MCQ
What is the $pH$ of solution obtained by mixing $5.076\ gm$ of methyl ammonium nitrate $(CH_3NH_3NO_3)$ to $120\ ml$, $0.225\ M$ methylamine $(CH_3NH_2$ ; $K_b$ = $4 × 10^{-4})$.
- A$3.7$
- B$4.3$
- ✓$10.3$
- D$11$
✓
Answer
Correct option: C.
$10.3$
c
m. moles of salt $=\frac{5.076}{94} \times 1000=54$
m. moles of salt $=\frac{5.076}{94} \times 1000=54$
m. moles of weak base $=120 \times 0.225=27$
$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]} \Rightarrow 4-\log 4+\log \frac{54}{27}$
$\mathrm{pOH}=3.7$
$\mathrm{pH}=10.3$
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