MCQ
$0.1\, M$ acetic acid solution is titrated against $0.1\, M\, NaOH$ solution. What would be the difference in $pH$ between $1/4$ and $3/4$ stages of neutralisation of acid ?
- A$2\,\log \,3/4$
- B$2\, log\, 1/4$
- C$ log\, 1/3$
- ✓$2\,log\,3$
✓
Answer
Correct option: D.
$2\,log\,3$
d
$CH _3 COOH + NaOH \longrightarrow CH _3 COONa + H _2 O$
$CH _3 COOH + NaOH \longrightarrow CH _3 COONa + H _2 O$
$\frac{0.3}{4} \quad\quad\quad\quad\quad \quad-\quad \quad\quad\quad\frac{0.1}{4}(1 / 4$ neutralization $)$
$\therefore \quad p H_{1}= p K_a+\log \frac{1}{3}$
$CH _3 COOH + NaOH \longrightarrow CH _3 COONa + H _2 O$
$\frac{0.1}{4}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{0.3}{4}$
$\therefore pH _2= pK _2+\log 3$
$\therefore pH _1 \sim pH _2=\log \frac{1}{3} \sim \log 3$
$=2 \log 3$
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