Question

Calculate the pH of each of the following solutions.
a. 0.100 M propanoic acid (HC₃H₅0₂ K_a = 1.3 x 10^-5)
b. 0.100 M sodium propanoate (NaC₃H₅O₂)
c. pure H₂O
d. 0.100 M HC₃H₅O₂ and 0.100 M NaC₃H₅O₂

Short answer

ThepH of each of the following solutions can be calculated.

a. The table of ICE can be represented as,

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

Then,

$\begin{array}{l}{K}_a=\frac{\left[H^{+}\right]\left[C_3{H}_5{O_2}^{-}\right]}{\left[H{C}_3{H}_5{O}_2\right]}\\ 1.3\times {10}^{-5}=\frac{x^2}{0.1}\\ x^2=1.3\times {10}^{-6}\\ x=1.1\times {10}^{-3}M\\ pH=-\log \left[H^{+}\right]\\ pH=-\log (1.1\times {10}^{-3})\\ pH=2.96\end{array}$

b. 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,

$\begin{array}{l}{K}_a=\frac{\left[O{H}^{-}\right]\left[H{C}_3{H}_5{O}_2\right]}{\left[C_3{H}_5{O_2}^{-}\right]}\\ 7.7\times {10}^{-10}=\frac{x^2}{0.1}\\ x^2=7.7\times {10}^{-11}\\ x=8.8\times {10}^{-6}M\\ pOH=-\log \left[O{H}^{-}\right]\\ pOH=-\log (8.8\times {10}^{-6})\\ pOH=5.06\\ pH+pOH=14\\ pH+5.06=14\\ pH=8.94\end{array}$

c.The table of ICE can be represented as,

$2H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$

Then,

$\begin{array}{l}{K}_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right]\\ 1\times {10}^{-14}=x^2\\ x=\left[O{H}^{-}\right]=\left[H_3{O}^{+}\right]\\ pH=-\log \left[H_3{O}^{+}\right]\\ pH=-\log (1\times {10}^{-7})\\ pH=7\end{array}$

d. It can be represented as,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}\\ pH=-\log (1.3\times {10}^{-5})+\log \left(\frac{0.1}{0.1}\right)\\ pH=-\log (1.3\times {10}^{-5})\\ pH=4.89\end{array}$

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

The table of ICE can be represented as,

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

Then,

$\begin{array}{l}{K}_a=\frac{\left[H^{+}\right]\left[C_3{H}_5{O_2}^{-}\right]}{\left[H{C}_3{H}_5{O}_2\right]}\\ 1.3\times {10}^{-5}=\frac{x^2}{0.1}\\ x^2=1.3\times {10}^{-6}\\ x=1.1\times {10}^{-3}M\\ pH=-\log \left[H^{+}\right]\\ pH=-\log (1.1\times {10}^{-3})\\ pH=2.96\end{array}$

pH of the solution for 0.100 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,

$\begin{array}{l}{K}_a=\frac{\left[O{H}^{-}\right]\left[H{C}_3{H}_5{O}_2\right]}{\left[C_3{H}_5{O_2}^{-}\right]}\\ 7.7\times {10}^{-10}=\frac{x^2}{0.1}\\ x^2=7.7\times {10}^{-11}\\ x=8.8\times {10}^{-6}M\\ pOH=-\log \left[O{H}^{-}\right]\\ pOH=-\log (8.8\times {10}^{-6})\\ pOH=5.06\\ pH+pOH=14\\ pH+5.06=14\\ pH=8.94\end{array}$

pH of the solution for pure H2O

The table of ICE can be represented as,

$2H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$

Then,

$\begin{array}{l}{K}_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right]\\ 1\times {10}^{-14}=x^2\\ x=\left[O{H}^{-}\right]=\left[H_3{O}^{+}\right]\\ pH=-\log \left[H_3{O}^{+}\right]\\ pH=-\log (1\times {10}^{-7})\\ pH=7\end{array}$

pH of the solution for 0.100 M HC3H5O2 and 0.100 M NaC3H5O2 

It can be represented as,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[base\right]}{\left[acid\right]}\\ pH=-\log (1.3\times {10}^{-5})+\log \left(\frac{0.1}{0.1}\right)\\ pH=-\log (1.3\times {10}^{-5})\\ pH=4.89\end{array}$