[fobbz]

18. Consider the solubility equilibrium:
SrCO₃ (s)     Sr²⁺ (aq) + CO₃^2- (aq)
The addition of which of the following substances will cause the equilibrium to shift right?
A. HCl (aq)
B. SrCO₃ (s)
C. Na₂CO₃ (aq)
D. Sr(NO₃)₂ (aq)

Why does adding more SrCO₃ not shift the equilibrium to the right? I do not understand, would it not need to balance out through LeChatlier's?

[lillybeans]

SrCO3 is a solid, and does not affect the position of equilibrium!! (only gases and aqueous solutions have an effect, that's why we don't include pure solids and liquids in equilibrium expressions)

The answer would be A.

The H from HCl combines with CO3^2- to form HCO3^2-, thus it removes carbonate ions away. Equilibrium always tries to restore what it lost, so then it will shift the equilibrium to the right, to regenerate the lost carbonate ions.

[fobbz]

awesome! thank you very much.

[maivu]

18. Consider the solubility equilibrium:
SrCO₃ (s)     Sr²⁺ (aq) + CO₃^2- (aq)
The addition of which of the following substances will cause the equilibrium to shift right?
A. HCl (aq)
B. SrCO₃ (s)
C. Na₂CO₃ (aq)
D. Sr(NO₃)₂ (aq)

Why does adding more SrCO₃ not shift the equilibrium to the right? I do not understand, would it not need to balance out through LeChatlier's?

Case A: HCl(aq)    H⁺  + Cl^-
H⁺ ions will react with CO₃^2- in the initial solution to form HCO₃^- or H₂CO₃, therefore, reduce the concentration of CO₃^2-. When the CO₃^2- ion is removed, the system will try to produce more of this ion. Hence the equation shifts to the right.

Case B: SrCO₃. This compound is in solid state, therefore it cannot make a change in initial solution. SrCO₃ is known as sparingly soluble salt. Therefore the solubility product for this reaction is:
K_sp = [Sr²⁺]*[CO₃^2-]
You can only make more SrCO₃ be dissolved by either change the K_sp or use the common ion effect. Adding more SrCO₃ cannot produce the change neither in K_sp nor common ion effect.

Case C: Na₂CO₃ 2Na⁺  + CO₃^2-
In this case, there is a common ion effect. SrCO₃ and Na₂CO₃ have CO₃^2- as a common ion. The equilibrium will shift to the left as followed the Le Chatelier's Principle.

Case D: Sr(NO₃)₂  Sr²⁺ + 2NO₃^-
There is NO₃^- acts as a common ion. Excess of NO₃^- will shift the equilibrium to the left.

Conclusion: Only case A can satisfy the question. Answer: A.

This is my answer and explanation for this question. If you find any mistake or have question on it. Please let me know.