Question

$\left(\text{CH}\right)_{3} \left(\text{NH}\right)_{2} \left(0.12 \, \text{mole} , \, \left(\text{pK}\right)_{\text{b}} = 3.3\right)$ is added to $0.08$ $moles$ of $HCl$ and the solution is diluted to one litre, resulting $pH$ of solution is:

• A. $10.7$
• B. $3.6$
• C. $10.4$
• D. $11.3$

Answer 1

Answer: (C)

$pOH=pK_{b}+log\frac{\left[C H_{3} N H_{3}^{+}\right]}{\left[\left(C H_{3} N H_{2}\right)\right]}$
$=3.3+log\frac{0.08}{0.04}=3.6$
$\therefore pH=10.4$