Q101 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Chemistry
The Basic buffer solution is made with $0.1\ \text{M}\ \text{NH}_4\text{OH}$ and $0.2\ \text{M}\ \text{NH}_4\text{Cl}$. If $K_b$ for the base is $1.8 \times 10^{-5}$, what is the pH of the buffer?
- A$4.74$
- B$9.26$
- Correct answerC$8.96$
- D$5.05$
Solution
Given:
$[\text{NH}_4\text{OH}] = 0.1\ \text{M}$
$[\text{NH}_4\text{Cl}] = 0.2\ \text{M}$
$K_b = 1.8 \times 10^{-5}$
First, calculate $\text{p}K_b$:
$\text{p}K_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 4.74$
Using the Henderson-Hasselbalch equation for a basic buffer:
$\text{pOH} = \text{p}K_b + \log \left( \dfrac{[\text{Salt}]}{[\text{Base}]} \right)$
Substituting the given values:
$\text{pOH} = 4.74 + \log \left( \dfrac{0.2}{0.1} \right)$
$\text{pOH} = 4.74 + \log 2$
$\text{pOH} = 4.74 + 0.30 = 5.04$
Now, calculate the pH:
$\text{pH} = 14 - \text{pOH}$
$\text{pH} = 14 - 5.04 = 8.96$
Answer: $8.96$
$[\text{NH}_4\text{OH}] = 0.1\ \text{M}$
$[\text{NH}_4\text{Cl}] = 0.2\ \text{M}$
$K_b = 1.8 \times 10^{-5}$
First, calculate $\text{p}K_b$:
$\text{p}K_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 4.74$
Using the Henderson-Hasselbalch equation for a basic buffer:
$\text{pOH} = \text{p}K_b + \log \left( \dfrac{[\text{Salt}]}{[\text{Base}]} \right)$
Substituting the given values:
$\text{pOH} = 4.74 + \log \left( \dfrac{0.2}{0.1} \right)$
$\text{pOH} = 4.74 + \log 2$
$\text{pOH} = 4.74 + 0.30 = 5.04$
Now, calculate the pH:
$\text{pH} = 14 - \text{pOH}$
$\text{pH} = 14 - 5.04 = 8.96$
Answer: $8.96$
Hence, the correct answer is option: 'C'.