The Ka of an acid is 3.2× 10-5. The degree of dissociation of the acid at concentration of 0.2 M is:

Question

The Ka of an acid is 3.2× 10-5. The degree of dissociation of the acid at concentration of 0.2 M is: (A)  6.0× 10-2 (B)  1.26× 10-2 (C)  40× 10-4 (D)  0.04× 10-3

Tardigrade Admin · Accepted Answer

From Ostwald's dilution law α =√(K a/c) =√(3.2 × 10-5/0.2) =1.26 × 10-2

Q. The $K a$ of an acid is $3.2 \times 1 0^{- 5}$ . The degree of dissociation of the acid at concentration of $0.2 M$ is:

$6.0 \times 1 0^{- 2}$

$1.26 \times 1 0^{- 2}$

$40 \times 1 0^{- 4}$

$0.04 \times 1 0^{- 3}$

From Ostwald's dilution law
$α = \frac{K a}{c}$
$= \frac{3.2 \times 1 0 ^{- 5}}{0.2}$
$= 1.26 \times 1 0^{- 2}$

Q. The Ka of an acid is 3.2×10−5. The degree of dissociation of the acid at concentration of 0.2M is:

6.0×10−2

1.26×10−2

40×10−4

0.04×10−3