The dissociation constant of acetic acid at a given temperature is 1.69 × 10-5. The degree of dissociation of 0.01 M acetic acid in the presence of 0.01 M HCl is equal to

Question

The dissociation constant of acetic acid at a given temperature is 1.69 × 10-5. The degree of dissociation of 0.01 M acetic acid in the presence of 0.01 M HCl is equal to (A) 0.41 × 10-2 (B) 0.18 × 10-7 (C) 0.169 × 10-2 (D) 0.013 × 10-4

Tardigrade Admin · Accepted Answer

Given, K=1.69 × 10-5 ∵ C α<<0.01 ∴ K=(C α(C α+0.01)/C(1-α))=0.01 α or α=(1.69 × 10-5/0.01)=0.169 × 10-2

Q. The dissociation constant of acetic acid at a given temperature is $1.69 \times 1 0^{- 5}$ . The degree of dissociation of $0.01 M$ acetic acid in the presence of $0.01 M H Cl$ is equal to

$0.41 \times 1 0^{- 2}$

$0.18 \times 1 0^{- 7}$

$0.169 \times 1 0^{- 2}$

$0.013 \times 1 0^{- 4}$

Given, $K = 1.69 \times 1 0^{- 5}$
$∵ C α << 0.01$
$∴ K = \frac{C α ( C α + 0.01 )}{C ( 1 - α )} = 0.01 α$
or $α = \frac{1.69 \times 1 0 ^{- 5}}{0.01} = 0.169 \times 1 0^{- 2}$

Q. The dissociation constant of acetic acid at a given temperature is 1.69×10−5. The degree of dissociation of 0.01M acetic acid in the presence of 0.01MHCl is equal to

0.41×10−2

0.18×10−7

0.169×10−2

0.013×10−4