Question
A 0.010 M solution of a weak monoprotic acid is 3.0% dissociated. What is the equilibrium constant, Ka, for this acid?
Answers
HA = H^+ + A^-
Ka = (H^+)(A^-)/(HA)
initial:
HA = 0.01
H^+ = 0
A^- = 0
change:
H^+ = x
A^- = x
HA = 0.01 - x
equilibrium:
x = 0.01 x 0.03 = ??
Substitute and solve for Ka.
Ka = (H^+)(A^-)/(HA)
initial:
HA = 0.01
H^+ = 0
A^- = 0
change:
H^+ = x
A^- = x
HA = 0.01 - x
equilibrium:
x = 0.01 x 0.03 = ??
Substitute and solve for Ka.
3.0 * 10^-5
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