Question

The indicator constant of phenolphthalein is approximately ${10}^{-8}.$ A solution is prepared by adding $100.01c.c.$ of $0.01N$ sodium hydroxide to $100.00c.c$ of $0.01N$ hydrochloric acid. If a few drops of phenolphthalein are now added, what fraction of the indicator is converted to its coloured form ?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{1}{3}$

NaOH is present in slight excess.
$\underset{100.01\times 0.01}{\mathrm{NaO}H}+\underset{100\times 0.01}{HCl}\to NaCl+H_{2}O$
$0.01c.c$ of N $NaOH$ is left unreacted.
The dissociation equilibrium of the indicator is $HPh\rightleftharpoons H^{+}+P{h}^{-}$
The expression for the dissociation constant is $K_a=\frac{\left[H^{+}\right]\left[P{h}^{-}\right]}{\left[HPh\right]}$
Also, $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
Substittue values in the above expression
${10}^{-8}=\frac{K_w\left[Ph\right]}{\left[O{H}^{-}\right]\left[HPh\right]}=\frac{{10}^{-14}\times \left[Ph\right]}{\frac{0.0001}{200}\left[HPh\right]},\frac{\left[P{h}^{-}\right]}{\left[HPh\right]}=\frac{1}{2}$
Hence, the fraction of the indicator that is converted to coloured form is
$\frac{\left[P{h}^{-}\right]}{\left[P{h}^{-}\right]+\left[HPh\right]}=\frac{1}{3}$