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A weak monoprotic acid of 0.1 M, ionizes to 1% in solution. What will be the pH of solution

Answer»

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Solution :`[H^(+)] = C. alpha`.
`[H^(+)] = 0.1 xx (1)/(100) = 10^(-3)`.
`pH = -LOG [H^(+)] = - log 10^(-3) = 3`.


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