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A certain weak acid has K(a), = 10^(-5)....

A certain weak acid has `K_(a), = 10^(-5)`. If the equilibrium constant for is its reaction with a strong base is repesented as `yxx 10^(y)` then find the value of y.

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The correct Answer is:
9
` HA+ OH^(-) hArr A^(-) +H_2O ,K`
` K= (1)/(K_b) = (K_a) /(K_w) =(10 ^(-5))/( 10^(-14))=10 ^(9) `
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