The pKa 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

Question

The pKa 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

Tardigrade Admin · Accepted Answer

For buffer solution pH=pKa+log ([Salt]/[Acid]) =4.5+log ([Salt]/[Acid]) as HA is 50 % ionized so [Salt] = [Acid] pH = 4.5 pH + pOH = 14 ⇒ pOH = 14 - 4.5 = 9.5 Hence (3) is correct.

Q. The $p K_{a}$ 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

4.5

2.5

9.5

7.0

For buffer solution
$p H = p K_{a} + l o g \frac{[ S a lt ]}{[ A c i d ]}$
$= 4.5 + l o g \frac{[ S a lt ]}{[ A c i d ]}$
as $H A$ is $50%$ ionized so $[S a lt] = [A c i d]$
$p H = 4.5$
$p H + pO H = 14$
$\Rightarrow pO H = 14 - 4.5 = 9.5$
Hence (3) is correct.

Q. The pKa​ 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

4.5

2.5

9.5

7.0

For buffer solution pH=pKa​+log[Acid][Salt]​ =4.5+log[Acid][Salt]​ as HA is 50% ionized so [Salt]=[Acid] pH=4.5 pH+pOH=14 ⇒pOH=14−4.5=9.5 Hence (3) is correct.

Q. The $p K_{a}$ 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

For buffer solution
$p H = p K_{a} + l o g \frac{[ S a lt ]}{[ A c i d ]}$
$= 4.5 + l o g \frac{[ S a lt ]}{[ A c i d ]}$
as $H A$ is $50%$ ionized so $[S a lt] = [A c i d]$
$p H = 4.5$
$p H + pO H = 14$
$\Rightarrow pO H = 14 - 4.5 = 9.5$
Hence (3) is correct.