Question #170501

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.

Expert's answer

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.

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