PLEASE ANSWER THE FOLLOWING POST LAB QUESTIONS GIVEN BELOW BY READING THE FOLLOW

Question

PLEASE ANSWER THE FOLLOWING POST LAB QUESTIONS GIVEN BELOW BY READING THE FOLLOWING BACK GROUND INFORMATION:

This experiment studied the equilibrium of a cobalt compound.

Each test tube contained 7mL of [CoCl4]^2- and each tube had different powder or liquids added to them. All six test tubes that contained [CoCl4]^2- were light purple and clear before anything was added to them. After adding a little water to test tube 1, it had turned light pink and remained clear. Test tube 2 turned light blue and still clear after adding six drops of 6M HCl. Test tube 3 turned light pink and precipitate was present after adding five drops of 1M AgNO3. Test tube 4 remained light purple while the two scoops of Ca(NO3)2 remained at the bottom. Test tube 5 turned blue while the bottom was a darker blue when 2 scoops of CaCl2 was added. Test tube 6 turned light pink when it was heated up to 75.2 degree Celsius then cooled down to .13 degree Celsius of ice water.

1. How does the reaction quotient, Q compare to K (Q > K, Q < K, or Q = K) immediately after the following things are done (before equilibrium is re-established). Explain your answers (“it is __ because ___”).

a. Addition of HCl(aq)

b. Addition of CaCl2(s)

c. Addition of AgNO3

2. Write the net ionic equation for the reaction that occurred after you added the silver nitrate? (test tube 3)

Test tube Observation(before Observation(after) 1 light purple, clear light pink, clear 2 light purple, clear light blue,clear 3 light purple, clear light pink with precipitate formed 4 light purple, clear light purple with Ca(NO3)2 sitting at bottom 5 light purple, clear dark blue 6 light purple, clear light pink Equilibrium equation is as follows: [Co(H2O), (aq) +4CI-(aq) [Coa,(aq) + 6H20(1) - 2+ Equilibrium constant, K = [Co (H2O), +][a

Explanation / Answer

The equilibrium reaction is

[Co(H2O)6]2+ (aq) + 4 Cl- (aq) <=======> [CoCl42-] (aq) + 6 H2O (l)

Kc = [CoCl42-]/[Co(H2O)62+][Cl-]4

(a) The reaction quotient Q is simply expressed as Kc, but the concentrations of the aqueous species is different from the equilibrium concentrations in the expression for Kc. When HCl is added to the aqueous solution, HCl ionizes to produce Cl-. The denominator in the expression for Q increases and consequently, Q < Kc.

(b) CaCl2 ionizes in water producing Cl-. As before, the denominator in the expression for Q is higher than the denominator in the expression for Kc. Consequently, Q < Kc.

(c) AgNO3 ionizes to produce Ag+ ions which combine with the free Cl- ions to form a precipitate of AgCl. Therefore, the concentration of Cl- falls, i.e, [Cl-] decreases and hence, the denominator in the expression for Q is lower than that in the expression for Kc. Consequently, Q > Kc.

2) The ionization of AgNO3 is given as

AgNO3 (aq) --------> Ag+ (aq) + NO3- (aq)

4 Ag+ (aq) + [CoCl42-] (aq) ----------> 4 AgCl (s) + Co2+ (aq)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.