm (D) 0.45 gm.

(b)

A student wants to prepare a saturated solution of Ag+ ion . He has got three samples AgCl (Ksp = 10 !10), AgBr (Ksp = 1.6 × 10 !13) and Ag2CrO4 (Ksp = 3.2 × 10 !11) . Which of the above compound will be used by him using minimum weight to prepare 1 lit. of saturated solution. (A) AgCl (B) AgBr (C) Ag2 CrO4 (D) all the above .

4.(a) (b)

5.(a)

(b)

If the solubility of Ag2SO4 in 10 –2 M Na2SO4 solution be 2 × 10 –8 M then Ksp of Ag2SO4 will be: (A) 32 × 10 –24 (B) 16 × 10 –18 (C) 32 × 10 –18 (D) 16 × 10 –24 The solubility of CaF2 in water at 1518ºC is 2 × 10 –4 mole/litre. Calculate Ksp of CaF2 and its solubility in 0.1M NaF solution. (A) 3.5 × 108 mole/litre (B) 3.0 × 109 mole/litre. (C) 3.3 × 10 –9 mole/litre (D) 4.0 × 107 mole/litre Calculate F— in a solution saturated with respect of both MgF2 and SrF2. Ksp(MgF2) = 9.5 x 10-9, Ksp(SrF2) = 4 x 10-9. (A) 3 × 10 –3 M. (B) 4 × 10 –2 M. (C) 3.5 × 10 –3 M

(D) 1 × 10 –3 M.

A solution is saturated with respect to SrCO3  & SrF2. The [CO32!] was found to be 1.2 x 10!3  M. The concentration of F! in the solution would be : K sp (SrCO3) = 10 –9, Ksp(SrF2) = 3 × 10 –11. (A) 3 x 10!3 M (B) 2 x 10!2 M (C) 6 x 10!2 M (D) 6 x 10!7 M

6.

Calculate the solubility of AgCl (s) in (a) pure water (b) 0.1 M NaCl (c) 0.01 M CaCl2 at 25º C . !10 Ksp (AgCl) = 2.56 ) 10 . Comment on the influence of [Cl !] on the solubility of AgCl.

7.

Find the solubility of CaF2 in 0.5 M solution of CaCl2 and water. How many times in solubility in the second case greater than in the first ? Ksp (CaF2) = 3.2 × 10 –11.

8.

If you place the amounts given below in pure water, will all of the salt dissolve before equilibrium can be established, or will some salt remain undissolved ? (a) 4.96 mg of MgF2 in 125 ml of pure water, Ksp = 3.2 x 10-8 (b) 3.9 mg of CaF2 in 100 ml of pure water, Ksp = 4 x 10-12 Also find the percentage saturation in each case. ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303

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DPP No.# 57  Solubility in presence of common ion  1. (i)

Solubility product of AgCl is 1.8 × 10 –10, then find out solubility of AgCl in 0.1M NaCl (A) 3.6 × 10 –10 (B) 1.8 × 10 –10 (C) 3.6 × 10 –9 (D) 1.8 × 10 –9

(ii)

0.2M AgNO3 (A) 3 × 10 –10

(B) 0.36 × 10 –10

(C) 9 × 10 –10

(D) 2.6 × 10 –10

2M CaCl2 (A) 9 × 10 –10

(B) 4.5 × 10 –11

(C) 1.5 × 10 –11

(D) 6 × 10 –11

(iv) pure water. (A) 1.34 × 10 –5

(B) 4.34 × 10 –5

(C) 2.87 × 10 –5

(D) 1.89 × 10 –5

(iii)

(iv)

2.

Ksp of PbCl2 is 4 ×10 –15 calculate its solubility in :

(i)

pure water

(ii)

0.2 M AlCl3 solution

(iii)

5 × 10 –3 M Pb3(PO4)2 solution

(iv)

0.02 M NaCl solution

(v)

0.4 M Pb(NO3)2 solution

3.

The solubility of AgCl will be minimum in  – (A) 0.01 M AgNO3 (B) pure water

(C) 0.01 M CaCl2

(D) 0.1 M NaCl

4.

The solubility product of AgCl is 4 × 10 –10 at 298 K. The solubility of AgCl in 0.04 M CaCl2 will be – (A) 2 × 10 –5 M (B) 1 × 10 –4 M (C) 5 × 10 –9 M (D) 2.2 × 10 –4 M

5.

The molar solubility of PbI2 in 0.2 M Pb(NO 3)2 solution in terms of solubility product, Ksp of PbI2 is – 1 / 2

(A)

( K sp % && ## 0 . 2 '   $

1 / 2

(B)

( K sp % && ## 0 . 8 '   $

1 / 3

( K sp  % ## (C) && 0 . 8 '   $

1 / 2

( K sp % ## (D) && 0 . 4 '   $

6.

