Question

If very small amount of phenolphthalein is added to $0.15mol{litre}^{-1}$ solution of sodium benzoate, what fraction of the indicator will exist in the coloured form? State any assumption that you make.
$K_a$ (Benzoic acid) $=6.2\times {10}^{-5},K_w\left(H_{2}O\right)=1\times {10}^{-14}$
$K_{ln}$ (Phenolphthalein) $3.16\times {10}^{-10}$

Answer 1

Formula for pH of salt hydrolysis:
$pH=\frac{1}{2}[p{K}_w+p{K}_a+\log C]$
$=\frac{1}{2}[14+{\log }_{10}6.2\times {10}^{-5}+\log 0.15]=8.6918$
Formula for pH of indicator:
$pH=p{K}_{ln}+{\log }_{10}\frac{\left[{In}^{-}\right]}{\left[HIn\right]}$
$8.6918=-{\log }_{10}3.16\times {10}^{-10}+{\log }_{10}\frac{\left[{In}^{-}\right]}{\left[HIn\right]}$
$0.16=\frac{\left[{In}^{-}\right]}{\left[HIn\right]}=$ Fraction of indicator in coloured form.