Question

(a) Use the following half-reaction and their reduction potentials to calculate the ${\mathit{K}}_{\mathbf{S}\mathbf{P}}$ of $\mathit{A}\mathit{g}\mathit{B}\mathit{r}$.

$\mathit{A}{\mathit{g}}^{\mathbf{+}}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{A}\mathit{g}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7996}\mathit{V}$

$\mathit{A}\mathit{g}\mathit{B}\mathit{r}\left(s\right)\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{A}\mathit{g}\left(s\right)\mathbf{+}\mathit{B}{\mathit{r}}^{\mathbf{-}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0713}\mathit{V}$

(b) Estimate the solubility of $\mathit{A}\mathit{g}\mathit{B}\mathit{r}$ in 0.10M $\mathit{N}\mathit{a}\mathit{B}\mathit{r}\left(aq\right)$.

Short answer

1. The solubility product$\left({\text{K}}_{\text{sp}}\right)\text{= 4.98}\times {\text{10}}^{\text{- 13}}$
2. The solubility of $\text{AgBr} \text{=} \text{4.98}\times {\text{10}}^{\text{- 12}}$.

Step-by-step solution

Step (1):- Calculation of KSP of AgBr.

$Ag\left(s\right)\to A{g}^{+}\left(aq\right)+e^{-}$$E^0=0.7996V$(1)

$AgBr\left(s\right)+e^{-}\to Ag\left(s\right)$$E^0=0.0713V$(2)

Reverse equation (1) and add with equation (2), then we will get,

$$\begin{gathered} Ag\left(s\right)\to A{g}^{+}\left(aq\right)+e^{-} \\ \frac{AgBr\left(s\right)+e^{-}\to Ag\left(s\right)+B{r}^{-}\left(aq\right)}{AgBr\left(s\right)\to A{g}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)} \end{gathered}$$

$\begin{array}{rcl}{{\text{E}}^{\text{o}}}_{\text{cell}}& =& {{\text{E}}^{\text{o}}}_{\text{cathode}}\text{-}{{\text{E}}^{\text{o}}}_{\text{anode}}\\ & =& \text{0.0713} \text{V -} \text{0.7966V}\\ & =& \text{- 0.7253} \text{V}\end{array}$

$$\begin{gathered} K_{eq}=\left[A{g}^{+}\right]\left[B{r}^{-}\right] \\ {K}_{eq}=K_{SP} \end{gathered}$$

Hence, the solubility product can be evaluated by the following equation:-

$\begin{array}{rcl}log{K}_{SP}& =& \frac{n}{0.0592V}{E}^0\\ log{K}_{SP}& =& \frac{1}{0.0592V}\left(-0.7253V\right)\\ log{K}_{SP}& =& -12.25\\ K_{Sp}& =& 4.98\times {10}^{-13}\end{array}$

Step (2):- Estimation of solubility of AgBr.

We have to prepare the ICE table to find the expression for the equilibrium condition.

	${\text{Ag}}^{\text{+}}\left(\text{aq}\right)$	${\text{Br}}^{\text{-}}\left(\text{aq}\right)$
initial	0.0	0.0
change	s	s
equilibrium	s	0.10+s

$\begin{array}{rcl}{K}_{sp}& =& \left[A{g}^{+}\right]\left[B{r}^{-}\right]\\ 4.98\times {10}^{-13}& =& s\times \left(0.10+s\right)\\ \text{0.1s}& =& \text{4.98}\times {\text{10}}^{\text{- 13}}\\ \text{s}& =& \text{4.98}\times {\text{10}}^{\text{- 12}}\end{array}$

The solubility of $\text{AgBr} \text{=} \text{4.98}\times {\text{10}}^{\text{- 12}}$.