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The `pK_(a)` of weak acid (HA) is 4.5 . The pOH of an aqueous buffered solution of HA is which 50% of the acid is ionized isA. `7.0`B. `4.5`C. `2.5`D. `9.5` |
Answer» Correct Answer - D When HA is 50% ionized , [HA] = `[A^(-)]` `pH = pK_(a)+log .([A^(-)])/([HA])=pK_(a)=4.5 ` (Given) `pOH = 14- pH = 14 - 4.5 = 9.5. |
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