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If \(0.561 \mathrm{~g}\) of \(\mathrm{KOH}\) is dissolved in water to give \(200 \mathrm{~mL}\) of solution at \(298 \mathrm{~K}\). Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its pH ?
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Verified Answer
\([\mathrm{KOH}]=0.561 / 56 \times 1000 / 200=0.05 \mathrm{M}\)
\(\mathrm{As} \mathrm{KOH} \longrightarrow \mathrm{K}^{+}+\mathrm{OH}^{-}\)
\(\left[\mathrm{K}^{+}\right]=\left[\mathrm{OH}^{-}\right]=0.05 \mathrm{M}\)
\(\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{w}} /\left[\mathrm{OH}^{-}\right]=10^{-14} / 0.05=10^{-14} /\left(5 \times 10^{-2}\right)\)
\(=2.0 \times 10^{-13} \mathrm{M}\)
\(\mathrm{pH}=-\log 2 \times 10^{-13}\)
\(=13-\log 2=12.69\)
\(\mathrm{As} \mathrm{KOH} \longrightarrow \mathrm{K}^{+}+\mathrm{OH}^{-}\)
\(\left[\mathrm{K}^{+}\right]=\left[\mathrm{OH}^{-}\right]=0.05 \mathrm{M}\)
\(\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{w}} /\left[\mathrm{OH}^{-}\right]=10^{-14} / 0.05=10^{-14} /\left(5 \times 10^{-2}\right)\)
\(=2.0 \times 10^{-13} \mathrm{M}\)
\(\mathrm{pH}=-\log 2 \times 10^{-13}\)
\(=13-\log 2=12.69\)
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