Question

A 50.0 mL sample of 0.0413 M $\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\left(aq\right)$is added to 50.0 mL of 0.100 M$\mathit{N}\mathit{A}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$. Calculate the $\mathbf{\left[}{\mathbf{Ag}}^{\mathbf{+}}\mathbf{\right]}$at equilibrium in the resulting solution.$\mathbf{[}{\mathbf{K}}_{\mathbf{sp}}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{AglO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{17}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{]}$

Short answer

Answer

The $\left[A{g}^{+}\right]$at equilibrium in the resulting solution is $\mathbf{1}\mathbf{.}\mathbf{08}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{M}$.

Step-by-step solution

Definition of concentration

The amount of a substance in a defined space is referred to as the concentration.

Calculations

For $\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\left(aq\right)$

$$\begin{gathered} \mathit{n}\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{\times }\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathit{i}\mathit{n}\mathbf{}\mathit{m}\mathit{L} \\ \mathit{n}\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0413}\mathbf{\times }\mathbf{50}\mathbf{.}\mathbf{0}\mathit{m}\mathit{L} \\ \mathit{n}\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{07}\mathbf{}\mathit{m}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

For $\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\left(aq\right)$

$$\begin{gathered} \mathit{n}\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{\times }\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathit{i}\mathit{n}\mathbf{}\mathit{m}\mathit{L} \\ \mathit{n}\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{\times }\mathbf{50}\mathbf{.}\mathbf{0}\mathit{m}\mathit{L} \\ \mathit{n}\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

The reaction on mixing $\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\left(aq\right)$and$\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

$\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathit{N}\mathit{o}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathit{A}\mathit{g}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathit{N}\mathit{a}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The formation of $\mathit{A}\mathit{g}\mathit{l}{\mathit{O}}_{\mathbf{3}}$as a precipitate takes place due to small ${\mathit{K}}_{\mathbf{s}\mathbf{p}}$value

Total volume of the mixture is 100 mL

Molarity of $\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\left(aq\right)$is calculated as

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{m}\mathbf{i}\mathbf{l}\mathbf{l}\mathbf{i}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{N}\mathbf{a}\mathbf{l}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{v}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{m}\mathbf{L}} \\ \mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{93}\mathbf{}\mathbf{m}\mathbf{m}\mathbf{o}\mathbf{l}}{\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{L}} \\ \mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0293}\mathit{M} \end{gathered}$$

$\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$is a strong electrolyte and it will dissociate completely into its ions as

$\mathit{N}\mathit{a}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathit{N}{\mathit{a}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The molarity of $\mathit{l}{{\mathit{O}}_{\mathbf{3}}}^{\mathbf{-}}\left(aq\right)$is $\mathbf{0}\mathbf{.}\mathbf{0293}\mathit{M}$.

The equilibrium equation for the solubility of $\mathit{A}\mathit{g}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

$\mathit{A}\mathit{g}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{A}{\mathit{g}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathit{l}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The expression for solubility constant is

${\mathit{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{\left[}{\mathbf{Ag}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{lO}}_{\mathbf{3}}\mathbf{\right]}$

Determining the concentration at equilibrium

	$\mathit{A}\mathit{g}\mathit{l}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{A}{\mathit{g}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathit{l}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$
Initially(M):	0	0.0293
At Equilibrium(M):	s	S+0.0293

Substituting the values, the $\mathbf{\left[}{\mathbf{Ag}}^{\mathbf{+}}\mathbf{\right]}$at equilibrium in the resulting solution is

$$\begin{gathered} \mathbf{3}\mathbf{.}\mathbf{17}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\left(s\right)\left(0.02993M\right) \\ \mathit{s}\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{17}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}{\mathbf{0}\mathbf{.}\mathbf{0293}\mathbf{M}} \\ \mathit{s}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{08}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{M} \end{gathered}$$