Question

A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.

Answer 1

PKa= 4.9 therefore ka= 10^-4.9=  1.259x10^-5

$ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}$

$[CH3CH2COO^-]$= 0.05
$[CH3CH2COOH]$= 0.10

Therefore 1.259x10^-5 = $\frac{[H^+][0.05]}{[0.1]}$
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore $[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}$

Therefore $[H^+]= 2.513*10^-5$

pH= -log [$H^+$] = -log(2.513*10^-5)= 4.59.