The value of kw in a neutral soln is 5-474-10-14 at 50 degrees- Calculate the pH of the soln at this temperature-

Given k_w = 5.474 × 10^ 14 pH=p k_w Hence pH= log(5.474 × 10^ 14)=13.26 ...

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pH the Power of H

The value of ...

Question

The value of $k w$ in a neutral soln is $5.474 \times 10^{- 14}$ at $50$ degrees. Calculate the $p H$ of the soln at this temperature.

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Solution

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k_{w}

1.08 \times 10^{- 14}

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K_{w} f o r H_{2} O i s 9.62 \times 10^{- 14} a t 60^{\circ} C

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