Question

The acid dissociation Ka of propionic acid C2H5CO2H is 1.310x10−5. Calculate the pH of a 3.010x10−4M aqueous solution of propionic acid. Round your answer to 2 decimal places.

Answer 1

Answer : The pH of the solution is, 4.25

Solution :  Given,

Concentration (C) = $3.010\times 10^{-4}M$

Acid dissociation constant = $k_a=1.310\times 10^{-5}$

The equilibrium reaction for dissociation of $C_2H_5CO_2H$ (weak acid) is,

                           $C_2H_5CO_2H\rightleftharpoons C_2H_5CO_2^-+H^+$

initially conc.        c                       0                0

At eqm.             $c(1-\alpha)$                $c\alpha$                $c\alpha$

where, $\alpha$ is degree of dissociation

First we have to calculate the value of $\alpha$

$K_a=\frac{(c\alpha)\times (c\alpha)}{c(1-\alpha)}$

$K_a=\frac{c(\alpha)^2}{(1-\alpha)}$

Now put all the given values in this formula, we get:

$1.310\times 10^{-5}=\frac{(3.010\times 10^{-4})(\alpha)^2}{(1-\alpha)}$

$\alpha=0.188$

Now we have to calculate the hydrogen ion concentration.

$[H^+]=c\alpha=(3.010\times 10^{-4})\times (0.188)=5.66\times 10^{-5}M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (5.66\times 10^{-5})$

$pH=4.25$

Therefore, the pH of the solution is, 4.25