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The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` isA. `log 2`B. `-log0.2`C. `1.0`D. `2.0` |
Answer» Correct Answer - C `[H^(+)] = (V_(a)N_(a) - V_(b) N_(b))/(V_(a) + V_(b))` `[H^(+)] = (50 xx 0.4 - 50 xx 0.2)/(100)` `[H^(+)] = (20 - 10)/(100) = 10^(-1)` `pH = -log 10^(-1) , pH = 1` |
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