NCERT Class XI Chemistry Equilibrium Solutions

Question : 65

Total: 73

Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?

Solution:

We know that
[H+] = √Kw = √2.7×10−14 = 1.643 × 10−7M
∴ pH = - log [H+] = - log (1.643 × 10−7) = 7 - 0.2156 = 6.78

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