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  4. mathrmCH3 mathrmNH2(0.12. mole, . mathrmpK mathrmb=3.3) is added to 0.08 moles of HCl and the solution is diluted to one litre, resulting pH of solution is:

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Q. $\mathrm{CH}_3 \mathrm{NH}_2\left(0.12\right.$ mole, $\left.\mathrm{pK}_{\mathrm{b}}=3.3\right)$ is added to $0.08$ $moles$ of $HCl$ and the solution is diluted to one litre, resulting $pH$ of solution is:

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

Solution

$pOH=pK_{b}+log\frac{\left[C H_{3} N H_{3}^{+}\right]}{\left[\left(C H_{3} N H_{2}\right)\right]}$

$=3.3+log\frac{0.08}{0.04}=3.6$

$\therefore pH=10.4$