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The ionic product of water at \(310 \mathrm{~K}\) is \(2.7 \times 10^{-14}\). What is the pH of neutral water at this temperature?
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As \(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}\)
For neutral water \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)
\(\therefore\left[\mathrm{H}^{+}\right]^2=2.7 \times 10^{-14}\)
or \(\sqrt{\left[\mathrm{H}^{+}\right]^2}=\sqrt{2.7 \times 10^{-14}}\)
\(\therefore \quad\left[\mathrm{H}^{+}\right]=1.643 \times 10^{-7}\)
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1.643 \times 10^{-7}\right)=7-0.2156=6.78\)
For neutral water \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)
\(\therefore\left[\mathrm{H}^{+}\right]^2=2.7 \times 10^{-14}\)
or \(\sqrt{\left[\mathrm{H}^{+}\right]^2}=\sqrt{2.7 \times 10^{-14}}\)
\(\therefore \quad\left[\mathrm{H}^{+}\right]=1.643 \times 10^{-7}\)
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1.643 \times 10^{-7}\right)=7-0.2156=6.78\)
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