Question

The copper (I) ion forms a complex ion with ${\mathbf{CN}}^{\mathbf{-}}$according to the following equation:
${\mathbf{Cu}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{3}\mathbf{CN}{\left(\mathrm{aq}\right)}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Cu}{\left(\mathrm{CN}\right)}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}\right)$${\mathbf{K}}_{\mathbf{f}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}$

a. Calculate the solubility of$\mathbf{CuBr}\left(s\right)$$\left(K_{\mathrm{sp}}=1.0\times {10}^{-5}\right)$in$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{L}$of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{MNaCN}$.
b. Calculate the concentration of${\mathbf{Br}}^{\mathbf{-}}$ at equilibrium.
c. Calculate the concentration of${\mathbf{CN}}^{\mathbf{-}}$ at equilibrium.

Short answer

1. The solubility of$\mathbf{CuBr}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ in$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{L}$ of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{M}\mathbf{}\mathbf{NaCN}$is $\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$.
2. The concentration of${\mathbf{Br}}^{\mathbf{-}}$ at equilibrium is $\mathbf{0}\mathbf{.}\mathbf{33}\mathbf{M}$.
3. The concentration of${\mathbf{CN}}^{\mathbf{-}}$ at equilibrium is $\mathbf{0}\mathbf{.}\mathbf{0048}\mathbf{M}$.

Step-by-step solution

Introduction to the Concept.

The solubility product constant is a constant that determines the solubility equilibrium that occurs between a substance in its solid-state and its ions in the aqueous state. It has the following general expression.

${\mathbf{A}}_{\mathbf{x}}{\mathbf{B}}_{\mathbf{y}}\left(S\right)\mathbf{\rightleftharpoons }{\mathbf{xA}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{yB}}^{\mathbf{-}}\left(\mathrm{aq}\right)$

${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{A}}^{+}\right]}^{\mathrm{x}}\times {\left[{\mathrm{B}}^{-}\right]}^{\mathrm{Y}}$

The solubility product constant is ${\mathrm{K}}_{\mathrm{sp}}$, and the concentration is represented by square brackets.

Determination of formation constant.

Here, the disassociation reaction of CuBr is

$\mathrm{CuBr}\to {\text{Cu}}^{+}+{\mathrm{Br}}^{-}-----\left(1\right)$

From the given Solubility constant expression is given as,

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Cu}}^{+}\right]\times \left[{\mathrm{Br}}^{-}\right]$

And the formation constant expression,

${\text{Cu}}^{+}\left(\mathrm{aq}\right)+3{\mathrm{CN}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons \mathrm{Cu}{\left(\mathrm{CN}\right)}_3^{2-}\left(\mathrm{aq}\right)-----\left(2\right)$

${\mathrm{K}}_{\mathrm{f}}=\frac{\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_3^2\right]}{\left[{\mathrm{Cu}}^{+}\right]\times {\left[{\mathrm{CN}}^{-}\right]}^3}$

Determination of solubility ofCuBr

By adding equation (1) and (2),

$\text{Cu}\mathrm{Br}+3\mathrm{CN}\rightleftharpoons \mathrm{Cu}{\left(\mathrm{CN}\right)}_3^{2-}+{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)$

$\mathrm{K}={\mathrm{K}}_{\mathrm{sp}}\times {\mathrm{K}}_{\mathrm{f}}$

$\mathrm{K}=\left(1.0\times {10}^{-5}\right)\times \left(1.0\times {10}^{11}\right)$

$\mathrm{K}=1.0\times {10}^6$

Therefore, the Solubility constant of CuBr is $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$.

Determination of concentration of Br−

From the given,

$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{M}\left[\mathrm{NaCN}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{M}\left[{\mathrm{CN}}^{-}\right]$

The ICE table for the below reaction is given as,

$\mathbf{CuBr}\mathbf{+}\mathbf{3}{\mathbf{CN}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Cu}{\left(\mathrm{CN}\right)}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{+}{\mathbf{Br}}^{\mathbf{-}}$

Initial(M)	1	0	0
Change(M)	-1	$\frac{1}{3}$(1)	$\frac{1}{3}$
Equillibrium(M)	0	0.33	0.33

Therefore, the concentration of ${\mathbf{Br}}^{\mathbf{-}}$is $\mathbf{0}\mathbf{.}\mathbf{33}\mathbf{M}$.

Determination of concentration of CN−

From the given,

Let’s take${\mathbf{CN}}^{\mathbf{-}}$as x,

The ICE table for the below reaction is given as,

$\mathbf{CuBr}\mathbf{+}\mathbf{3}{\mathbf{CN}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Cu}{\left(\mathrm{CN}\right)}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{+}{\mathbf{Br}}^{\mathbf{-}}$

Initial(M)	0	0.33	0.33
Change(M)	x	$-\frac{1}{3}\left(x\right)$	$-\frac{1}{3}\left(x\right)$
Equilibrium(M)	x	$0.33-\frac{1}{3}\left(x\right)$	$0.33-\frac{1}{3}\left(x\right)$

Here the equilibrium constant expression is written as,

$\mathrm{K}=\frac{\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_3^2\right]\times \left[{\mathrm{Br}}^{-}\right]}{\left[{\mathrm{CN}}^{-}\right]}$

By putting the values except $\frac{1}{3}$ ,

$1.0\times {10}^6=\frac{{\left(0.33\right)}^2}{{\left(x\right)}^3}$

$\mathrm{x}=0.0048\text{M}$

Therefore, the concentration of ${\mathbf{CN}}^{\mathbf{-}}$is $\mathbf{0}\mathbf{.}\mathbf{0048}\mathit{M}$

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