Chapter 16, Problem 74E

A solution is prepared by mixing 100.0 mL of 1.0 × 10⁻⁴ M Be(NO₃)₂ and 100.0 mL of 8.0 M NaF.

$\begin{array}{ll}{\text{Be}}^{\text{2+}}(aq)+F^{-}(aq)\rightleftharpoons Be{F}^{+}(aq)\hfill & K_1=7.9\times {10}^4\hfill \\ {\text{BeF}}^{\text{+}}(aq)+F^{-}(aq)\rightleftharpoons Be{F}_2(aq)\hfill & K_2=5.8\times {10}^3\hfill \\ {\text{BeF}}_{\text{2}}(aq)+F^{-}(aq)\rightleftharpoons Be{F}_3{}^{-}(aq)\hfill & K_3=6.1\times {10}^2\hfill \\ {\text{BeF}}_{\text{3}}{}^{-}(aq)+F^{-}(aq)\rightleftharpoons Be{F}_4{}^{2-}(aq)\hfill & K_4=2.7\times {10}^1\hfill \end{array}$

Calculate the equilibrium concentrations of F⁻, Be²⁺, BeF⁺, BeF₂, BeF₃⁻, and BeF₄²⁻ in this solution.

Interpretation Introduction

Interpretation: The value of different equilibrium constant for the reaction involving the formation of ${\text{BeF}}_4{}^{2-}$ in a stepwise manner is given. The equilibrium concentration of ${\text{BeF}}_3{}^{-}$, ${\text{BeF}}_4{}^{2-}$, ${\text{BeF}}_4{}^{2-}$, ${\text{BeF}}_2$, ${\text{BeF}}^{\text{+}}$, ${\text{Be}}^{\text{2+}}$ is to be calculated.

Concept introduction: The equilibrium that is present between the species that is unionized and its ions is called ionic equilibrium.

Answer to Problem 74E

The equilibrium concentration of ${\text{BeF}}_4{}^{2-}$ is $\underset{\_}{5.0\times 10^{-5}}$ and equilibrium concentration of ${\text{F}}^{-}$ is $\underset{\_}{4.0\text{}M}$.

The equilibrium concentration of ${\text{BeF}}_3{}^{-}$ is $\underset{\_}{4.6\times 10^{-7}M}$.

The equilibrium concentration of ${\text{BeF}}_2$ is. $\underset{\_}{1.9\times 10^{-10}M}$.

The equilibrium concentration of ${\text{BeF}}^{\text{+}}$ is. $\underset{\_}{8.2\times 10^{-15}M}$.

The equilibrium concentration of ${\text{Be}}^{\text{2+}}$ is. $\underset{\_}{2.6\times 10^{-10}M}$.

Explanation of Solution

Explanation

To determine: The equilibrium concentration of ${\text{BeF}}_3{}^{-}$, ${\text{BeF}}_4{}^{2-}$, ${\text{BeF}}_4{}^{2-}$, ${\text{BeF}}_2$, ${\text{BeF}}^{\text{+}}$, ${\text{Be}}^{\text{2+}}$ in the given solution.

Explanation

The equilibrium concentration of ${\text{BeF}}_4{}^{2-}$ is $5.0\times {10}^{-5}$ and equilibrium concentration of ${\text{F}}^{-}$ is $4.0\text{}\text{M}$.

Given

The concentration of $\text{Be}{\left({\text{NO}}_3\right)}_2$ is $1.0\times {10}^{-4}\text{M}$.

The concentration of $\text{NaF}$ is $8.0\text{M}$.

The volume of $\text{Be}{\left({\text{NO}}_3\right)}_2$ is $100.0\text{mL}$.

The volume of $\text{NaF}$ is $100.0\text{mL}$.

The value of $K_1$ is $7.9\times {10}^4$.

The value of $K_2$ is $5.8\times {10}^3$.

The value of $K_3$ is $6.1\times {10}^2$.

The value of $K_4$ is $2.7\times {10}^1$.

