The pK a of a weak acid, HA, is equal to the pK b of a weak base, BOH. The pH of an aqueous solution of the corresponding salt, BA, will be.

Question

The pK a of a weak acid, HA, is equal to the pK b of a weak base, BOH. The pH of an aqueous solution of the corresponding salt, BA, will be. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The salt is of weak acid and weak base. ∴ pH =(1/2) pK w +(1/2) pK a -(1/2) pK b =(1/2) × 14+(1/2) × x-(1/2) × x =7

Q. The $p K_{a}$ of a weak acid, $H A$ , is equal to the $p K_{b}$ of a weak base, $BO H$ . The $p H$ of an aqueous solution of the corresponding salt, $B A$ , will be______.

The salt is of weak acid and weak base.
$∴ p H = \frac{1}{2} p K_{w} + \frac{1}{2} p K_{a} - \frac{1}{2} p K_{b}$
$= \frac{1}{2} \times 14 + \frac{1}{2} \times x - \frac{1}{2} \times x$
$= 7$

Q. The pKa​ of a weak acid, HA, is equal to the pKb​ of a weak base, BOH. The pH of an aqueous solution of the corresponding salt, BA, will be______.

The salt is of weak acid and weak base. ∴pH=21​pKw​+21​pKa​−21​pKb​ =21​×14+21​×x−21​×x =7

Q. The $p K_{a}$ of a weak acid, $H A$ , is equal to the $p K_{b}$ of a weak base, $BO H$ . The $p H$ of an aqueous solution of the corresponding salt, $B A$ , will be______.

The salt is of weak acid and weak base.
$∴ p H = \frac{1}{2} p K_{w} + \frac{1}{2} p K_{a} - \frac{1}{2} p K_{b}$
$= \frac{1}{2} \times 14 + \frac{1}{2} \times x - \frac{1}{2} \times x$
$= 7$