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`pH` of a solution made by mixing `200mL` of `0.0657M NaOH, 140 mL` of `0.107M HC1` and `160mL` of `H_(2)O` isA. `3.04`B. `2.43`C. `2.74`D. `2.27` |
Answer» Correct Answer - B 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`. |
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