If a weak acid, such as acetic acid (CH3COOH) is added to the above solution, wo
ID: 781161 • Letter: I
Question
If a weak acid, such as acetic acid (CH3COOH) is added to the above solution, would you expect to see more or less precipitate (CaC2O4) formed than when hydrochloric acid is added to the solution? Explain your reasoning by showing the effect of the addition of acetic acid on the equilibrium for the reaction and how this differs from the effect of hydrochloric acid on the equilibrium. The three reactions below are all at equilibrium simultaneously in solution. Ca 2+(aq) + C2O2-4 (aq) doubelarrow CaC2O4(s) H2C2O4(aq) doublearrow H+(aq) + HC2O4(aq) HC2O-4(aq) doublearrow H-(aq) + C2O42-(aq) If hydrochloric acid is added to this solution, would you expect to see more or less of the precipitate (CaC20 4) formed in the solution? Explain your reasoning by showing the effect of the addition of hydrochloric acid on the equilibrium for the reaction. If sodium hydroxide is added to this solution, would you expect to see more or less of the precipitate (CaC20 4) formed in the solution? Explain your reasoning by showing the effect of the addition of sodium hydroxide on the equilibrium for the reaction.Explanation / Answer
a. If you added HCl inside the solution, the concentration of H+ will increase, which will shifted to toward the formation of H2C2O4 or HC2O4-, thus reduce the concentration of C2O4(2-) Less CaC2O4 will be formed at this case.
HCL <--> H+ + Cl -
b. When NaOH is added to these solution, it is similar to input Na+ and OH- ions into the solution. OH- will neutralize with H+, and more C2O4(2-) will be released. This will simulate more precipitate to be produced.
c.
CH3COOH <--> H+ + Ch3COO-
Ch3COOH is a weak acid. For the equation above, the equalibrium is more toward to the left side. So, when you add weak acid into the solution, the high concentration of H+ ions in solution will stimulate forming of weak acid. Thus the concentration of H+ ions reduce and more HC2O4- will react and produce C2O4(2-). Then, more precipitate will form as C2O4(2-) react with Ca2+. Same as question b.
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