Question

Barium sulfate is only sparingly soluble in water. When excess solid barium sulfate is placed in contact with water to form a saturated solution, the following equilibria are established:

1) BaSO4(s) <---> Ba 2+(aq) + SO4 2-(aq)

2) SO4 2-(aq) + H2O(l) <---> HSO4-(aq) +OH- (aq)

3) 2H2O (l) <---> H3O+(aq) +OH-(aq)

Will the concentration of Ba2+ ion increase, decrease, or remain the same when each of the following is done? Explain your reasoning.

a) solid sodium sulfate is added to the solution

b) several drops of concentrated sodium hydoxide solution are added to the solution (ignore the effect of dilution)

c) additional solid barium sulfate is placed in contact with the solution

Answer 1

Ans

a)

When sodium sulfate is added into the solution , the amount of sulfate ion in the solution will increase , which will cause less dissociation of barium sulfate in the solution , hence the concentration of Ba2+ will decrease.

b)

When sodium hydroxide is added in the solution , the concentration of hydroxide ions in the solution will increase, which will cause the reaction 2 to move backwards and hence sulfate concentration increases in the solution , causing less dissociation of barium sulfate and then decrease in concentration of Ba2+ ions.

c)

When additional solid barium sulfate is placed in contact with the solution , the equillibrium is moved forward and the Ba2+ concentration in the solution increases.