Asked by Anonymous
A 0.108 M sample of a weak acid is 4.16% ionized in solution. what is the hydroxide concentration of this solution?
I know that i set up an ice table. The equation would be weak acid + H20 goes to OH + the acid. What do i do with the 4.16 %? How does that fit into the problem. Do i use that to figure out the Kb value?
I know that i set up an ice table. The equation would be weak acid + H20 goes to OH + the acid. What do i do with the 4.16 %? How does that fit into the problem. Do i use that to figure out the Kb value?
Answers
Answered by
DrBob222
Let's let HA stand for the weak acid.
HA + H2O ==> H3O^+ + A^-
You also know that
H2O ==> H^+ + OH^- and
Kw = (H^+)(OH^-)
So if the solution is 0.108 M and the solution is 4.16% ionized, then the (H^+)= 0.108 x 0.0416 = ??
Now use (H^+) and Kw to calculate (OH^-). Check my work.
HA + H2O ==> H3O^+ + A^-
You also know that
H2O ==> H^+ + OH^- and
Kw = (H^+)(OH^-)
So if the solution is 0.108 M and the solution is 4.16% ionized, then the (H^+)= 0.108 x 0.0416 = ??
Now use (H^+) and Kw to calculate (OH^-). Check my work.
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