Chapter 20, Problem 20.80P

Interpretation Introduction

Interpretation:

The $\text{pH}$ value of vitamin C has to be calculated.

Concept Introduction:

The $\text{pH}$ value of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion or hydrogen ion concentration.

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

Answer to Problem 20.80P

The $\text{pH}$ value of solution is $3.3$.

Explanation of Solution

Given: $\text{0}\text{.20 M}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(aq\right)$.

First moles of vitamin C present in $\text{500}\text{.0 mg}$ is calculated.

    $\begin{array}{c}\text{Moles of vitamin C = }\frac{\text{vitamin C mass}}{\text{vitamin C molar mass}}\\ \\ =\frac{500.0\times {10}^{-3}\text{ g}}{176.12\text{ g/mol}}\\ \\ =\text{ 2}\text{.84}\times {\text{10}}^{\text{-3}}\text{ moles}\end{array}$

Now, molarity of the solution is determined as shown below,

    $\begin{array}{c}\text{Molairty = }\frac{\text{Vitamin C moles}}{\text{Volume of solution in L}}\\ \\ =\frac{2.84\times {10}^{-3}\text{ moles}}{1\text{L}}\\ \\ =2.84\times {10}^{-3}\text{M}\end{array}$

The given acid is diprotic acid therefore, the concentration of ${\text{H}}^{\text{+}}$ present in the solution is determined as follows,

For first dissociation the concentration of hydronium ion is shown as below,

    $\begin{array}{l}{\text{ Ascorbic acid +H}}_{\text{2}}\text{O}\rightleftharpoons {\left[\text{Ascorbic acid}\right]}^{-}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \\ Initial2.84\times {10}^{-3}\text{ 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium2.84\times {10}^{-3}\text{ -x x x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{{\left[\text{Ascorbic acid}\right]}^{-}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[\text{Ascorbic acid}\right]}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{{\left[\text{Ascorbic acid}\right]}^{-}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[\text{Ascorbic acid}\right]}\\ \\ 9.12\times {10}^{-5}=\frac{x^2}{2.84\times {10}^{-3}-x}\\ \\ x=\text{ 5}\text{.09}\times {10}^{-4}={\left[\text{Ascorbic acid}\right]}^{-}\text{ = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\end{array}$

For second dissociation the concentration of hydronium ion is shown as below,

  $\begin{array}{l}{\left[\text{Ascorbic acid}\right]}^{-}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\left[\text{Ascorbic acid}\right]}^2{}^{-}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \\ Initial\text{5}\text{.09}\times {10}^{-4}\text{ 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{5}\text{.09}\times {10}^{-4}\text{ -x x x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{Ascorbic acid}}^{2-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{{\left[\text{Ascorbic acid}\right]}^{-}}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{Ascorbic acid}}^{2-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{{\left[\text{Ascorbic acid}\right]}^{-}}\\ \\ 1.99\times {10}^{-12}=\frac{x^2}{5.09\times {10}^{-4}-x}\\ \\ x=\text{ 3}\text{.18}\times {10}^{-8}=\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\end{array}$

Therefore, the total concentration of hydronium ions present in solution is $\text{5}\text{.09}\times {10}^{-4}\text{ + 3}\text{.18}\times {10}^{-8}\text{ = 5}\text{.09}\times {10}^{-4}\text{ M}$.

The $\text{pH}$ value of vitamin C solution is as follows,

    $\begin{array}{c}\text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =\text{ -log}\left[\text{5}\text{.09}\times {10}^{-4}\right]\\ \\ =\text{ 3}\text{.3}\end{array}$