Question

How much of 0.3 M $NH_{4}OH$ should be mixed with 30 mL of 0.2 M solution of $NH_{4}Cl$ to give buffer solution of pH 8.65?
(Given: $\text{pK}_{\text{b}}$ of $\text{NH}_{\text{3}} = 4.75$ )

• A. 2 mL
• B. 3 mL
• C. 5 mL
• D. 7 mL

Answer 1

Answer: (C)

It is a basic buffer
$pH=8.65$ ,
pOH = 14 - 8.65 = 5.35
$pOH=pK_{b}+log\frac{\left[\right. \overset{\oplus}{N} H_{4} \left]\right.}{\left[\right. N H_{4} O H \left]\right.}$
5.35 = 4.75 + log $\frac{\left[\right. \overset{\oplus}{N} H_{4} \left]\right.}{\left[\right. N H_{4} O H \left]\right.}$
$\frac{\left[\right. \overset{\oplus}{N} H_{4} \left]\right.}{\left[\right. N H_{4} O H \left]\right.}=$ antilog (5.35 - 4.75)
= antilog (0.6) $\approx4.0$
So $\frac{\left[\right. \overset{\oplus}{N} H_{4} \left]\right.}{\left[\right. N H_{4} O H \left]\right.}=\frac{30 \times 0.2}{0.3 \times V}\approx4.0$
So V = 5 mL.