H2CO3 (aq) ⇌ H+ (aq) + HCO3 - (aq) Ka = 4.5 x 10 -7
A. The students needed to prepare 500 ml of 0.00045M H 2 CO 3 solution. Determine the concentration of the H+ ions when the solution can reach equilibrium.
B. Then, 100 ml of 0.00075M (NH4) HCO3 solution was added. Determine the resulting equilibrium
concentration of the carbonate ion.
CH3COOH ==> H^+ + CH3COO^-
Ka = (H^+)(CH3COO^-)/(CH3COOH)
If I substitute Ka = 1.8 x 10^-5 (your text may have a different number) and (H^+) from the problem as 2.56E-13 along with (CH3COO^-) = 0.05 from the problem and solve for (CH3COOH), I get something like 7.11E-10 which is, indeed, quite small. Or we could look at it another way, CH3COO^- is a base and hydrolyzes to produce
CH3COO^- + HOH ==> CH3COOH + OH^-
Kb = (Kw/Ka) = (1E-14/1.8E-5) = 5.55E-10 = (CH3COOH)(OH^-)/(CH3COO^-).
If I substitute 0.039 for OH and 0.05 for CH3COO^- and solve for CH3COOH, I get 7.11E-10.