The concentration of any species in the solution before any reaction takes place is given as,

${\text{M}}_{\text{i}}{\text{V}}_{\text{i}}={\text{M}}_{\text{f}}{\text{V}}_{\text{f}}$

Where,

• ${\text{M}}_{\text{i}}$ is the initial molarity.

• ${\text{M}}_{\text{f}}$ is the final molarity.

• ${\text{V}}_{\text{i}}$ is the initial volume of the solution.

• ${\text{V}}_{\text{f}}$ is the volume of solution after mixing.

Substitute the value of ${\text{M}}_{\text{i}}$, ${\text{V}}_{\text{i}}$ and ${\text{V}}_{\text{f}}$ of ${\text{Be}}^{2+}$ in the above equation as,

$\begin{array}{c}{\text{M}}_{\text{i}}{\text{V}}_{\text{i}}={\text{M}}_{\text{f}}{\text{V}}_{\text{f}}\\ 1.0\times {10}^{-4}\text{M}\times 100{\text{mL=M}}_{\text{f}}\times 200\text{mL}\\ {\text{M}}_{\text{f}}=5.0\times {10}^{-5}\text{M}\end{array}$

Similarly substitute the value of ${\text{M}}_{\text{i}}$, ${\text{V}}_{\text{i}}$ and ${\text{V}}_{\text{f}}$ of ${\text{F}}^{-}$ in the above equation as,

$\begin{array}{c}{\text{M}}_{\text{i}}{\text{V}}_{\text{i}}={\text{M}}_{\text{f}}{\text{V}}_{\text{f}}\\ 8.0\text{M}\times 100\text{mL}={\text{M}}_{\text{f}}\times 200\text{mL}\\ {\text{M}}_{\text{f}}=4.00\text{M}\end{array}$

As it is seen from the given values of equilibrium constants that the reaction almost goes towards the completion. The final equation is therefore given as,

$\begin{array}{cccccc}& {\text{Be}}^{2+}& +& 4{\text{F}}^{-}& \to & {\text{BeF}}_4{}^{2-}\\ \text{Before reaction}\text{(M) }& 5.0\times {10}^{-5}& & 4.00& & 0\\ \text{Change (M) }& -5.0\times {10}^{-5}& & -5.0\times {10}^{-5}& & +5.0\times {10}^{-5}\\ \text{After reaction (M)}& 0& & 4.00-4\left(5.0\times {10}^{-5}\right)& & 5.0\times {10}^{-5}\end{array}$

This gives the equilibrium concentration of ${\text{BeF}}_4{}^{2-}$ as $\underset{\_}{5.0\times 10^{-5}}$ and equilibrium concentration of ${\text{F}}^{-}$ as $\underset{\_}{4.0\text{}M}$.

The equilibrium concentration of $\left[{\text{BeF}}_3{}^{-}\right]$ is $\underset{\_}{4.6\times 10^{-7}M}$.

The equilibrium constant for the reaction involving the formation of ${\text{BeF}}_4{}^{2-}$ is given as,

$K_4=\frac{\left[{\text{BeF}}_4{}^{2-}\right]}{\left[{\text{BeF}}_3{}^{-}\right]\left[{\text{F}}^{-}\right]}$

Substitute the values of $K_4$, $\left[{\text{BeF}}_4{}^{2-}\right]$ and $\left[{\text{F}}^{-}\right]$ in the above equation as,

$\begin{array}{c}{K}_4=\frac{\left[{\text{BeF}}_4{}^{2-}\right]}{\left[{\text{BeF}}_3{}^{-}\right]\left[{\text{F}}^{-}\right]}\\ 2.7\times {10}^1=\frac{5.0\times {10}^{-5}\text{M}}{\left[{\text{BeF}}_3{}^{-}\right]\left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ \left[{\text{BeF}}_3{}^{-}\right]=\frac{5.0\times {10}^{-5}\text{M}}{2.7\times {10}^1\times \left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ =\underset{\_}{4.6\times 10^{-7}M}\end{array}$

This gives the concentration of $\left[{\text{BeF}}_3{}^{-}\right]$.

