Q.

At 25°C, Ksp for PbBr2 is equal to 8 × 10–5. If the salt is 80% dissociated, what is the solubility of PbBr2 in mol/litre ?

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a

{[\frac{{{{10}^{ - 5}}}}{{1.6 \times 1.6}}]^{\frac{1}{2}}}

b

{[\frac{{{{10}^{ - 4}}}}{{0.8 \times 0.8}}]^{\frac{1}{3}}}

c

{[\frac{{{{10}^{ - 5}}}}{{1.6 \times 1.6}}]^{\frac{1}{3}}}

d

\large {[\frac{{{{10}^{ - 4}}}}{{1.6 \times 1.6}}]^{\frac{1}{3}}}

answer is A.

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Detailed Solution

S = (10-4) / (1.6 × 1.6)1/3

Given data:

  • Solubility product constant, Ksp = 8 × 10-5
  • Degree of dissociation = 80% (0.80)

The dissociation of PbBr2 is represented by the equation:

PbBr2(s) ⇌ Pb2+(aq) + 2Br-(aq)

Using the solubility product expression:

Ksp = [Pb2+] × [Br-]2

Let the solubility of PbBr2 be 'S'. Since the dissociation is 80%, the concentration of Pb2+ will be 0.80S, and the concentration of Br- will be 2 × 0.80S = 1.6S.

The expression for Ksp becomes:

Ksp = (0.80S) × (1.6S)2

Substitute the value of Ksp (8 × 10-5) and solve for S:

8 × 10-5 = (0.80S) × (1.6S)2

After simplifying the equation, the solubility (S) is:

S = (10-4) / (1.6 × 1.6)1/3

Concept Behind:

In this question, we are using the concept of solubility product (Ksp) and the degree of dissociation of the salt. The solubility of PbBr2 is calculated by considering the dissociation and the concentrations of the ions formed.

The degree of dissociation is applied as a factor to express the concentration of ions in equilibrium with the solid salt, leading to the final expression for solubility.

PbB{r_2}\left( s \right) \rightleftharpoons \mathop {PbB{r_2}\left( {aq} \right)}\limits_{S \times \left( {0.80} \right)} \xrightarrow{{}}\mathop {P{b^{ + 2}}\left( {aq} \right)}\limits_{S \times \left( {0.80} \right)} + \mathop {2B{r^ - }\left( {aq} \right)}\limits_{2S \times \left( {0.80} \right)}

{K_{sp}} = \left[ {P{b^{ + 2}}} \right] \times {\left[ {B{r^ - }} \right]^2}

\Rightarrow 8 \times {10^{ - 5}} = \left( {0.8} \right)S \times {\left\{ {\left( {1.6} \right) \times S} \right\}^2}

\Rightarrow S = {\left( {\frac{{8 \times {10}^{ - 5}}}{{8 \times 1.6 \times 1.6}}} \right)^{\frac{1}{3}}}

\Rightarrow S = {\left( {\frac{{{10}^{ - 4}}}{{1.6 \times 1.6}}} \right)^{\frac{1}{3}}}

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