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 equivalents of `H^(+)`
`= 20 xx 1/2 +30 xx1/3 = 20 `
Total milliequivalents 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 `