Chemistry - Electrochemistry Question with Solution | TestHub

The problem asks for the limiting molar conductivity of B⁻ ions, given the Ksp values for AgA and AgB, the conductivity of a saturated solution containing both, and the limiting molar conductivities of Ag⁺ and A⁻.

First, let's write down the dissociation reactions and Ksp expressions for AgA and AgB:

AgA(s) $\rightleftharpoons$ Ag⁺(aq) + A⁻(aq)

K${}_{sp}$(AgA) = [Ag⁺][A⁻] = 3 $\times$ 10⁻¹⁴

AgB(s) $\rightleftharpoons$ Ag⁺(aq) + B⁻(aq)

K${}_{sp}$(AgB) = [Ag⁺][B⁻] = 1 $\times$ 10⁻¹⁴

Let the solubility of AgA be $s_1$ and the solubility of AgB be $s_2$.

Then, [A⁻] = $s_1$ and [B⁻] = $s_2$.

The total concentration of Ag⁺ ions in the saturated solution is [Ag⁺] = $s_1+s_2$.

Substituting these into the Ksp expressions:

K${}_{sp}$(AgA) = ($s_1+s_2$)$s_1$ = 3 $\times$ 10⁻¹⁴ (Equation 1)

K${}_{sp}$(AgB) = ($s_1+s_2$)$s_2$ = 1 $\times$ 10⁻¹⁴ (Equation 2)

Divide Equation 1 by Equation 2:

$\frac{s_1}{s_2}$ = 3 $⟹$$s_1=3s_2$

Substitute $s_1=3s_2$ into Equation 2:

($3s_2+s_2$)$s_2$ = 1 $\times$ 10⁻¹⁴

$4s_2^2$ = 1 $\times$ 10⁻¹⁴

$s_2^2$ = 0.25 $\times$ 10⁻¹⁴

$s_2$ = $\sqrt{0.25\times 10^{-14}}$ = 0.5 $\times$ 10⁻⁷ M

Now find $s_1$:

$s_1$ = 3$s_2$ = 3 $\times$ 0.5 $\times$ 10⁻⁷ M = 1.5 $\times$ 10⁻⁷ M

The concentrations of the ions are:

[A⁻] = $s_1$ = 1.5 $\times$ 10⁻⁷ M

[B⁻] = $s_2$ = 0.5 $\times$ 10⁻⁷ M

[Ag⁺] = $s_1+s_2$ = 1.5 $\times$ 10⁻⁷ + 0.5 $\times$ 10⁻⁷ = 2.0 $\times$ 10⁻⁷ M

The conductivity ($\kappa$) of the solution is given by Kohlrausch's Law:

$\kappa$ = $\sum c_i{\lambda }_i$

where $c_i$ is the concentration of ion $i$ in mol cm⁻³ and ${\lambda }_i$ is the limiting molar conductivity of ion $i$ in S cm² mol⁻¹.

Given $\kappa$ = 375 $\times$ 10⁻¹⁰ S cm⁻¹

Given ${\lambda }_{A{g}^{+}}$ = 60 S cm² mol⁻¹

Given ${\lambda }_{A^{-}}$ = 80 S cm² mol⁻¹

Let ${\lambda }_{B^{-}}$ be the limiting molar conductivity of B⁻.

First, convert concentrations from M (mol L⁻¹) to mol cm⁻³:

1 M = 1 mol L⁻¹ = 1 mol / 1000 cm³ = 10⁻³ mol cm⁻³

[Ag⁺] = 2.0 $\times$ 10⁻⁷ M = 2.0 $\times$ 10⁻⁷ $\times$ 10⁻³ mol cm⁻³ = 2.0 $\times$ 10⁻¹⁰ mol cm⁻³

[A⁻] = 1.5 $\times$ 10⁻⁷ M = 1.5 $\times$ 10⁻⁷ $\times$ 10⁻³ mol cm⁻³ = 1.5 $\times$ 10⁻¹⁰ mol cm⁻³

[B⁻] = 0.5 $\times$ 10⁻⁷ M = 0.5 $\times$ 10⁻⁷ $\times$ 10⁻³ mol cm⁻³ = 0.5 $\times$ 10⁻¹⁰ mol cm⁻³

Now, apply Kohlrausch's Law:

$\kappa$ = [Ag⁺]${\lambda }_{A{g}^{+}}$ + [A⁻]${\lambda }_{A^{-}}$ + [B⁻]${\lambda }_{B^{-}}$

375 $\times$ 10⁻¹⁰ = (2.0 $\times$ 10⁻¹⁰)(60) + (1.5 $\times$ 10⁻¹⁰)(80) + (0.5 $\times$ 10⁻¹⁰)(${\lambda }_{B^{-}}$)

Divide by 10⁻¹⁰:

375 = (2.0)(60) + (1.5)(80) + (0.5)(${\lambda }_{B^{-}}$)

375 = 120 + 120 + 0.5${\lambda }_{B^{-}}$

375 = 240 + 0.5${\lambda }_{B^{-}}$

0.5${\lambda }_{B^{-}}$ = 375 - 240

0.5${\lambda }_{B^{-}}$ = 135

${\lambda }_{B^{-}}$ = $\frac{135}{0.5}$ = 270 S cm² mol⁻¹

The final answer is $\boxed{270}$.