Question

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

• A. $K_a=1.73\times {10}^{-5}$, $pH=9.34$
• B. $K_a=1.73\times {10}^{-7}$, $pH=6.73$
• C. $K_a=1.73\times {10}^{-4}$, $pH=9.73$
• D. None of these

Answer 1

The correct option is A $K_a=1.73\times {10}^{-5}$, $pH=9.34$

Let the molarity of the weak acid be x

$pH=p{K}_a+\log \frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$

$4\cdot 16=p{K}_a+\log \frac{[0.1\times 10]}{[x\times 50-0.1\times 10]}$

$4\cdot 16=p{K}_a+\log \left[\frac{1}{50x-1}\right]$ (1)

$4.76=p{K}_a+\log \frac{[25\times 0.1]}{[50x-25\times 0.1]}$

$4\cdot 76=p{K}_a+\log \left[\frac{2\cdot 5}{50x-2\cdot 5}\right]$ (2)

Subtracting 1 & 2

$4\cdot 76-4\cdot 16=\log \left[\frac{2\cdot 5}{50x-2\cdot 5}\right]-\log \left(\frac{1}{50x-1}\right]$

$0.6=\log (\frac{2\cdot 5}{50x-2.5}\times 50x-1]$

$3.98=\frac{135x-2.5}{50x-2.5}$

$\Rightarrow 199x-9.95=125x-2.5$

$x=0.1$ Substituting in (1)

$p{K}_a=4.76\Rightarrow K_a=1.73\times {10}^{-5}$

$pH=14-4.76=9.24$