Which of the following mixture gives a buffer solution with $\mathrm{pH}=9.25$ ?

Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.75$

Question

Which of the following mixture gives a buffer solution with $\mathrm{pH}=9.25$ ?

Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.75$

Accepted Answer

B. 1. For a basic buffer of $\mathrm{NH_4OH}$ and its salt $\mathrm{NH_4Cl}$:
$$
\mathrm{pOH}=\mathrm{p}K_b+\log\left(\frac{[	ext{salt}]}{[	ext{base}]}ight)
$$

2. Given $\mathrm{p}K_b=4.75$ and required pH = 9.25:
$$
\mathrm{pOH}=14-9.25=4.75
$$

3. Substituting:
$
4.75=4.75+\log\left(\frac{[	ext{salt}]}{[	ext{base}]}ight) \;\Rightarrow\; \log\left(\frac{[	ext{salt}]}{[	ext{base}]}ight)=0 \;\Rightarrow\; \frac{[	ext{salt}]}{[	ext{base}]}=1
$

4. So we need **equal moles of salt and base** after the reaction $\mathrm{NH_4OH+HClightarrow NH_4Cl+H_2O}$. If $n_b$ (base) and $n_a$ (acid) are the initial moles, the condition for equal salt and remaining base is:
$$
n_a=n_b-n_a \;\Rightarrow\; n_b=2n_a
$$

5. Checking **Option B**: $n_b=0.2	imes0.5=0.10$ mol, $n_a=0.1	imes0.5=0.05$ mol. Indeed $n_b=2n_a$ ($0.10=2	imes0.05$).

6. After reaction: salt = 0.05 mol, base left = $0.10-0.05=0.05$ mol — equal, confirming this is the required buffer with $\mathrm{pOH}=\mathrm{p}K_b=4.75$ and $\mathrm{pH}=9.25$.

Hence, the answer is **Option B**.

Which of the following mixture gives a buffer solution with pH=9.25\mathrm{pH}=9.25pH=9.25 ? Given : pKb(NH4OH)=4.75\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.75pKb​(NH4​OH)=4.75

Which of the following mixture gives a buffer solution with $\mathrm{pH}=9.25$ ?

Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.75$