ENCE 433 Dr. Alba Torrents

ENCE 433 Dr. Alba Torrents

EXAMINATION 3 "SHOW ALL WORK" Fall 1996

Thursday, November 21

No Partial credit for YES/NO; HIGH/LOW without explanation.

State assumption; remember to check them

Clearly indicate intermediate and final results

System	Add	How would vary ?
CaCO₃(s) + H₂O closed	H₂O	[Ca⁺²] --- as solid is present C_T --- as solid is present
CaCO₃(s) + H₂O open	H₂O	[Ca⁺²] --- C_T ---
CaCO₃(s) + H₂O closed	NaHCO₃	[Ca⁺²] ¯ common ion C_T
Fe(OH)₂(s) + H₂O	NaOH	Fe⁺² ¯ common ion or pH pH ¯ add base
Ag⁺, Cl^- , H₂O IAP = K_so	AgCl(s)	Ag⁺ --- already at equilibrium
Ag⁺, Cl^- , H₂O IAP = K_so	H₂O	Ag⁺ ¯ add water, dilution
Ag⁺, Cl^- , H₂O AgCl(s)	H₂O	Ag⁺ --- solid will dissolve

To answer a question like this, we must imagine the "instant of mixing" , and instant before any reaction that may happen has had a chance to take place. We then calculate Q at the instant of mixing, and compare it to the equilibrium constant. Bear in mind than when two solutions are mixed, dilution always occurs.

Because NaNO3 is a soluble, strong electrolyte the only reaction that can possible occur when these two solutions are mixed is the formation of insoluble BaF2. Let us calculate both [Ba+2] and [F-] at the instant of mixing.

No. mmol Ba+2 added = (70 mL) (0.050 mmol L-1) = 3.50 mmol

Total volume after mixing = 100.00 mL

Hence at the instant of mixing:

This is the reaction whose equilibrium constant is Ksp of BaF2. The reaction quotient is:

Since Q < Ksp the reaction does not proceed to the left.

The solution is undersaturated and no BaF2 will form.

(a)

The pH will decrease and more Mn⁺² could dissolve if the hydroxide was the only solid phase.

As MnCO₃ is another solid phase, some may precipitate as carbonate and thus [Mn⁺²] decrease.

(c)

(d)

[Mn⁺²] from hydroxide = 10^-12.8/(10^-7)² = 10^1.2 M very large, so not controlling

[Mn⁺²] from carbonate = 10^-10.4 / [CO₃^-2]

So, carbonate controls the solubility