Chapter 18, Problem 112A

Interpretation Introduction

Interpretation:

The quantity of water that must be added to increase the $pH$ is to be calculated.

Concept introduction:

$k_a$ value remains constant at constant temperature. To calculate the amount water that should be added, firstly volume of solution is to be calculated using $pH$ value.

Formulas to be used are:

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

Answer to Problem 112A

$Volumeofwater=30.1mL$

Explanation of Solution

$HX$dissociates as:

$HX\to H^{+}+X^{-}$

Original solution:

$\begin{array}{l}{k}_a=2.14\times {10}^{-6}\\ pH=3.8\end{array}$

$\begin{array}{l}\left[H^{+}\right]=anti\log (-3.8)\\ =1.58\times {10}^{-4}M\end{array}$

$\left[X^{-}\right]=\left[H^{+}\right]$

$\begin{array}{l}{k}_a=\frac{\left[H^{+}\right]\times \left[X^{-}\right]}{\left[HX\right]}\\ 2.14\times {10}^{-6}=\frac{1.58\times {10}^{-4}\times 1.58\times {10}^{-4}}{\left[HX\right]}\\ \left[HX\right]=0.0117M\end{array}$

Dilute solution:

$\begin{array}{l}{k}_a=2.14\times {10}^{-6}\\ pH=4\end{array}$

$\begin{array}{l}\left[H^{+}\right]=anti\log (-4.0)\\ =1\times {10}^{-4}M\end{array}$

$\left[X^{-}\right]=\left[H^{+}\right]$

$\begin{array}{l}{k}_a=\frac{\left[H^{+}\right]\times \left[X^{-}\right]}{\left[HX\right]}\\ 2.14\times {10}^{-6}=\frac{1\times {10}^{-4}\times 1\times {10}^{-4}}{\left[HX\right]}\\ \left[HX\right]=0.00467M\end{array}$

Now, number of moles of solute will remain same after dilution.

So,

$\begin{array}{l}0.0117\times 20=0.00467\times V_{final}\\ V_{final}=\frac{0.0117\times 20}{0.00467}\\ =50.1mL\end{array}$

$\begin{array}{c}Volumeofwater=50.1-20\\ =30.1mL\end{array}$

Conclusion