Question

Calculate the $pH$ after the addition of $90ml$ and $100ml$ respectively of $0.1NNaOH$ to $100ml0.1NC{H}_{3}COOH$.

(Given $p{K}_a$ for $C{H}_{3}COOH=4.74$)

Answer 1

$1)$ Addition of $90ml$ base:

$m$ moles of acid $=100\times 0.1=10$

$m$ moles of base added $=90\times 0.1=9$

$\therefore m$ moles of salt formed $=9$

$\left[salt\right]=\frac{9}{90+100}=\frac{9}{190}M,\left[acid\right]=\frac{10-9}{190}=\frac{1}{190}M$

$pH=pKa+\log \frac{\left[salt\right]}{\left[base\right]}=4.74+\log \frac{9}{1}$

$pH=5.69$

$2)$ Addition of $100ml$ base:

$m$ moles of base added $=100\times 0.1=10$

$\therefore$ All of the acid reacts with base to give salt of strong base and weak acid.

$\left[salt\right]=\frac{10}{100+100}=0.05M$

$pH=7+\frac{1}{2}[pKa+\log c]=7+\frac{1}{2}[4.74+\log 0.05]$

$pH=8.72$