Question

The $\text{pH}$ of a buffer solution having $0.001\text{M}N{H}_{4}0H\text{and}0.01\text{M}N{H}_{4}Cl\text{is}(K_b\text{of base is}0.01$):

• A. $8$
• B. $11$
• C. $3$
• D. $13$

Answer 1

The correct option is B $11$

The expression for the $\text{pOH}$ of the basic buffer solution is as given below:
$\text{pOH}=\text{p}{K}_b+\log \frac{\left[\text{salt}\right]}{\text{[base}]}$
$\text{p}{K}_b=-\log K_b=-\log 0.01=2$
By substituting values in the above equation, we get
$\Rightarrow \text{pOH}=2+\log \frac{0.01}{0.001}=3$
$\Rightarrow \text{pH}=14-\text{pOH}=14-3=11$.