Question

A weak acid (50.0 mL) was titrated with 0.1 M NaOH. The pH values when 10.0 mL and 25.0 mL of base have been added are found to be 4.16 and 4.76 respectively. Calculate $K_p$ of the acid and pH at the equivalence point.

Answer 1

Weak acid $\left(50ml\right)$ $+$ $0.1M$ $NaOH$ form buffer (strong base)

$M$ be molarity of weak acid

We know Henderson equation

$pH=pKa+log\frac{\left(salt\right)}{\left(acid\right)}$

(1) $4.16=pKa+log\frac{[10\times 0.1]}{[50\times M-10\times 0.1]}$

(2) $4.76=pKa+log\left[\frac{25\times 0.1}{50\times M-25\times 0.1}\right]$

From $\left(1\right)$ & $\left(2\right)$

$4.76-4.16=log(\frac{2.5}{50M-2.5}\times \frac{50M-1}{1})$

$0.6=log\left(\frac{125M-2.5}{30M-2.5}\right)$

$3.98=\frac{125M-2.5}{50M-2.5}$

$199.05M-9.95=125M-2.5$

$74.05M=7.45$

$M\approx 0.1$

$\Rightarrow pKa=4.76K_a=1.73\times {10}^{-5}!!$ $PH=4-PKa=14-4.76=9.24$