1) DILUTION EFFECT ON THE PERCENT IONIZATION OF A WEAK ACID a. A weak acid, HX,

Question

1) DILUTION EFFECT ON THE PERCENT IONIZATION OF A WEAK ACID
a. A weak acid, HX, is 1.3 % ionized in 0.20 M solution. What percent of HX is ionized
in a 0.030 M solution? Show the complete setup.
percent ionization__________%
b. From your result in (a) above answer the following questions:
i) How did the percent of ionization change upon dilution?
_____________
(increased, or decreased)
ii) How did the [H3O +] concentration of the above weak acid change upon
dilution?
_______________
(increased, or decreased)

2) In 0.45 M benzoic acid, HC7H5O2, the [H3O +] is 5.4 x 10-3 M. Calculate the value of
the equilibrium constant, Ka. Show the complete set up.
Ka = _________

Explanation / Answer

(a)

Given, 1.3% ionized in 0.20 M solution

So, % ionization =[[H+] / 0.20 ] x 100 = 1.3

[H+] = 2.6 x 10^-3 M

So, Ka = [H+][X-] / [HX] = (2.6 x 10 ^-3)^2 (0.20 - 2.6 x 10^-3) = 6.76 x 10^-6 / 0.1974 = 3.42 x 10^-5

Thus, Ka = 3.42 x 10^-5

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If [HX] = 0.030 M

Ka = [H+][X-] / [HX] = x^2 / 0.030-x

3.42 x 10^-5 = x^2 / 0.030

x^2 = 1.03 x 10^-6

x = 1.013 x 10^-3

% ionization =[(1.013 x 10^-3) / 0.030 ] x 100 = 3.37%

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(b)

The percent of ionization increases upon dilution.

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(c)

The [H3O+] concentration of the above weak acid decreases upon dilution

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(2)

HC7H5O2 <----------------> C7H5O2- + H+

[H3O+] = [C7H5O2-] = 5.4 x 10^-3 M

[HC7H5O2] = 0.45 -x = 0.4446

Ka = [H3O+] [C7H5O2-] / [HC7H5O2] = (5.4 x 10^-3)^2 / 0.4446 = 6.56 x 10^-5

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