Question

Calculate the pH after 0.020 moles of HCl is added to 1.00 L of each of the four solutions in Exercise 21.

Short answer

ThepH of each of the following solutions can be calculated.

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}\\ pH=4.89+\log \left(\frac{0.08}{0.12}\right)\\ pH=4.71\end{array}$

Step-by-step solution

Concept of pH of the solution

The pH of the solution can be found using,

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

Here, hydrogen ion concentration is [H⁺].

pH of the solution for 0.100 M propanoic acid after 0.02 mole of HCl is added to 1 L of 0.1 M 

The table of ICE can be represented as,

$H{C}_3{H}_5{O}_2\rightleftharpoons H^{+}+C_3{H}_5{O_2}^{-}$

Then,

$\begin{array}{l}{K}_a=\frac{\left[H^{+}\right]\left[C_3{H}_5{O_2}^{-}\right]}{\left[H{C}_3{H}_5{O}_2\right]}\\ 1.3\times {10}^{-5}=\frac{(0.02+x)x}{(0.1-x)}\\ x=0.00006M\\ \left[H^{+}\right]=0.020+x\\ \approx 0.02M\\ pH=-\log \left(0.020\right)\\ pH=1.7\end{array}$

pH of the solution for 0.100 M sodium propanoate after 0.02 mole of HCl is added to 1 L of 0.1 M

The table of ICE can be represented as,

$C_3{H}_5{O_2}^{-}+H^{+}\rightleftharpoons H{C}_3{H}_5{O}_2$

Then,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}\\ pH=4.89+\log \left(\frac{0.08}{0.02}\right)\\ pH=5.49\end{array}$

pH of the solution after 0.02 mole of HCl is added to 1L of pure waterfor0.100 M sodium propanoate 

The strong acid in water is HCl.

$\begin{array}{l}\left[H^{+}\right]=0.02M\\ pH=-\log \left[H^{+}\right]\\ pH=-\log \left(0.02\right)\\ pH=1.7\end{array}$

pH of the solution after 0.02 mole of HCl is added to 1L of pure water for 0.100 M HC3H5O2 and 0.100 M NaC3H5O2 

It can be denoted as,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}\\ pH=4.89+\log \left(\frac{0.08}{0.12}\right)\\ pH=4.71\end{array}$