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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 65 of 73
Marks: +1, -0
Ionic product of water at 310 K is 2.7 × 10−1410^{-14}. What is the pH of neutral water at this temperature?
Solution:  
We know that
[H+][\mathrm{H}^{+}] = Kw\sqrt{K_w} = 2.7×10−14\sqrt{2.7 \times 10^{-14}} = 1.643 × 10−7 M10^{-7}\,\mathrm{M}
∴ pH = - log [H+][\mathrm{H}^{+}] = - log (1.643 × 10−710^{-7}) = 7 - 0.2156 = 6.78
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