InterviewSolution
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InterviewSolution
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The concentration of `H^(+)` ion in a `0.2 M` solution of `HCOOH` is `6.4xx10^(-3) "mole" L^(-1)`. To this solution `HCOONa` is added so as to adjust the concentration of `HCOONa` to one mole per litre. What will be the `pH` of this solution? `K_(a)` for `HCOOH` is `2.4xx10^(-4)` and the degree of dissociation of `HCOONa` is `0.75`A. `3.19`B. `4.19`C. `5.19`D. `6.19` |
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Answer» Correct Answer - B Assuming that the addition of `HCOONa` suppresses the ionization of `HCOOH`, we can use the expression `pH=pK_(a)+log``(["Salt"])/([Acid])` to compute `pH` of the solution, since salt is `75%` dissociated we will get, `pH= -log(2.4xx10^(-4))+log``(0.75)/(0.2)` `=3.62+0.57=4.19` |
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