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At 25°C, K_sp 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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${[\frac{{{{10}^{ - 5}}}}{{1.6 \times 1.6}}]^{\frac{1}{2}}}$

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

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

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

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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