A. Effect of lowering the pH on the solubility of an insoluble salt. Obtain 5.0

Question

A. Effect of lowering the pH on the solubility of an insoluble salt.

Obtain 5.0 mL each of 1.0 M calcium chloride and 0.25 M sodium oxalate solutions

Pour both solutions into a 50 mL beaker (mixing well). What reaction has occurred?

Add approximately 10 mL of 6 M nitric acid and stir. What reaction has occurred? What do you observe? Dispose of the solution in the appropriate waste receptacle and thoroughly clean the beaker.

Write the net ionic equation for the reaction (which involves one of the species in the reaction that you've just written) that occurs when nitric acid is added to the beaker in Part A. Examine the two reactions shown above for part A and explain, using lessthanorequalto Chatelier's principle, why the changes occurred in the beaker after adding nitric acid.

Explanation / Answer

The reaction taking place in the beaker upon the addition of calcium chloride (CaCl2) and sodium oxalate (Na2C2O4) is as below:

CaCl2 (aq) + Na2C2O4 (aq) ------> CaC2O4 (s) + 2 NaCl (aq)

Decompose the aqueous species into ions and write the ionic equation as

Ca2+ (aq) + 2 Cl- (aq) + 2 Na+ + C2O42- (aq) -----> CaC2O4 (s) + 2 Na+ (aq) + 2 Cl- (aq)

Cancel out the common terms from both sides and write the net ionic equation as

Ca2+ (aq) + C2O42- (aq) ------> CaC2O4 (s)

If the above reaction is at equilibrium, then we can write the equilibrium constant for the reaction as

K = [CaC2O4]/[Ca2+][C2O42-] ….(1)

Calcium oxalate (CaC2O4) is insoluble in water and acetic acids, but is soluble in mineral acids. Therefore, the reaction that takes place when HNO3 is added to the system at equilibrium is

CaC2O4 (s) + 2 HNO3 (aq) -------> Ca(NO3)2 (aq) + H2C2O4 (aq)

Decompose the aqueous species into ions and write

CaC2O4 (s) + 2 H+ (aq) + 2 NO3- (aq) -----> Ca2+ (aq) + 2 NO3- (aq) + 2 H+ (aq) + C2O42- (aq)

Cancel out common terms from the expression to get the net ionic equation as

CaC2O4 (s) ------> Ca2+ (aq) + C2O42- (aq) (ans)

The equilibrium constant for the above reaction can be written as

K’ = [Ca2+][C2O42-]/[CaC2O4] = 1/K

Therefore, when HNO3 is added to the solution in part A above, decomposition of calcium oxalate to the corresponding ions take place. Therefore, the denominator in expression (1) above increases. K is equilibrium constant and hence, depends only on the temperature. Since the temperature remains unchanged, hence K must remain constant. Therefore, to balance the effect of increased ionic concentration, more CaC2O4 must be formed from CaCl2 and NaC2O4 and first reaction essentially goes to completion (ans).

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