`pH` of a solution made by mixing `200mL` of `0.0657M NaOH, 140 mL` of `0.107M HC1` and `160mL` of `H_(2)O` is

Options

A. `3.04`
B. `2.43`
C. `2.74`
D. `2.27`

Answer

B

Explanation

mmol of `NaOH = 200 xx 0.0657 = 13.14`
mmol of `HC1 = 140 xx 0.107 = 14.98`
mmol of `NaOH` left `= 14.98 - 13.14 = 1.84`
Total volume `= (200 + 140 + 160) mL = 500 mL`
`[H^(o+)] = (1.84 m mol)/(500 mL) = 0.00368 = 368 xx 10^(-5)`
`pH =- log (368 xx 10^(-5)) =- 2.5658 +5`
`= 2.4343 ~~ 2.43`
`pH = 2.43`.