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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

`99.0` %

B

`1.00` %

C

`99.9` %

D

`0.100` %

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(a) = 1xx10^(-5)`
`{:(HA,hArr,H^(+),+,A^(-)),(0.1m//L,,0,,0),(0.1-x,,x,,x):}`
`10^(-5) = (x^(2))/(0.1-x)`
Assume `x lt lt 0.1 `
`x^(2) = 10^(-6) " or " x = 10^(-3)`
% dissociation = `(10^(-3))/(0.1) xx100 = 1 % `
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