Ksp of Zn(OH)2 is 4.5 × 10 –17 then its solubility in a solution having pH = 10 will be  – (A) 4.5 × 10 –10 (B) 1.4 × 10 –10 (C) 6.7 × 10 –10 (D) 7.6 × 10 –10

7.

Solubility of Mg(OH)2 having Ksp equal to 8.9 × 10 –13, in a solution containing 500 ml of 0.2 M NH4OH and 500 ml of 0.4 M Ca(OH)2 is – (A) 3.4 × 10 –19 (B) 55.63 × 10 –13 (C) 2.34 × 10 –9 (D) 8.34 × 10 – 13

8.

Molar solubility of As2S3 in 0.3 M Al2S3 solution in terms of solubility product Ksp of As2S3 is – (A)

K sp 2.918

(B) 3

K sp 2.916

(C)

K sp 1. 2

(D) 3

K sp 1. 2

9.

Ksp of PbI2 is 8 × 10 –12. A solution contains 2 × 10 –3 M Pb+2 ions and 2 × 10 –4 M I – ions. Predict ppt of PbI2 will form or not.

11.

The precipitate of CaF2 (Ksp = 1.7 × 10 –10) is obtained when equal volumes of the following are mixed – (A) 10 –4 Ca2+ + 10 –4 M F – (B) 10 –2 M Ca2+ + 10 –3 M F – (C) 10 –5 M Ca2+ + 10 –3 M F – (D) 10 –3 M Ca2+ and 10 –3 M F – ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303

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12.

To Ag2CrO4 solution over its own precipitate, CrO4 –2 ions are added. This result in (A) increase in Ag+ concentration (B) decrease in Ag+ concentration (C) increase in solubility product (D) None of these

13.

What is the minimum concentration of SO42 –  required to precipitate BaSO4  in a solution containing 1 × 10 –4 mole of Ba2+ ? Ksp for BaSO4 = 4 × 10 –10 (A) 4 × 10 –10 M (B) 4 × 10 –6 M (C) 2 × 10 –7 M (D) 2 × 10 –3 M

14.

What would be the pH at which Fe(OH) 2 begins to precipitate from a solution containing 0.009 M Fe+2 ions. Ksp of Fe(OH)2 is 1.8 × 10 –15. (A) 8.37 (B) 6.35 (C) 7.65 (D) 9.28

15.

How many grams of CaBr2 (MM = 200) can be added to 250 ml of 0.01 M solution of silver nitrate to just start the precipitation of silver bromide. Ksp of AgBr is 5.0 × 10 –13.

16.

A solution contains 0.1 M Zn+2 ions and is saturated with H2S. Calculatee amount of HCl which should be added in 500 ml of solution to prercipitate ZnS. Ksp of ZnS = 13.5 × 10 –23. K1 of H2S = 10 –7  and K2 of H2S = 1.5 × 10 –14 concentration of H2S in saturated solutin is 0.1N.

17.

A solution has 0.02 M A+2 and 0.1M NH4OH calculate the concentration of NH4Cl required to prevent the formation of A(OH)2 in solution. Ksp of A(OH)2 is 2 × 10 –12 and Kb of NH4OH is 10 –5.

ANSWER KEY  DPP No.# 55  2. 5. 10. 15.

(i) (A) (ii) (B) Br –= 4 × 10 –7 m/l (B) 11. (B)  –3 10 m/l

3. 6. 12.

Ksp of AB2 = 25.6 × 10 –8, A = 4 × 10 –3 m/l 4. (C) 7. (B) 8. (A) 9.  –17 (i) (A) (ii) (D) 13. 3.34 × 10 14.

3.2 ×10 –11 (C) 26.24 × 10 –12

DPP No.# 56  1.(a) 4.(a) 6. 7. 8.

(D) (b) (B) 2.(a) (B) (b) (A) 3.(a) (B) (B) (b) (C) 5.(a) (A) (b) (C)  -5  -9 (a) 1.6 × 10   mol/lit. (b) 2.56 × 10 mol/lit. (c) 1.28 × 10 –8 mol/lit. 4 × 10 –6 M, 2 × 10 –4 M, 50 times. (a) Will dissolve, 32% saturation (b) will not dissolve, 100% saturation.

(b)

(B)

DPP No.# 57  1. 2. 3.

(i) (D) 1.8 × 10 –9 (i)10 –5 M (ii) (C) 4. (C)

11. 17.

(B) 0.1 M

12.

(B)

Solubility in presence of common ion  (ii) (C) (iii) (B) (iv) (A)  –14  –7 1.11 × 10  M (iii) 2.58 ×10  M (iv) 5. (B) 6. (A) 7. (B) 13.

(B)

14.

(C)

15.

10 –11 M (v) 8. (A)

1.25 × 10 –9 gm 16.

5 × 10 –8 M 9. Yes 6.01 gm

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