Question

A certain weak acid, HA, has a *K*a value of
9.5×10^{−7}.

1. Calculate the percent ionization of HA in a 0.10 *M*
solution

2.Calculate the percent ionization of HA in a 0.010 *M*
solution.

Answer #1

Ka for weak acid = 9.5 x 10^-7

1. [HA] = 0.10 M

HA + H2O <==> H3O+ + A-

let x amount of HA has dissociated

then,

Ka = [H3O+][A-]/[HA]

9.5 x 10^-7 = x^2/0.10

x = [H3O+] = 3.1 x 10^-4 M

percent ionization = (3.1 x 10^-4) x 100/0.10 = 0.31%

--

2. [HA] = 0.010 M

HA + H2O <==> H3O+ + A-

let x amount of HA has dissociated

then,

Ka = [H3O+][A-]/[HA]

9.5 x 10^-7 = x^2/0.010

x = [H3O+] = 9.75 x 10^-5 M

percent ionization = (9.75 x 10^-5) x 100/0.010 = 1.0%

--

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Include units, Thank you!

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