If solubility product for CaF 2 is 1 . 7 × 10 - 10 at 298 K , calculate the solubility in mol L - 1 .

The solubility equilibrium for CaF 2 is represented as: CaF 2 ( s ) ⇌ Ca 2 + ( aq ) + 2 F - ( aq ) Let the solubility of CaF 2 be s moles / litre . Then the solution will contain Ca 2 + = s moles / litre And, F - = 2 s moles / litre ∴ K sp for CaF 2 = [ Ca 2 + ] [ F - ] 2 = ( s ) ( 2 s ) 2 = 4 s 3 Hence, s = 1 . 7 × 10 - 10 3 4 = 3 . 5 × 10 - 4 mol L - 1

How many moles of AgBr ( K sp = 5 × 10 - 13 mol 2 L - 2 ) will dissolve in 0 . 01 M NaBr solution?

As NaBr dissociates completely, Therefore, in 0 . 01 M NaBr solution, [ Br - ] = 0 . 01 M . If solubility of AgBr in 0 . 01 M NaBr solution is s mol L - 1 , then from AgBr that dissolves, [ Ag + ] = [ Br - ] = s mol L - 1 ∴ Total [ Br - ] = 0 . 01 + s ≃ 0 · 01 M ( ∵ s ≪ 0 . 01 ) Now, K sp for AgBr = [ Ag + ] [ Br - ] = s × 0 . 01 = 0 . 01 s ∴ 0 . 01 s = 5 × 10 - 13 Or, s = 5 × 10 - 11 mol L - 1 .

The values of K sp of two sparingly soluble salts, Ni ( OH ) 2 and AgCN are 2 . 0 × 10 - 15 and 6 . 0 × 10 - 17 , respectively. Which salt is more soluble? Explain.

The solubility equilibrium for Ni ( OH ) 2 may be represented as: Ni ( OH ) 2 ⇌ Ni 2 + + 2 OH - Let the solubility of Ni ( OH ) 2 be s moles / litre . Then the solution will contain s moles of Ni 2 + ions and 2 s moles of OH - ions respectively per litre. Hence, the solubility product, K sp of Ni ( OH ) 2 , would be given by the expression, K sp = [ Ni 2 + ] [ OH - ] 2 = ( s ) × ( 2 s ) 2 = 4 s 3 ∴ 2 × 10 - 15 = 4 s 3 ( ∵ K sp = 2 × 10 - 15 ) Or s 3 = 0 . 5 × 10 - 15 ∴ s = 5 · 8 × 10 - 5 molL - 1 Similarly, for AgCN , the solubility equilibrium may be represented as: AgCN ⇌ Ag + + CN - Let the solubility of AgCN be s moles / litre . Then the solution will contain s moles of Ag + ions and s moles of CN - ions respectively per litre. Hence, the solubility product, K sp of AgCN , would be given by the expression, K sp = [ Ag + ] [ CN - ] = s x s = s 2 ∴ 6 . 0 × 10 - 17 = s 2 ( ∵ K sp = 6 . 0 × 10 - 17 ) Or s = 7 . 8 × 10 - 9 molL - 1 The solubility of Ni ( OH ) 2 is 5 · 8 × 10 - 5 molL - 1 and that of AgCN is 7 . 8 × 10 - 9 molL - 1 . Hence, Ni ( OH ) 2 is more soluble.

Given that the solubility product of radium sulphate ( RaSO 4 ) is 4 × 10 - 11 . Calculate the solubility in pure water.

The solubility equilibrium for RaSO 4 may be represented as: RaSO 4 ⇌ Ra 2 + + SO 4 2 - Let the solubility of RaSO 4 be s moles / litre . Then the solution will contain s moles of Ra 2 + ions and s moles of SO 4 2 - ions respectively per litre. Hence, the solubility product, K sp of RaSO 4 , would be given by the expression, K sp = [ Ra 2 + ] [ SO 4 2 - ] = ( s ) × ( s ) = s 2 But the value of K sp = 4 × 10 - 11 (Given) ∴ 4 × 10 - 11 = s 2 Or s = 6 . 66 × 10 - 6 mol L - 1 .

