Question

A $0.1$ M solution of weak acid $HA$ is 1% dissociated at 298 K. The new $pH$ when $0.2$ M of $NaA$ is added to it is $530\times {10}^{-x}$, then what is the value of $x$?

Answer 1

For weak acid HA : ${\alpha }_{HA}=\frac{1}{100}=0.01,\left[HA\right]=0.1M$

$\therefore K_a=c{\alpha }^2=0.1\times (0.01{)}^2={10}^{-5}$

Now 0.2 M NaA, a salt of HA, is added to it resulting a buffer solution of [HA] = 0.1 M and [NaA] = 0.2 M

$pH=-\left(log{10}^{-5}\right)+log\frac{0.2}{0.1}$

$pH=5.3010=530\times {10}^{-2}$

$\Rightarrow x=2$