Question

Calculate the degree of ionisation and the pH of a 0.01 M aqueous solution of formic acid at 298K. [Given, Ka (HCOOH) = 1.7  10−4at 298K]. 

Answer 1

$HCOOH \leftrightarrows H^+ +HCOO^-$

When equilibrium is reached the concentrations of each form will be:

c(HCOOH) = 0.01 - x M,

c(H⁺) = x M,

c(HCOO^-) = x M, where x is the dissociated portion of the acid

$Ka(HCOOH)=1.7*10^{-4}=\dfrac{c(H^+)c(HCOO^-)}{c(HCOOH)}=\dfrac{x^2}{0.01-x}$

This equation gives $x =0.0012$ M

Then the degree of ionisation: $\alpha=\dfrac{c(H^+)}{c(HCOOH)}=\dfrac{0.0012}{0.01}=0.12$

$pH=-log_{10}c(H^+)= -log_{10}0.0012=2.92$