The dissociation constant of a substituted benzoic acid is 1.0× 10- 4 at 25oC . The pH of 0.01 M solution of its sodium salt is

Question

The dissociation constant of a substituted benzoic acid is 1.0× 10- 4 at 25oC . The pH of 0.01 M solution of its sodium salt is (A) 7 (B) 6 (C) 8 (D) 4

Tardigrade Admin · Accepted Answer

For a salt of a weak acid and strong base pH=7+(1/2)pKa+(1/2)logC =7+(1/2)(- log 1 × 10-4)+(1/2) log (0.01) =8.

Q. The dissociation constant of a substituted benzoic acid is $1.0 \times 1 0^{- 4}$ at $2 5^{o} C$ . The pH of 0.01 M solution of its sodium salt is

7

6

8

4

For a salt of a weak acid and strong base
$p H = 7 + \frac{1}{2} p K a + \frac{1}{2} l o g C$
$= 7 + \frac{1}{2} (- lo g 1 \times 1 0^{- 4}) + \frac{1}{2} lo g (0.01)$
$= 8.$

Q. The dissociation constant of a substituted benzoic acid is 1.0×10−4 at 25oC . The pH of 0.01 M solution of its sodium salt is

7

6

8

4

For a salt of a weak acid and strong base pH=7+21​pKa+21​logC =7+21​(−log1×10−4)+21​log(0.01) =8.

Q. The dissociation constant of a substituted benzoic acid is $1.0 \times 1 0^{- 4}$ at $2 5^{o} C$ . The pH of 0.01 M solution of its sodium salt is

For a salt of a weak acid and strong base
$p H = 7 + \frac{1}{2} p K a + \frac{1}{2} l o g C$
$= 7 + \frac{1}{2} (- lo g 1 \times 1 0^{- 4}) + \frac{1}{2} lo g (0.01)$
$= 8.$