Question

1.) Explain the following observations in terms of shifting equilibria by Le Chatelier's Principle. Write the equation for the equilibrium involved in each case.

a.) barium chromate, BaCrO4, is insoluble in basic solution but dissolves readily in acidic solution.

b.) The addition of NaOH to the iron (III)- thiocyanate equilibrium causes the red color to disappear and a rust colored precipitate to form.

c.) The solubility of lead chloride, PbCl2, in 0.10 M HCl is much lower than in pure water.

d.) Limestone, CaCO3, is an insoluble salt but will dissolve in acid.

e.) For part C, how does the rate of the reaction vary as the temperature is increased? Explain.

Answer 1

a)BaCrO₄ is a slightly soluble salt
The equilibrium is
BaCrO₄ ↔ Ba²⁺ + CrO₄^2- K_sp = 1.2 x 10^-10
When H⁺ is added to a mixture containing BaCrO₄ and CrO₄^2- the equilibrium will shift on the right ( Le Chatelier's principle ) since this reaction occurs :
CrO₄^2- + 2 H⁺ ↔ Cr₂O₇^2- + H₂O
Therefore the equilibrium reaction is :
2 BaCrO₄ ( s) + 2 H⁺ (aq) ↔ 2 Ba²⁺ (aq) + Cr₂O₇^2- (aq) + H₂O(l)
BaCrO₄ will dissolve in acidic solution.

b) NaOH + Fe(SCN)₃ ↔ Fe(OH)₃(s) + 3Na⁺(aq) + 3SCN^-(aq).

This reaction should go to completion because the iron (III) hydroxide is insoluble and removes it from equilibrium.

c) PbCl₂(aq) ↔ Pb²⁺(aq) + 2Cl^-(aq)

If HCl is added to this closed system it will cause the system to shift to the left, therefore less soluble.

d) Limestone, CaCO₃, is an insoluble salt but will dissolve in acid.

CaCO₃(s) + 2HCl(aq) ↔ CaCl₂(aq)+CO₂(g)+H₂O(l)

Here the gas escapes from the solution and a reverse reaction is impossible in this case.