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The pH of a solution obtained by mixing 50 ml of 1 N HCl and 30 ml of 1 N NaOH is [log 2.5 = 0.3979]A. 3.979B. 0.6021C. 12.042D. 1.2042 |
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Answer» Correct Answer - B No. of milli equivalent of `NaOH = 30 xx 1 = 30` No. of milli equivalent of `HCl = 50 xx 1 = 50` `:.` No. of milli equivalent of HCl left after titration = 50 - 30 = 20. Total volume of the mixture = 50 + 30 = 80 ml i.e. 20 milli equivalent or 0.02 equivalent of HCl are present in 80 ml. `:.` 250 milli equivalent or 0.25 equivalent of HCl are present in 1000 ml or 1 litre. i.e., `0.25 N HCl ~~ 0.25 N NaOH` (Monobasic) So, `[H^(+)] = -log_(10)[2.5 xx 10^(-1)] rArr pH = 1 - log_(10)[0.25]` `pH = -log_(10)[2.5 xx 10^(-2) rArr pH = 1 - log_(10)2.5` `pH = 1 - 0.3979 rArr pH = 0.6021`. |
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