Given that the solubility product of radium sulphate ( RaSO 4 ) is 4 × 10 - 11 . Calculate the solubility in 0 . 10 M Na 2 SO 4 .

Suppose the solubility of RaSO 4 in 0 . 10 M Na 2 SO 4 be s mol L - 1 . RaSO 4 in the solution dissociates as : RaSO 4 ⇌ Ra 2 + + SO 4 2 - Thus, s mol L - 1 of RaSO 4 in the solution gives s mol L - 1 of Ra 2 + ion and s mol L - 1 of SO 4 2 - ions. But, SO 4 2 - ions are also produced from Na 2 SO 4 . As Na 2 SO 4 dissociates completely, SO 4 2 - from Na 2 SO 4 = 0 . 10 M . Hence, [ Ra 2 + ] = s mol L - 1 And, [ SO 4 2 - ] = s + 0 . 1 mol L - 1 ≃ 0 . 1 mol L - 1 ( ∵ s is very small as compared to 0 . 1 ) K sp of RaSO 4 = [ Ra 2 + ] [ SO 4 2 - ] ∴ s ( 0 . 1 ) = 4 × 10 - 11 Or s = 4 × 10 - 10 .

E M B I B E

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE

S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Ionic Equilibrium in Solution, Exercise 38: PROBLEMS FOR PRACTICE

Author:S C Kheterpal, S N Dhawan & P N Kapil

S C Kheterpal Chemistry Solutions for Exercise - S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Ionic Equilibrium in Solution, Exercise 38: PROBLEMS FOR PRACTICE

Attempt the practice questions on Chapter 9: Ionic Equilibrium in Solution, Exercise 38: PROBLEMS FOR PRACTICE with hints and solutions to strengthen your understanding. Pradeep's Chemistry Vol 1 solutions are prepared by Experienced Embibe Experts.

Questions from S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Ionic Equilibrium in Solution, Exercise 38: PROBLEMS FOR PRACTICE with Hints & Solutions

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 1

Calculate the solubility of silver chloride in water at room temperature if the solubility product of $AgCl$ is $1.6 \times 10^{- 10}$ .

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 2

If solubility product for ${CaF}_{2}$ is $1.7 \times 10^{- 10}$ at $298 K$ , calculate the solubility in $mol L^{- 1}$ .

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 3

How many moles of $AgBr (K_{sp} = 5 \times 10^{- 13} {mol}^{2} L^{- 2})$ will dissolve in $0.01 M NaBr$ solution?

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 4

Calculate the solubility of $A_{2} X_{3}$ in pure water, assuming that neither kind of ion reacts with water. The solubility product of $A_{2} X_{3}, K_{sp} = 1.1 \times 10^{- 23}$ .

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 5

The values of $K_{sp}$ of two sparingly soluble salts, $Ni (OH)_{2}$ and $AgCN$ are $2.0 \times 10^{- 15}$ and $6.0 \times 10^{- 17}$ , respectively. Which salt is more soluble? Explain.

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 6

Calculate the molar solubility of $Ni (OH)_{2}$ in $0.10 M NaOH$ . The ionic product of $Ni (OH)_{2}$ is $2.0 \times 10^{- 15}$ .

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 7

Given that the solubility product of radium sulphate $({RaSO}_{4})$ is $4 \times 10^{- 11}$ . Calculate the solubility in pure water.

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 9 - Ionic Equilibrium in Solution>PROBLEMS FOR PRACTICE>Q 7

Given that the solubility product of radium sulphate $({RaSO}_{4})$ is $4 \times 10^{- 11}$ . Calculate the solubility in $0.10 M {Na}_{2} {SO}_{4}$ .

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