7.12 Common-Ion Effect

Revisiting Le Chatelier

Multiple Choice

2SO_2(g)+O_2(g) ⇌ 2SO_3(g) + Heat

Adding SO_2(g) will

• shift equilibrium right

• shift equilibrium left

• increase rate of reaction

• have no change

Multiple Choice

CoCI₄^2- +6H₂O →Co(H₂O)₆²⁺ + 4CI^- + Heat

What will happen when the temperature is increased?

• Position of equilibrium will shift to left and become more blue

• Position of equilibrium will shift to right and become more pink

• No change in position of equilibrium

• Position of equilibrium will shift to left and become more pink

Multiple Choice

What happens when sodium ions (Na⁺) are added?

• The reaction shifts left, towards the reactants

• The reaction shifts right, towards the products

• There is no shift

Multiple Choice

What happens when fluoride ions (F^-) are removed?

• The reaction shifts left, towards the reactants

• The reaction shifts right, towards the products

• There is no shift

Possible Quiz Questions

Multiple Choice

A saturated aqueous solution of CdF₂ is prepared. The equilibrium in the solution is represented above. In the solution, [Cd²⁺]_eq = 0.0585 and [F^-]_eq = 0.117. Some 0.090M NaF is added to the saturated solution. Which of the following identifies the molar solubility of in pure water and explains the effect that the addition of has on this solubility?

• The molar solubility of CdF₂ in pure water is 0.0585M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF₂.

• The molar solubility of CdF₂ in pure water is 0.0585M, and adding NaF has no effect on the solubility because only changes in temperature can increase or decrease the molar solubility of an ionic solid.

• The molar solubility of CdF₂ in pure water is 0.117M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF₂.

• The molar solubility of CdF₂ in pure water is 0.176M, and adding NaF increases this solubility because the Na⁺ ions displace the Cd²⁺ ions, causing the equilibrium to shift to favor the products.

Multiple Choice

Shown above is information about the dissolution of AgCl (s) in water at 298K. In a chemistry lab a student wants to determine the value of K_sp, the molar solubility of AgCl, by measuring [Ag⁺] in a saturated solution prepared by mixing excess AgCl and distilled water. How would the results of the experiment be altered if the student mixed excess with tap water (in which [Cl^-] = 0.010) instead of distilled water, and the student did not account for the [Cl^-] in the tap water?

• The value obtained for K_sp would be too small because Cl^- (aq) ions would be attracted to the Ag⁺ ions in the AgCl crystals, thus preventing water molecules from reaching the crystals.

• The value obtained for K_sp would be too small because less AgCl (s) would dissolve because of the common ion effect due to the Cl^- (aq) already in the water.

• The value obtained for K_sp would be too large because more AgCl (s) would dissolve because of the attractions between Ag⁺ ions in the AgCl crystals and the Cl^- (aq) ions in the water.

• The results of the experiment would not be altered because 0.010M is such a small concentration of Cl^- (aq) ions and thus has no effect on the dissolution of AgCl(s).

Multiple Choice

Shown above is the chemical equation for the dissolution of the slightly soluble salt CuBr (s). Its K_sp value in pure water was experimentally determined. was found to be much less soluble in a 0.001M NaBr solution than in pure water. Which of the following correctly explains the decrease in solubility of CuBr in NaBr?

• The concentration of water is much less in the NaBr solution than in pure water, reducing the rate of dissolution of the salt.

• The presence of Na⁺ ions in the solution inhibits the dissolution of salts.

• The presence of additional Br^- already in the solution decreases the value of the K_sp for CuBr, causing the solubility to decrease.

• The presence of additional Br^- ions already in the solution means equilibrium will be reached when much less CuBr has dissolved.