Question

Calculate the $pH$ of ${10}^{-6}M$ ${CH}_{3}COOH$
Take $K_a=2\times {10}^{-5}$

Answer 1

We have,

The concentration of $C{H}_{3}COOH$ is ${10}^{-6}M$

and $K_a=2\times {10}^{-5}$

Now we have,

$K_a=\frac{\left[H^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

let the equilibrium concentration of $\left[H^{+}\right]$ and $\left[C{H}_{3}COOH\right]$ be $x$.

$\therefore 2\times {10}^{-5}=\frac{x^2}{{10}^{-6}}$

$x^2=2\times {10}^{-11}$

$\therefore x=4.47\times {10}^{-12}$

Now $pH=-lo{g}_{10}\left[H^{+}\right]$

$\therefore pH=-log(4.47\times {10}^{-11})$

$pH=11.35$