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Assuming complete dissociation, calculate the pH of the following solutions:
(a) \(0.003 \mathrm{M}\) HCl
(b) \(0.005 \mathrm{M} \mathrm{NaOH}\)
(c) \(0.002 \mathrm{M} \mathrm{HBr}\)
(d) \(\mathbf{0 . 0 0 2} \mathrm{M} \mathrm{KOH}\)
(a) \(0.003 \mathrm{M}\) HCl
(b) \(0.005 \mathrm{M} \mathrm{NaOH}\)
(c) \(0.002 \mathrm{M} \mathrm{HBr}\)
(d) \(\mathbf{0 . 0 0 2} \mathrm{M} \mathrm{KOH}\)
Solution:
2112 Upvotes
Verified Answer
(a) \(\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}\),
\(\therefore \quad\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=3 \times 10^{-3} \mathrm{M}\),
\(\mathrm{pH}=-\log \left(3 \times 10^{-3}\right)=2.52\)
(b) \(\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}\)
\(\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]\)
\(\therefore \quad\left[\mathrm{OH}^{-}\right]=5 \times 10^{-3} \mathrm{M}\)
\(\left[\mathrm{H}^{+}\right]=10^{-14} /\left(5 \times 10^{-3}\right)\)
\(=2 \times 10^{-12} \mathrm{M}\)
\(\mathrm{pH}=-\log \left(2 \times 10^{-12}\right)=11.699\)
\(\mathrm{HBr}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}+\mathrm{Br}^{-},\left[\mathrm{H}^{+}\right]=[\mathrm{HBr}]\)
(c) \(\mathrm{HBr}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}+\mathrm{Br}^{-}\), \(\therefore \quad\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M}\),
\(\begin{aligned}
&\therefore \quad\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M}, \\
&\mathrm{pH}=-\log \left(2 \times 10^{-3}\right)=2.699
\end{aligned}\)
(d) \(\mathrm{KOH}(\mathrm{aq}) \longrightarrow \mathrm{K}^{+}+\mathrm{OH}^{-},\left[\mathrm{OH}^{-}\right]=[\mathrm{KOH}]\)
\(\begin{aligned}
&\therefore\left[\mathrm{OH}^{-}\right]=2 \times 10^{-3} \mathrm{M},\left[\mathrm{H}^{+}\right] \\
&=10^{-14} /\left(2 \times 10^{-3}\right)=5 \times 10^{-12}
\end{aligned}\)
\(\mathrm{pH}=-\log \left(5 \times 10^{-12}\right)=11.30\)
\(\therefore \quad\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=3 \times 10^{-3} \mathrm{M}\),
\(\mathrm{pH}=-\log \left(3 \times 10^{-3}\right)=2.52\)
(b) \(\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}\)
\(\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]\)
\(\therefore \quad\left[\mathrm{OH}^{-}\right]=5 \times 10^{-3} \mathrm{M}\)
\(\left[\mathrm{H}^{+}\right]=10^{-14} /\left(5 \times 10^{-3}\right)\)
\(=2 \times 10^{-12} \mathrm{M}\)
\(\mathrm{pH}=-\log \left(2 \times 10^{-12}\right)=11.699\)
\(\mathrm{HBr}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}+\mathrm{Br}^{-},\left[\mathrm{H}^{+}\right]=[\mathrm{HBr}]\)
(c) \(\mathrm{HBr}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}+\mathrm{Br}^{-}\), \(\therefore \quad\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M}\),
\(\begin{aligned}
&\therefore \quad\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M}, \\
&\mathrm{pH}=-\log \left(2 \times 10^{-3}\right)=2.699
\end{aligned}\)
(d) \(\mathrm{KOH}(\mathrm{aq}) \longrightarrow \mathrm{K}^{+}+\mathrm{OH}^{-},\left[\mathrm{OH}^{-}\right]=[\mathrm{KOH}]\)
\(\begin{aligned}
&\therefore\left[\mathrm{OH}^{-}\right]=2 \times 10^{-3} \mathrm{M},\left[\mathrm{H}^{+}\right] \\
&=10^{-14} /\left(2 \times 10^{-3}\right)=5 \times 10^{-12}
\end{aligned}\)
\(\mathrm{pH}=-\log \left(5 \times 10^{-12}\right)=11.30\)
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