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30 cc of (M)/(3) HCl, 20 cc of (M)/(2) HNO_(3) and 40 cc of (M)/(4) NaOH solutions are mixed and the volumes was made up to 1dm^(3). The pH of the resulting solution is

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Solution :Total MILLI equivalent of `h^(+) = 20 xx (1)/(2) + 30 xx (1)/(3) = 20`
Total milliequivalent of `OH^(-) = 40 xx (1)/(4) = 10`
Thus, milliequivalents of `H^(+)` LEFT `= 20 - 10 = 10`
Total volume of solution `= 1 dm^(3) = 1000 ml`
`:. [H^(+)] = (10)/(1000) = 10^(-2)`
`rArr PH = -log[H^(+)] = -log [10^(-2)] = 2`.


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