Question

What would be the pH of a solution in which the molar ratio of propanoic acid: sodium propanoate is 1:10 ka for propanoic acid is 1.34×10^-5 mol/dm^3

Answer 1

$[H^+_{(aq)}]=\frac{Ka[Acid_{(aq)}]}{salt_{(aq)}}$

$=1.34×10^{-5}×\frac{1}{10}$

$=1.33×10^{-6}$ mol dm^-3

pH= $-log[H^+_{(aq)}]=- log (1.34×10^{-6})$

$=5.87$