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The `pK_(a)` of a weak acid `(HA)` is `4.5`. The `pOH` of an aqueous buffered solution of `HA` in which `50%` of the acid is ionized is:A. `4.5`B. `2.5`C. `9.5`D. `7.0` |
Answer» Correct Answer - C For buffer solution : `pH= pK_(a)+log (["Salt"])/(["Acid"])= 4.5+log (["Salt"])/(["Acid"])` As HA is `50%` ionised, so [Salt] = [Acid] `pH = 4.5` `pH+pOH=14` `:. pOH= 14-4.5 = 9.5` |
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