The pK a of a weak acid, HA, is 4.80 and the pK b of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be:

Question

The pK a of a weak acid, HA, is 4.80 and the pK b of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be: (A) 8.23 (B) 9.41 (C) 7.01 (D) 5.91

Tardigrade Admin · Accepted Answer

Let, the salt of weak acid and base like

C H_{3} COON H_{4}

Now,
Salt hydrolysis formula is:

K_{h} = \frac{K _{w}}{K _{a} K _{b}}

h = \frac{K _{w}}{K _{a} K _{b}}

p H = \frac{1}{2} [p K_{w} + p K_{a} - p K_{b}]

So,

p H

of Salt is,

p H = \frac{1}{2} (14 + 4.80 - 4.78)

= 7.01

This is the required solution.

Q. The pKa​ of a weak acid, HA, is 4.80 and the pKb​ of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be:

8.23

9.41

7.01

5.91

Let, the salt of weak acid and base like CH3​COONH4​ Now, Salt hydrolysis formula is: Kh​=Ka​Kb​Kw​​ h=Ka​Kb​Kw​​​ pH=21​[pKw​+pKa​−pKb​] So, pH of Salt is, pH=21​(14+4.80−4.78) =7.01 This is the required solution.

Q. The $p K_{a}$ of a weak acid, $H A$ , is $4.80$ and the $p K_{b}$ of a weak base, $BO H$ , is $4.78$ . The $p H$ of an aqueous solution of the corresponding salt, $B A$ , will be: