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$40 \mathrm{~mL}$ of $0.1 \mathrm{M}$ ammonia solution is mixed with $20 \mathrm{~mL}$ of $0.1 \mathrm{MHCl}$. What is the $\mathrm{pH}$ of the mixture? ( $\mathrm{p} K_b$ of ammonia solution is 4.74)
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Verified Answer
The correct answer is:
9.26
$40 \mathrm{~mL}$ of $0.1 \mathrm{M}$ ammonia solution $=40 \times 0.1$ $=4$ milliequivalent ammonia solution $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}=20 \times 0.1$ $=2$ milli equivalent of $\mathrm{HCl}$
$$
\begin{array}{lccc}
& \mathrm{NH}_4 \mathrm{OH}+ & \mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \\
\text { Initial milli-eqv. } & 4 & 2 & 0 \\
\text { Milli-eqv. after } & 4-2 & 0 & 2 \\
\text { reaction } & =2 & &
\end{array}
$$
$$
\begin{gathered}
\therefore \mathrm{pOH}=\mathrm{pK} K_b+\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}=4.74+\log \frac{2}{2} \\
=4.74+\log 1=4.74 \\
\therefore \mathrm{pH}=14-4.74=9.26
\end{gathered}
$$
$$
\begin{array}{lccc}
& \mathrm{NH}_4 \mathrm{OH}+ & \mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \\
\text { Initial milli-eqv. } & 4 & 2 & 0 \\
\text { Milli-eqv. after } & 4-2 & 0 & 2 \\
\text { reaction } & =2 & &
\end{array}
$$
$$
\begin{gathered}
\therefore \mathrm{pOH}=\mathrm{pK} K_b+\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}=4.74+\log \frac{2}{2} \\
=4.74+\log 1=4.74 \\
\therefore \mathrm{pH}=14-4.74=9.26
\end{gathered}
$$
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