Question
A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is
Answer: Option A
:
A
K=α2C1−α;α=0.01100
∴K=α2C=[0.01100]2×1=1×10−8
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:
A
K=α2C1−α;α=0.01100
∴K=α2C=[0.01100]2×1=1×10−8
Was this answer helpful ?
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