Answer to Question #168232 in Inorganic Chemistry for Arya Prakash

Question

Calculate the pH of 0.100moldm^-3 propanoic acid

Answer 1

First we need to write the Ka expression

CH₃CH₂COOH_(aq) $\iff$ CH₃CH₂COO^- _(aq) + H₃O⁺_(aq)

Ka $=$ $\dfrac{[H3O^+][CH3CH2COO^-]}{[CH3CH2COOH]}$

$Since [H3O^+] =[CH3CH2COO^-]$

Then Ka $=\dfrac{[H3O^+]^2}{[CH3CH2COOH]}$

Now, Ka for Propanoic acid is 1.34x10^-5 and concentration is 0.1M

Therefore, 1.34x10^-5 $=$ $\dfrac{[H3O^+]^2}{0.1M}$

$[H3O^+]^2 = 1.34x10^-3 x 0.1M$

$[H3O^+] =$ $\sqrt{1.34x10^-3x0.1M}$

$[H3O^+] = 1.15758x10^-3$

$pH =-log(1.15758x10^-3)=2.94$

pH = 2.9