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Question

Chemistry Question on Equilibrium

The pKapK_a of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50%50\% of the acid is ionized is

A

4.5

B

2.5

C

9.5

D

7

Answer

9.5

Explanation

Solution

For buffer solution pH=pKa+log[Salt][Acid]pH=pK_{a}+log \frac{\left[Salt\right]}{\left[Acid\right]} =4.5+log[Salt][Acid]=4.5+log \frac{\left[Salt\right]}{\left[Acid\right]} as HAHA is 50%50\% ionized so [Salt]=[Acid]\left[Salt\right] = \left[Acid\right] pH=4.5pH = 4.5 pH+pOH=14pH + pOH = 14 pOH=144.5=9.5\Rightarrow pOH = 14 - 4.5 = 9.5 Hence (3) is correct.