Question

Calculate the pH after 0.020 moles of NaOH 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.

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ pH=-\log (1.3\times {10}^{-5})+\log \left(\frac{0.12}{0.08}\right)\\ pH=5.07\end{array}$

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 

It can be determined as,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}\\ pH=4.89+\log \left(\frac{0.02}{0.08}\right)\\ pH=4.29\end{array}$

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

The table of ICE can be represented as,

$C_3{H}_5{O_2}^{-}+H_{2}O\rightleftharpoons O{H}^{-}+H{C}_3{H}_5{O}_2$

Then,

$\begin{array}{l}pOH=0.02+x\\ pOH=-\log \left[O{H}^{-}\right]\\ pOH=-\log \left(0.02\right)\\ pOH=1.7\\ pH=14-1.7\\ pH=12.30\end{array}$

pH of the solution after 0.02 mole of NaOH is added to 1L of pure water  

The strong acid in water is NaOH.

$\begin{array}{l}pOH=0.02+x\\ pOH=-\log \left[O{H}^{-}\right]\\ pOH=-\log \left(0.02\right)\\ pOH=1.7\\ pH=14-1.7\\ pH=12.30\end{array}$

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,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ pH=-\log (1.3\times {10}^{-5})+\log \left(\frac{0.12}{0.08}\right)\\ pH=5.07\end{array}$