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What is the kw of pure water at 50.0°c, if the ph is 6.630? 5.50 × 10-14 2.13 × 10-14 1.00 × 10-14 2.34 × 10-7 there is not enough information to calculate the kw?

Bethany Evans

in Chemistry

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Donald Ward on July 6, 2018

The answer is: the Kw of pure water at 50.0°C is from 5.50 × 10-14.pH = 6.630.pH = -log[H⁺].[H⁺] = 10∧(-pH).[H⁺] = 10∧(-6.63) = 2.34·10⁻ ⁷ M[H⁺] · [OH⁻] = x.Kw = ?. Kw = [H⁺] · [OH⁻]. Kw = x2.Kw = (2.34·10⁻⁷ M)2.Kw = 5.50·10⁻1⁴ M2.Kw is the ion product of water.


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