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

Options

A. 2
B. 1
C. 3
D. 8

Answer

A

Explanation

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`.