InterviewSolution
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InterviewSolution
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A volume of `50.00 mL` of a weak acid of unknown concentration is titrated with `0.10 M` solution of `NaOH`. The equivalence point is reached after `39.30 mL` of `NaOH` solution has been added. At the half-equivalence point `(19.65 mL)`, the `pH` is `4.85`. Thus, initial concentration of the acid and its `pK_(a)` values areA. `[HA]` initial `=0.1 M`, `pK_(a)=4.85`B. `[HA]` initial `=0.079 M`, `pK_(a)=2.93`C. `[HA]` initial `=0.1 M`, `pK_(a)=3.70`D. `[HA]` initial `=0.079 M`, `pK_(a)=4.85` |
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Answer» Correct Answer - D `underset(Acid)(M_(1)V_(1))=underset(Base)(M_(2)V_(2)` `M_(1)xx50.0=0.10xx39.30` `:. M_(1)=(0.10xx39.30)/(50)=0.0786 M` When half-equivalence point is reached `[HA]=[A^(-)]` `pH=pK_(a)+log``([A^(-)])/([HA])` `:. pH=pK_(a)=4.85` |
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