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If `50ml` of `0.2 M KOH` is added to `40 ml` of `0.05 M HCOOH`, the `pH` of the resulting solution is (`K_(a)=1.8xx10^(-4)`)A. `3.4`B. `7.5`C. `5.6`D. `3.75` |
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Answer» Correct Answer - A `pH=-logK_(a)+log``(["Salt"])/([Acid])` `["Salt"]=(0.2xx50)/(1000)=0.01`, `[Acid]=(0.5xx40)/(1000)=0.02` `pH=-log(1.8xx10^(-4))+log``(0.01)/(0.02)` `pH=4-log(1.8)+log 0.5` `pH=4-log (1.8)-0.301` `pH=3.4` |
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