Chapter 8, Problem 131E

Interpretation Introduction

Interpretation:The equilibrium concentrations of ${\text{F}}^{\text{-}},{\text{ Be}}^{2+},{\text{ BeF}}^{+},{\text{ BeF}}_{\text{2}},{\text{ BeF}}_3^{-}$ and ${\text{BeF}}_4^{2-}$ needs to be calculated for a mixture formed by mixing 100.0 mL of each $1.0\times {10}^{-4}{\text{ M Be(NO}}_{\text{3}}{)}_2$ and $\text{8}\text{.0 M NaF}$ .

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 131E

  $\begin{array}{l}{\text{[F}}^{\text{-}}\text{] = 4}\text{.0 M}\\ {\text{[Be}}^{2+}\text{] = 2}{\text{.6×10}}^{\text{-20}}\text{ M }\\ {\text{[BeF}}^{+}]\text{ = 8}{\text{.2×10}}^{\text{-15}}\text{ M}\\ {\text{[BeF}}_{\text{2}}\text{] = 1}{\text{.9×10}}^{\text{-10}}\text{ M}\\ {\text{[BeF}}_3^{-}]=\text{ 4}{\text{.6×10}}^{\text{-7}}\text{ M}\\ {\text{[BeF}}_4^{2-}]=\text{5}{\text{.0×10}}^{\text{-5}}\text{ M}\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{\text{Be}}^{\text{2+}}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}^{+}\left(\text{aq}\right){\text{ K}}_1\text{ = 7}\text{.9}\times {\text{10}}^4\\ {\text{BeF}}^{+}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}_2\left(\text{aq}\right){\text{ K}}_2\text{ = 5}\text{.8}\times {\text{10}}^3\\ {\text{BeF}}_2\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}_3^{-}\left(\text{aq}\right){\text{ K}}_3\text{ = 6}\text{.1}\times {\text{10}}^2\\ {\text{BeF}}_3^{-}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}_4^{2-}\left(\text{aq}\right){\text{ K}}_4\text{ = 2}\text{.7}\times {\text{10}}^1\end{array}$

The total volume of the solution is:

  $100.0\text{ mL + 100}\text{.0 mL = 200}\text{.0 mL}$

Before any reaction to occur, the concentration of ${\text{Be}}^{\text{2+}}$ and ${\text{F}}^{\text{-}}$ is:

  $\text{molarity = }\frac{\text{number of moles}}{\text{Volume of solution in L}}$

Substituting the values:

  $\begin{array}{l}\left[{\text{Be}}^{\text{2+}}\right]\text{ = }\frac{\left(\text{100}\text{.0 mL}\right)\left(\text{1}{\text{.0×10}}^{\text{-4}}\text{ M}\right)}{\left(\text{200}\text{.0 mL}\right)}\\ \left[{\text{Be}}^{\text{2+}}\right]\text{ = 5}{\text{.0×10}}^{\text{-5}}\text{ M}\\ \left[{\text{F}}^{\text{-}}\right]\text{ = }\frac{\left(\text{100}\text{.0 mL}\right)\left(\text{8}\text{.0 M}\right)}{\left(\text{200}\text{.0 mL}\right)}\\ \left[{\text{F}}^{\text{-}}\right]\text{ = 4}\text{.00 M}\end{array}$

The values of given formation constant are large so, the reaction is assumed to undergo completion. The net reaction is written as:

  ${\text{ Be}}^{\text{2+}}\left(\text{aq}\right){\text{ + 4F}}^{\text{-}}\left(\text{aq}\right)\to {\text{BeF}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)$

The concentration before and after the reaction is:

  $\begin{array}{l}{\text{ Be}}^{\text{2+}}\left(\text{aq}\right){\text{ + 4F}}^{\text{-}}\left(\text{aq}\right)\to {\text{ BeF}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)\\ \text{Before reaction(M): 5}{\text{.0×10}}^{\text{-5}}\text{ 4}\text{.00 0 }\\ \text{After reaction(M): 0 }\left(\text{4}\text{.00-4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)\text{ 5}{\text{.0×10}}^{\text{-5}}\end{array}$

So, the ${\text{[BeF}}_4^{2-}]=\text{5}{\text{.0×10}}^{\text{-5}}\text{ M}$ .

Now, the $\left[{\text{BeF}}_{\text{4}}{}^{\text{2-}}\right]$ is calculated using equation:

  ${\text{BeF}}_3^{-}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}_4^{2-}\left(\text{aq}\right){\text{ K}}_4\text{ = 2}\text{.7}\times {\text{10}}^1$

The expression for equilibrium constant is:

  ${\text{K}}_{\text{4}}\text{=}\frac{\left[{\text{BeF}}_{\text{4}}{}^{\text{2-}}\right]}{\left[{\text{BeF}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{F}}^{\text{-}}\right]}$

Substituting the values:

  $\begin{array}{l}\text{2}{\text{.7×10}}^{\text{1}}\text{ = }\frac{\left[\text{5}{\text{.0×10}}^{\text{-5}}\right]}{[{\text{BeF}}_{\text{3}}{}^{\text{-}}]\left(\text{4}\text{.00 - 4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ [{\text{BeF}}_{\text{3}}{}^{\text{-}}]\text{ =}\frac{\left[\text{5}{\text{.0×10}}^{\text{-5}}\right]}{\text{2}{\text{.7×10}}^{\text{1}}\left(\text{4}\text{.00 - 4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ [{\text{BeF}}_{\text{3}}{}^{\text{-}}]\text{ =}\frac{\left[\text{5}{\text{.0×10}}^{\text{-5}}\right]}{\text{107}\text{.9946}}\\ [{\text{BeF}}_{\text{3}}{}^{\text{-}}]\text{ = 4}{\text{.6×10}}^{\text{-7}}\end{array}$

