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

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

A

1×10−81 \times 10 ^ { - 8 }

B

1×10−41 \times 10 ^ { - 4 }

C

1×10−61 \times 10 ^ { - 6 }

D

10−510 ^ { - 5 }

Answer

1×10−81 \times 10 ^ { - 8 }

Explanation

Solution

K=α2C1−α;α=0.01100≈1∴ K=α2C=[0.01100]2×1\mathrm { K } = \frac { \alpha ^ { 2 } \mathrm { C } } { 1 - \alpha } ; \alpha = \frac { 0.01 } { 100 } \approx 1 \therefore \mathrm {~K} = \alpha ^ { 2 } \mathrm { C } = \left[ \frac { 0.01 } { 100 } \right] ^ { 2 } \times 1

=1×10−8= 1 \times 10 ^ { - 8 }.