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The ka of a monoprotic weak acid is 4.97 × 10-3. what is the percent ionization of a 0.193 m solution of this acid?
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The ka of a monoprotic weak acid is 4.97 × 10-3. what is the percent ionization of a 0.193 m solution of this acid?

User Emmanuella
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1 Answer

3 votes
Given:

Ka of monoprotic weak acid = 4.97 x 10^-3
Concentration = 0.193 m

H2A ----> 2H2+ + A-
0.193 0 0
-x x x
------------------------------
0.193 -x x x

where x is the percent ionization

4.97 x 10^-3 = x^2 * x / 0.193 - x

solve for x

x = 0.082


User James Hart
by
8.8k points
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