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The pH of a solution obtained by mixing 100mL of a solution pH of `=3` with 400mL of a solution of pH `=4` isA. `3- log 2.8`B. `7- log 2.8`C. `4- log 2.8`D. `5-log 2.8 ` |
Answer» Correct Answer - C `N_(1)V_(1)+N_(2)V_(2)=N_(3)(100+400)` or `N_(3)=(0.1+0.04)/(500)=(0.14)/(500)=2.8 xx 10^(-4)M` `pH=-log (2.8 xx 10^(-4))=4=log 2.8` |
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