The concentration of ${\text{BeF}}_2$ is calculated using equation:

  ${\text{BeF}}_2\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}_3^{-}\left(\text{aq}\right){\text{ K}}_3\text{ = 6}\text{.1}\times {\text{10}}^2$

The expression for equilibrium constant is:

  ${\text{K}}_{\text{3}}\text{=}\frac{\left[{\text{BeF}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{BeF}}_{\text{2}}\right]\left[{\text{F}}^{\text{-}}\right]}$

Substituting the values:

  $\begin{array}{l}\text{6}{\text{.1×10}}^{\text{2}}\text{= }\frac{\left[\text{4}{\text{.6×10}}^{\text{-7}}\right]}{\left[{\text{BeF}}_{\text{2}}\right]\left(\text{4}\text{.00-4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ \left[{\text{BeF}}_{\text{2}}\right]\text{ = }\frac{\left[\text{4}{\text{.6×10}}^{\text{-7}}\right]}{\text{6}{\text{.1×10}}^{\text{2}}\left(\text{4}\text{.00-4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ \left[{\text{BeF}}_{\text{2}}\right]\text{ = }\frac{\left[\text{4}{\text{.6×10}}^{\text{-7}}\right]}{\text{2439}\text{.878}}\\ \left[{\text{BeF}}_{\text{2}}\right]\text{ = 1}{\text{.9×10}}^{\text{-10}}\end{array}$

The concentration of ${\text{BeF}}^{+}$ is calculated using equation:

  ${\text{BeF}}^{+}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}_2\left(\text{aq}\right){\text{ K}}_2\text{ = 5}\text{.8}\times {\text{10}}^3$

The expression for equilibrium constant is:

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

Substituting the values:

  $\begin{array}{l}\text{5}{\text{.8×10}}^{\text{3}}\text{= }\frac{\text{1}{\text{.9×10}}^{\text{-10}}}{\left[{\text{BeF}}^{\text{+}}\right]\left(\text{4}\text{.00 - 4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ \left[{\text{BeF}}^{\text{+}}\right]\text{ = }\frac{\text{1}{\text{.9×10}}^{\text{-10}}}{\text{5}{\text{.8×10}}^{\text{3}}\left(\text{4}\text{.00 - 4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ \left[{\text{BeF}}^{\text{+}}\right]\text{ = }\frac{\text{1}{\text{.9×10}}^{\text{-10}}}{23198.84}\\ \left[{\text{BeF}}^{\text{+}}\right]\text{ = 8}{\text{.2×10}}^{\text{-15}}\end{array}$

The concentration of ${\text{Be}}^{2+}$ is calculated using equation:

  ${\text{Be}}^{\text{2+}}\left(\text{aq}\right){\text{ + F}}^{\text{-}}\left(\text{aq}\right)\rightleftharpoons {\text{BeF}}^{+}\left(\text{aq}\right){\text{ K}}_1\text{ = 7}\text{.9}\times {\text{10}}^4$

The expression for equilibrium constant is:

  ${\text{K}}_{\text{1}}\text{ = }\frac{\left[{\text{BeF}}^{\text{+}}\right]}{\left[{\text{Be}}^{\text{2+}}\right]\left[{\text{F}}^{\text{-}}\right]}$

Substituting the values:

  $\begin{array}{l}\text{7}{\text{.9×10}}^{\text{4}}\text{ = }\frac{\text{8}{\text{.2×10}}^{\text{-15}}}{[{\text{Be}}^{\text{2+}}]\left(\text{4}\text{.00-4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ [{\text{Be}}^{\text{2+}}]\text{ = }\frac{\text{8}{\text{.2×10}}^{\text{-15}}}{\text{7}{\text{.9×10}}^{\text{4}}\left(\text{4}\text{.00-4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\right)}\\ [{\text{Be}}^{\text{2+}}]\text{ = }\frac{\text{8}{\text{.2×10}}^{\text{-15}}}{\text{31}{\text{.6×10}}^{\text{4}}}\\ [{\text{Be}}^{\text{2+}}]\text{ = 2}{\text{.6×10}}^{\text{-20}}\end{array}$

The concentration of ${\text{F}}^{-}$ is calculated using:

  $\begin{array}{l}\left[{\text{F}}^{\text{-}}\right]\text{= 4}\text{.00-4}\left(\text{5}{\text{.0×10}}^{\text{-5}}\right)\\ \left[{\text{F}}^{\text{-}}\right]\text{ = 4}\text{.0}\end{array}$

Hence, the concentration at equilibrium:

  $\begin{array}{l}{\text{[F}}^{\text{-}}\text{] = 4}\text{.0 M}\\ {\text{[Be}}^{2+}\text{] = 2}{\text{.6×10}}^{\text{-20}}\text{ M }\\ {\text{[BeF}}^{+}]\text{ = 8}{\text{.2×10}}^{\text{-15}}\text{ M}\\ {\text{[BeF}}_{\text{2}}\text{] = 1}{\text{.9×10}}^{\text{-10}}\text{ M}\\ {\text{[BeF}}_3^{-}]=\text{ 4}{\text{.6×10}}^{\text{-7}}\text{ M}\\ {\text{[BeF}}_4^{2-}]=\text{5}{\text{.0×10}}^{\text{-5}}\text{ M}\end{array}$