Question

A certain weak acid, HA, has a Ka value of 1.8×10−7.
Part A
Calculate the percent dissociation of HA in a 0.10 M solution.

Answer 1

Give me a second to do this..

Answer 2

Dissociation equation of the acid;

HA(aq) <-----> H+(aq) + A-(aq)
0.010-x M ........ x M ........ x M

Ka = [H+] [A-] / [HA]

1.3x10^-7 = (x)(x) / (0.010 - x)

Since the amount of x is very small compared to 0.010 M, it is neglected.
1.3x10^-7 = (x)(x) / (0.010)
x^2 = 1.3x10^-9
x = 3.6x10^-5 M

Since 3.6x10^-5 M is the amount dissociared (ionized)
and 0.010 M is the initial amount,
the percent ionization of the acid will be:
(3.6x10^-5 M / 0.010 M) x 100 = 0.36 %

Answer 3

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Answer 4

thanks !