A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is:

Question

A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is: (A)  1× 10-8  (B)  1× 10-4  (C)  1× 10-6  (D)  10-5

Tardigrade Admin · Accepted Answer

For weak electrolytes, according to Ostwalds dilution law α =√K.V here, α =0.01%=0.0001=1× 10-4 V=(1/C)=(1/1.0)=1litre ∴ Ka=(α 2/V)=((1× 10-4)2/1) =1× 10-8

Q. A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is:

$1 \times 10^{- 8}$

$1 \times 10^{- 4}$

$1 \times 10^{- 6}$

$10^{- 5}$

For weak electrolytes, according to Ostwalds dilution law $α = K . V$ here, $α = 0.01$ $V = \frac{1}{C} = \frac{1}{1.0} = 1 l i t re$ $∴$ $K_{a} = \frac{α ^{2}}{V} = \frac{( 1 \times 10 ^{- 4} ) ^{2}}{1}$ $= 1 \times 10^{- 8}$

Q. A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is:

1×10−8

1×10−4

1×10−6

10−5

For weak electrolytes, according to Ostwalds dilution law α=K.V​ here, α=0.01 V=C1​=1.01​=1litre ∴ Ka​=Vα2​=1(1×10−4)2​ =1×10−8

$1 \times 10^{- 8}$

$1 \times 10^{- 4}$

$1 \times 10^{- 6}$

$10^{- 5}$

For weak electrolytes, according to Ostwalds dilution law $α = K . V$ here, $α = 0.01$ $V = \frac{1}{C} = \frac{1}{1.0} = 1 l i t re$ $∴$ $K_{a} = \frac{α ^{2}}{V} = \frac{( 1 \times 10 ^{- 4} ) ^{2}}{1}$ $= 1 \times 10^{- 8}$