Question

Will a precipitate ofform if 1.0 mL of 1.0M $\mathit{C}\mathit{d}{\mathbf{\left(}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}$is added to 1.0 L of 5.0 M$\mathit{N}{\mathit{H}}_{\mathbf{3}}$ ?

$\begin{array}{l}\mathbf{C}{\mathbf{d}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{C}{\mathbf{d}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\\ \mathbf{C}\mathbf{d}{\left(\mathbf{OH}\right)}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{C}{\mathbf{d}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{O}{\mathbf{H}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}{\mathbf{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\end{array}$

Short answer

The precipitate of$Cd{\left(OH\right)}_2$will not form as the original amount of$Cd{\left(N{O}_3\right)}_2$is less than 0.41 M and$C{d}^{2+}$ will get accommodated as$Cd{\left(N{H}_3\right)}_4^{2+}$.

Step-by-step solution

Definition of precipitate

An insoluble solid that emerges from a liquid solution is defined as precipitate

Calculation

The expression for the equilibrium constant is

$\begin{array}{l}K=\frac{\left[Cd{\left(N{H}_3\right)}_4^{2+}\right]{\left[O{H}^{-}\right]}^2}{{\left[N{H}_3\right]}^4}\\ K=K_{sp}\times K_f\\ K=(5.9\times {10}^{-15})(1.0\times {10}^7)\\ K=5.9\times {10}^{-8}\end{array}$

The value of $K_{b}forN{H}_3$is$1.8\times {10}^{-5}$ .

$\begin{array}{l}1.8\times {10}^{-5}=\frac{x\times x}{5.0}\\ x^2=1.8\times {10}^{-5}\times 5\\ x=\sqrt{9\times {10}^{-5}}\\ x=9.5\times {10}^{-3}\end{array}$

The concentration of$O{H}^{-}$ ions.

$\left[O{H}^{-}\right]=9.5\times {10}^{-3}M$

The concentration of $\left[Cd{\left(N{H}_3\right)}_4^{2+}\right]$,

$5.9\times {10}^{-8}=\frac{\left[Cd{\left(N{H}_3\right)}_4^{2+}\right]{(9.5\times {10}^{-3})}^2}{{(5.0)}^4}$

$\begin{array}{l}\left[Cd{\left(N{H}_3\right)}_4^{2+}\right]=\frac{5.9\times {10}^{-8}\times 625}{90.25\times {10}^{-6}}\\ \left[Cd{\left(N{H}_3\right)}_4^{2+}\right]=40.86\times {10}^{-2}\\ \left[Cd{\left(N{H}_3\right)}_4^{2+}\right]=0.41M\end{array}$

Here,$C{d}^{2+}$ will get accommodated as$Cd{\left(N{H}_3\right)}_4^{2+}$ .

And the initial concentration of$Cd{\left(N{O}_3\right)}_2$ is .$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L}$