MCQ
One litre of an aqueous solution contain $0.15\, mole$ of $CH_3COOH$ $(pK_a = 4.8)$ and $0.15\, mole$ of $CH_3COONa$. After the addition of $0.05\, mole$ of solid $NaOH$ to this solution, the $pH$ will be
- A$4.5$
- B$4.8$
- ✓$5.1$
- D$5.4$
$pH = pK _{ a }+\log \frac{\text { [Conjugate Base }]}{[\text { Acid }]}$
After adding $NaOH$, it will react with acid and salt forms so now
$pH = pK _{ a }+\log \frac{\text { [Conjugate Base }]}{[\text { Acid }]}$
$=4.8+\log (0.15+0.05) /(0.15-0.05)$
$=4.8+\log (0.20 / 0.10)$
$=5.1$
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