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A weak acid, HA, has a K(a) of 1.00xx10^...

A weak acid, HA, has a `K_(a)` of `1.00xx10^(-5)`. If `0.100` mol of the acid is dissolved in 1 L of water, the percentage of the acid dissociated at equilibrium is the closed to

A

`1.00%`

B

`99.9%`

C

`0.100%`

D

`99.0%`

Text Solution

Verified by Experts

The correct Answer is:
A
`alpha=sqrt(K_a/c)=sqrt((1.0xx10^(-5))/10^(-1))=10^(-2)`
`therefore` Percentage of acid dissociated =`10^(-2)xx100` =1%
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