Question

What amount (in milli mole) of sodium acetate $\left(C{H}_{3}COONa\right)$ must be added to $2\text{L}$ of $0.1\text{M}$ acetic acid solution to given a solution of pH = 3 [Given: $K_{a\left(C{H}_{3}COOH\right)}={10}^{-5}]$

Answer 1

$\begin{array}{cccc}& C{H}_{3}COOH\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons & C{H}_{3}CO{O}^{-}\left(aq\right)+& H_3{O}^{+}\left(aq\right)\\ t=0& 0.1M& 0& 0\\ t=t& 0.1-x& x& x\end{array}$
$K_a=\frac{x^2}{0.1-x}$
As it is a weak acid then $(0.1-x)\approx 0.1$
$\left[H_3{O}^{+}\right]=x=\sqrt{K_a\times 0.1}={10}^{-3}$
$\therefore pH=3$
On addition of $C{H}_{3}COONa,$ pH will change. So, we don't have to add any $C{H}_{3}COONa,$ to form a solution of $pH=3$.
So, answer is $0$.