Question

A solution is prepared by mixing 100.0 mL of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M}\mathbf{ }\mathit{B}\mathit{e}\mathbf{(}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$ and 100.0 mL of 8.0 M NaF.

$$\begin{gathered} \mathbf{B}{\mathbf{e}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathbf{F}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{B}\mathbf{e}{\mathbf{F}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}} \\ \mathbf{B}\mathbf{e}{\mathbf{F}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathbf{F}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{B}\mathbf{e}{\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}} \\ \mathbf{B}\mathbf{e}{\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathbf{F}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{B}\mathbf{e}{{\mathbf{F}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{3}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}} \\ \mathbf{B}\mathbf{e}{{\mathbf{F}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathbf{F}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\rightleftarrows }\mathbf{B}\mathbf{e}{\mathbf{F}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{4}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{1}} \end{gathered}$$

Calculate the equilibrium concentrations of ${\mathit{F}}^{\mathbf{-}}\mathbf{,}\mathit{B}{\mathit{e}}^{\mathbf{2}\mathbf{+}}\mathbf{,}\mathit{B}\mathit{e}{\mathit{F}}^{\mathbf{+}}\mathbf{,}\mathit{B}\mathit{e}{\mathit{F}}_{\mathbf{2}}\mathbf{,}\mathit{B}\mathit{e}{\mathit{F}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{,}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{B}\mathit{e}{\mathit{F}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$ in this solution.

Short answer

The equilibrium concentrations are as follows

$\begin{array}{c}\left[F^{-}\right]=4.0 M\\ \left[B{e}^{2+}\right]=2.6\times {10}^{-20} M\\ \left[Be{F}^{+}\right]=8.2\times {10}^{-15} M\\ \left[Be{F}_2\right]=1.9\times {10}^{-10} M\\ \left[Be{F}_3^{-}\right]=4.6\times {10}^{-7} M\\ \left[Be{F_4}^{2-}\right]=5.0\times {10}^{-5} M\end{array}$

Step-by-step solution

Definition of equilibrium concentration

Equilibrium is a state of reaction where the rate of forward reaction equals to the rate of backward reaction.

Calculation for finding equilibrium concentration

Thetotal volume of the solution is 200 mL

$\begin{array}{c}Molarity=\frac{number of moles}{Volume of solution in L}\\ \left[B{e}^{2+}\right]=5.0\times {10}^{-5}M\\ \left[F^{-}\right]=\frac{\left(100.0 mL\right)\left(8.0 M\right)}{(200.0 mL)}\\ \left[F^{-}\right]=4.00 M\end{array}$

The expression for equilibrium constant for the reaction $Be{F_3}^{-}\left(aq\right)+F^{-}\left(aq\right)\rightleftharpoons Be{F}_4^{2-}\left(aq\right)$is $K_4=\frac{\left[Be{F}_4^{2-}\right]}{\left[Be{F_3}^{-}\right]\left[F^{-}\right]}$

$\left[Be{F}_4^{2-}\right]=5.0\times {10}^{-5} M$

$\begin{array}{l}\left[Be{F}_3^{-}\right]=\frac{\left[5.0\times {10}^{-5}\right]}{107.9946}\\ \left[Be{F}_3^{-}\right]=4.6\times {10}^{-7}\mathrm{M}\end{array}$

The expression for equilibrium constant for the reaction ${\mathrm{BeF}}_2\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons \mathrm{Be}{{\mathrm{F}}_3}^{-}\left(\mathrm{aq}\right)$is $K_3=\frac{\left[Be{F}_3^{-}\right]}{\left[Be{F}_2\right]\left[F^{-}\right]}$

$\begin{array}{l}\left[Be{F}_2\right]=\frac{\left[4.6\times {10}^{-7}\right]}{2439.878}\\ \left[Be{F}_2\right]=1.9\times {10}^{-10}\mathrm{M}\end{array}$

The expression for equilibrium constant for the reaction ${\mathrm{BeF}}^{+}\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{BeF}}_2\left(\mathrm{aq}\right)$is $K_2=\frac{\left[Be{F}_2\right]}{\left[Be{F}^{+}\right]\left[F^{-}\right]}$

$\begin{array}{l}\left[Be{F}^{+}\right]=\frac{\left[1.9\times {10}^{-10}\right]}{23198.84}\\ \left[Be{F}^{+}\right]=8.2\times {10}^{-15}\mathrm{M}\end{array}$

The expression for equilibrium constant for the reaction ${\mathrm{Be}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{BeF}}^{+}\left(\mathrm{aq}\right)$is$K_1=\frac{\left[Be{F}^{+}\right]}{\left[B{e}^{2+}\right]\left[F^{-}\right]}$

$\begin{array}{l}\left[B{e}^{2+}\right]=\frac{8.2\times {10}^{-15}}{31.6\times {10}^4}\\ \left[B{e}^{2+}\right]=2.6\times {10}^{-20}\mathrm{M}\end{array}$

The concentration of $F^{-}$is calculated as

$\begin{array}{l}\left[F^{-}\right]=4.00-4(5.0\times {10}^{-5})\\ \left[F^{-}\right]=4.0\mathrm{M}\end{array}$