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The pH of a solution obtained by mixing 10 mL of `0.1MHCI` and 40mL of `0.2MH_(2)SO_(4)` is:A. `1.4865`B. `0.4865`C. `0.4685`D. 3 |
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Answer» Correct Answer - C Milli-equivalent of `H^(+)` from `HCI=10xx0.1=1` Milli-equivalent of `H^(+)` form `H_(2)SO_(4)` `=40xx0.2xx2=16` Total Meq. Of `H^(+)` in solution `= 1+16=17` `:. [H^(+)]= (Meq.)/(V_(i nmL))=(17)/(50)=3.4xx10^(-1)` `pH= -log[H^(+)= -log 0.34=0.4685` |
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