Question 1571
CBSE Class 11 Chemistry · Equilibrium · Acid-base equilibria and pH
The ionization constant of propanoic acid is 1.32 × 10–5. Determine
(a)
the degree of ionization of the acid in its 0.05M solution
(b)
also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Solution to Question 1571
Complete workingFor propanoic acid alone use x≈√(KaC). In 0.010 M HCl, common H⁺ suppresses dissociation and α≈Ka/[H⁺].
\[\begin{aligned}x &= \sqrt{(1.32\times10^{-5})(0.050)}\\&= 8.12\times10^{-4}\end{aligned}\]
\[\begin{aligned}\alpha &= 0.0163,\quad pH\\&= 3.09\end{aligned}\]
\[\begin{aligned}\alpha_{HCl} &= \frac{1.32\times10^{-5}}{0.010}\\&= 1.32\times10^{-3}\end{aligned}\]
Final answers by part
(a)
α=0.0163 and pH=3.09 in pure acid
(b)
α≈1.32×10⁻³ in 0.010 M HCl.