Question

An acid-base indicator which is a weak acid has a $p{K}_{a}value=5.35$. At what concentration ratio of sodium acetate to acetic acid would the indicator show a colour half-way between those of its acid and conjugate base forms? $p{K}_a$ of acetic acid $=\mathrm{4.75.}[log2=0.3]$.

• A. 4:1
• B. 7:1
• C. 5:1
• D. 2:1

Answer 1

Answer: (A)

4:1

The indicator is a weak acid. its dissociation equilibrium is as given below.
$HIn\rightleftharpoons H^{+}+I{n}^{-}$
The expression for the dissociation constant $K_a$ is
$K_a=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]}$
Colour is observed at halfway of ionization. $\left[I{n}^{-}\right]=\left[HIn\right]$
Hence $K_a=\left[H^{+}\right]$ or $pH=p{K}_a=5.35$.
The expression for the pH of buffer containing acetic acid and sodium acetate is
$pH=p{K}_a+log\frac{\left[Salt\right]}{\left[Acid\right]}$
Substitute values in this expression.
$5.35=4.75+log\frac{\left[Salt\right]}{\left[Acid\right]}\Rightarrow \frac{\left[Salt\right]}{\left[Acid\right]}=\frac{4}{1}$
Hence, $5.35=4.75+log\frac{\left[Salt\right]}{\left[Acid\right]}\Rightarrow \frac{\left[Salt\right]}{\left[Acid\right]}=\frac{4}{1}$.