Consider the equilibrium reactions as below. CaCO 3 (s) -------> Ca 2+ (aq) + CO
ID: 699401 • Letter: C
Question
Consider the equilibrium reactions as below.
CaCO3 (s) -------> Ca2+ (aq) + CO32- (aq)
Ksp = [Ca2+][CO32-] ……(1)
CO32- is the conjugate base of the weak acid HCO3- and establishes equilibrium as
CO32- (aq) + H2O (l) <======> HCO3- (aq) + OH- (aq)
Kb = [HCO3-][OH-]/[CO32-] ……(2)
CO2 dissolves in water to form HCO3- as below.
CO2 (g) <======> CO2 (aq) + H2O (l) --------> H2CO3 (aq) -------> H+ (aq) + HCO3- (aq)
When CO2 is bubbled into water, CO2 combines with water producing carbonic acid, H2CO3, a weak acid which splits up into proton (H+) and bicarbonate anion (HCO3-). Since the reaction produces H+, the pH of the solution will decrease (a lower value of pH means higher H+ concentration and viceversa). Therefore, rule out options (c) and (d).
The dissolution reaction produces HCO3-; therefore, the numerator in expression (2) increases. However, the base ionization constant, Kb is constant at a particular temperature; therefore, to keep Kb constant, the denominator in (2) must increase, i.e, [CO32-] must increase. An increased [CO32-] means that [Ca2+] must decrease. This is definitely true, since Ksp is an equilibrium constant and must remain constant at a particular temperature. Therefore, (a) is the correct answer.
Explanation / Answer
11. A student puts some limestone (CaCO3(s)) to a beaker of water. A small portion of the added limestone is dissolved, and the rest remains as solid. What will happen to the solution if carbon dioxide (CO2) is bubbled into the beaker? (a) pH will decrease and calcium concentration will decrease, (b) pH will decrease and calcium concentration will increase, (c) pH will increase and calcium concentration will decrease, or (d) pH will increase and calcium concentration will increase.