Write a balanced net ionic equation to show why the solubility of NiCO3 (s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Consider only the FIRST STEP in the reaction with strong acid. Use the pull-down boxes to specify states such as (aq) or (s). + + + K =

Balanced net ionic equation :

Ag2CO3 (s) + H3O+ (aq) \rightleftharpoons 2 Ag+ (aq) + HCO3- (aq) + H2O (l)

K = 0.169

Explanation:

Reaction 1 : Ag2CO3 (s) \rightleftharpoons 2 Ag+ (aq) + CO32- (aq) : Ksp = 8.1 x 10-12

Reaction 2 : HCO3- (aq) + H2O (l) \rightleftharpoons CO32- (aq) + H3O+ (aq) : Ka2 = 4.8 x 10-11

If we reverse Reaction 2,

Reaction 3 : CO32- (aq) + H3O+ (aq) \rightleftharpoons HCO3- (aq) + H2O (l) : Kb1 = 1/Ka2

Kb1 = 1/(4.8 x 10-11)

Kb1 = 2.08 x 1010

If we combine Reaction 1 and Reaction 3, we get

Ag2CO3 (s) + H3O+ (aq) \rightleftharpoons 2 Ag+ (aq) + HCO3- (aq) + H2O (l)

K = (Ksp) * (Kb1)

K = (8.1 x 10-12) * (2.08 x 1010)

K = 0.169

the answer is na (sodium)