At 80ºC, pure distilled water has H3O+ concentration equal to 1 × 10– 6 mol L–1. The value of Kw at this temperature will be -

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Chemistry Foundation

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Question

At 80ºC, pure distilled water has H₃O⁺ concentration equal to 1 × 10^{– 6} mol L^–1. The value of K_w at this temperature will be -

1 × 10^–12

1 × 10^–14

1 × 10^{– 8}

1 × 10^{– 6}

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K_w of pure distilled water is - K_w = [H₃O⁺] [OH¯]
and in pure distilled water [H₃O⁺] = [OH¯]
At 80º C [H₃O⁺] =1 × 10^–6 mol L^–1
So value of K_w at this temperature will be -
K_w= [1 × 10^–6] [1 × 10^–6] = 1 × 10^–12M²