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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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Knowledge Check

  • A weak acid HA has a K_(a) of 1.00 xx 10^(-5) . If 0.100 mol of this acid is dissolved in one litre of water the percentage of acid dissociated at equilibrium is closed to

    A
    `99.0%`
    B
    `1.00%`
    C
    `99.9%`
    D
    `0.100%`

  • K_(a) of HA at 25^(@) is 10^(-5) . If 0.1mol of this acid is dissolved in 1L of aqueous solution, the percent dissociation at equilibrium will be closer to

    A
    `0.1%`
    B
    `1.0%`
    C
    `99.0%`
    D
    `99.9%`

  • A weak acid fix ( K_(a) = 1 xx 10^(-5)) on reaction with NaOH gives NaX. For 0.1 m aqueous solution of NaX, the percentage hydrolysis is

    A
    1.0E-5
    B
    0.0001
    C
    0.0015
    D
    0.01
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