The equilibrium concentration of $\left[{\text{BeF}}_2\right]$ is. $\underset{\_}{1.9\times 10^{-10}M}$.

The equilibrium constant for the reaction involving the formation of ${\text{BeF}}_3{}^{-}$ is given as,

$K_3=\frac{\left[{\text{BeF}}_3{}^{-}\right]}{\left[{\text{BeF}}_2\right]\left[{\text{F}}^{-}\right]}$

Substitute the values of $K_3$, $\left[{\text{BeF}}_3{}^{-}\right]$ and $\left[{\text{F}}^{-}\right]$ in the above equation as,

$\begin{array}{c}{K}_3=\frac{\left[{\text{BeF}}_3{}^{-}\right]}{\left[{\text{BeF}}_2\right]\left[{\text{F}}^{-}\right]}\\ 6.1\times {10}^2=\frac{4.6\times {10}^{-7}\text{M}}{\left[{\text{BeF}}_2\right]\left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ \left[{\text{BeF}}_2\right]=\frac{4.6\times {10}^{-7}\text{M}}{6.1\times {10}^2\times \left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ =\underset{\_}{1.9\times 10^{-10}M}\end{array}$

This gives the concentration of $\left[{\text{BeF}}_2\right]$.

The equilibrium concentration of $\left[{\text{BeF}}^{\text{+}}\right]$ is. $\underset{\_}{8.2\times 10^{-15}M}$.

The equilibrium constant for the reaction involving the formation of ${\text{BeF}}_2$ is given as,

$K_2=\frac{\left[{\text{BeF}}_2\right]}{\left[{\text{BeF}}^{\text{+}}\right]\left[{\text{F}}^{-}\right]}$

Substitute the values of $K_2$, $\left[{\text{BeF}}_2\right]$ and $\left[{\text{F}}^{-}\right]$ in the above equation as,

$\begin{array}{c}{K}_2=\frac{\left[{\text{BeF}}_2\right]}{\left[{\text{BeF}}^{\text{+}}\right]\left[{\text{F}}^{-}\right]}\\ 5.8\times {10}^3=\frac{1.9\times {10}^{-10}\text{M}}{\left[{\text{BeF}}^{\text{+}}\right]\left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ \left[{\text{BeF}}^{\text{+}}\right]=\frac{1.9\times {10}^{-10}\text{M}}{5.8\times {10}^3\times \left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ =\underset{\_}{8.2\times 10^{-15}M}\end{array}$

This gives the concentration of ${\text{BeF}}^{+}$

Explanation

The equilibrium concentration of $\left[{\text{Be}}^{\text{2+}}\right]$ is. $\underset{\_}{2.6\times 10^{-10}M}$.

The equilibrium constant for the reaction involving the formation of ${\text{BeF}}^{\text{+}}$ is given as,

$K_1=\frac{\left[{\text{BeF}}^{\text{+}}\right]}{\left[{\text{Be}}^{\text{2+}}\right]\left[{\text{F}}^{-}\right]}$

Substitute the values of $K_1$, $\left[{\text{BeF}}^{\text{+}}\right]$ and $\left[{\text{F}}^{-}\right]$ in the above equation as,

$\begin{array}{c}{K}_1=\frac{\left[{\text{BeF}}^{\text{+}}\right]}{\left[{\text{Be}}^{\text{2+}}\right]\left[{\text{F}}^{-}\right]}\\ 7.9\times {10}^4=\frac{8.2\times {10}^{-15}\text{M}}{\left[{\text{Be}}^{\text{2+}}\right]\left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ \left[{\text{Be}}^{\text{2+}}\right]=\frac{8.2\times {10}^{-15}\text{M}}{7.9\times {10}^4\times \left(4.00-4\left(5.0\times {10}^{-5}\right)\right)\text{}\text{M}}\\ =\underset{\_}{2.6\times 10^{-10}M}\end{array}$

This gives the concentration of ${\text{Be}}^{2+}$

Conclusion

The value of equilibrium constant determines the amount of product formed and the species that are present in the solution at equilibrium.