Chapter 13, Problem 107E

Interpretation Introduction

Interpretation: The $\text{pH}$ of the given solution, prepared by dissolving one vitamin C tablet in enough water to make $200.0\text{mL}$ of solution, is to be calculated.

Concept introduction: The $\text{pH}$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $\text{pH}$ of a solution is calculated by the formula, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Answer to Problem 107E

Answer

The $\text{pH}$ of the given solution is $\underset{\_}{2.97}$.

Explanation of Solution

Explanation

To determine: The $\text{pH}$ of the given solution, prepared by dissolving one vitamin C tablet in enough water to make $200.0\text{mL}$ of solution.

The equilibrium constant expression for the initial dissociation reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]}{\left[{\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6\right]}$

The initial dissociation reaction is,

${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\left(aq\right)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{a}}$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]}{\left[{\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6\right]}$ (1)

The number of moles of ${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6$ is $\underset{\_}{2.84\times 10^{-3}mol}$.

The mass of one vitamin C tablet is $500\text{mg}\left(0.500\text{g}\right)$.

The molar mass of ${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6$ $\begin{array}{l}=\text{6C}+8\text{H}+6\text{O}\\ =\left(\left(6\times 12\right)+\left(8\times 1\right)+\left(6\times 16\right)\right)\text{g/mol}\\ =176\text{g/mol}\end{array}$

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the value of the given mass and the molar mass of ${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6$ in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{0}\text{.500}\text{g}}{\text{176}\text{g/mol}}\\ =\underset{\_}{2.84\times 10^{-3}mol}\end{array}$

The initial $\left[{\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6\right]$ is $\underset{\_}{0.014M}$.

The calculated number of moles of ${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6$ is $2.84\times {10}^{-3}\text{mol}$.

The total volume is $200\text{mL}\left(0.200\text{L}\right)$.

The concentration is calculated by the formula,

$\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}$

Substitute the value of the number of moles of ${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6$ and the volume in the above expression.

$\begin{array}{c}\text{Concentration}=\frac{\text{2}\text{.84}\times {\text{10}}^{-3}\text{mol}}{\text{0}\text{.200}\text{L}}\\ =\underset{\_}{0.014M}\end{array}$

The $\left[{\text{H}}^{\text{+}}\right]$ from first reaction is $\underset{\_}{1.051\times 10^{-3}M}$.

The change in concentration of ${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6$ is assumed to be $x$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.014& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.014-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6\right]$ is $\left(0.014-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]$ is $x\text{M}$.

The $K_{\text{a}}$ for ${\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6$ is $7.9\times {10}^{-5}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{H}}_2{\text{C}}_6{\text{H}}_6{\text{O}}_6\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]$ in equation (1).

$\begin{array}{c}7.9\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{\left[0.014-x\right]}\\ 7.9\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.014-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.014$. Hence, it is ignored from the term $\left[0.014-x\right]$.

Simplify the above expression.

$\begin{array}{c}7.9\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.014\right]}\\ {\left[x\right]}^2=\left(1.106\times {10}^{-6}\right)\\ \left[x\right]=\underset{\_}{1.051\times 10^{-3}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ from the first reaction is $\underset{\_}{1.051\times 10^{-3}M}$.

The initial $\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]$ is $\underset{\_}{1.051\times 10^{-3}M}$.

According to the ICE table formed,

The $\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]$ is equal to the $\left[{\text{H}}^{\text{+}}\right]$,that is $1.051\times {10}^{-3}\text{M}$.

This is the initial $\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]$ for the further reaction.

The $\left[{\text{H}}^{\text{+}}\right]$ from second reaction is $\underset{\_}{1.6\times 10^{-12}M}$.

The change in concentration of ${\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}$ is assumed to be $y$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{C}}_6{\text{H}}_6{\text{O}}_6^{2-}\left(aq\right)\\ \text{Inititial}& 1.051\times {10}^{-3}& & 1.051\times {10}^{-3}& & 0\\ \text{Change}& -y& & +y& & +y\\ \text{Equilibrium}& \left(1.015\times {10}^{-3}\right)-y& & \left(1.051\times {10}^{-3}\right)+y& & y\end{array}$

The equilibrium concentration of $\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]$ is $\left(\left(1.015\times {10}^{-3}\right)-y\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $\left(1.051\times {10}^{-3}\right)+y\text{M}$.

The equilibrium concentration of $\left[{\text{C}}_6{\text{H}}_6{\text{O}}_6^{2-}\right]$ is $y\text{M}$.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_6{\text{H}}_6{\text{O}}_6^{2-}\right]}{\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]}$ (2)

The $K_{\text{a}}$ for this reaction is $1.6\times {10}^{-12}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{HC}}_6{\text{H}}_6{\text{O}}_6^{-}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{C}}_6{\text{H}}_6{\text{O}}_6^{2-}\right]$ in equation (2).

$1.6\times {10}^{-12}=\frac{\left[\left(1.051\times {10}^{-3}\right)+y\right]\left[y\right]}{\left[\left(1.051\times {10}^{-3}\right)-y\right]}$

The value of $y$ will be very small as compared to $1.051\times {10}^{-3}$. Hence, it is ignored from the term $\left[\left(1.051\times {10}^{-3}\right)-y\right]$ and from $\left[\left(1.051\times {10}^{-3}\right)+y\right]$

Simplify the above expression.

$\begin{array}{c}1.6\times {10}^{-12}=\frac{\left[\left(1.051\times {10}^{-3}\right)\right]\left[y\right]}{\left[1.051\times {10}^{-3}\right]}\\ \left[y\right]=\underset{\_}{1.6\times 10^{-12}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ from the second reaction is $\underset{\_}{1.6\times 10^{-12}M}$.

The total $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{1.05116\times 10^{-3}M}$.

The total $\left[{\text{H}}^{\text{+}}\right]$ is calculated by the formula,

$\text{Total}\left[{\text{H}}^{\text{+}}\right]=\left[{\text{H}}^{\text{+}}\right]\text{from}\text{first}\text{reaction}+\left[{\text{H}}^{\text{+}}\right]\text{from}\text{second}\text{reaction}$

Substitute the value of the $\left[{\text{H}}^{\text{+}}\right]$ from the first and second reaction in the above expression.

$\begin{array}{c}\text{Total}\left[{\text{H}}^{\text{+}}\right]=\left(\left(1.051\times {10}^{-3}\right)+\left(1.6\times {10}^{-12}\right)\right)\text{M}\\ =\underset{\_}{1.05116\times 10^{-3}M}\end{array}$

The required $\text{pH}$ value is $\underset{\_}{2.97}$.

The $\text{pH}$ of a solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[1.05116\times {10}^{-3}\right]\\ =\underset{\_}{2.97}\end{array}$

Conclusion

Conclusion

The $\text{pH}$ of the given solution, prepared by dissolving one vitamin C tablet in enough water to make $200\text{mL}$ of solution is $\underset{\_}{2.97}.$