Solubility Combining Ionic Equations,

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The discussion focuses on calculating the concentration of AgCl(aq) in equilibrium with undissolved AgCl(s) at 26°C. Participants discuss deriving the equilibrium constant K3 by combining K1 and K2, leading to the conclusion that K3 equals 3.8E-8. The correct approach involves recognizing that Ksp is based solely on the concentration of AgCl(aq) since solids do not factor into the equilibrium expression. The final concentration of AgCl(aq) is confirmed to be 3.8E-8 mol/L. The conversation highlights the importance of correctly applying equilibrium constants in solubility calculations.
Maharg
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Homework Statement



Consider the following equilibria at T = 26 0C, in which all ions are aqueous:
Ag+(aq) + Cl-(aq) ----> AgCl(aq) K1 = 2.0E2
AgCl(s) ----> Ag+(aq) + Cl-(aq) K2 = 1.9E-10

Consider the reaction
AgCl(s) ----> AgCl(aq)


and calculate the concentration of AgCl(aq) in equilibrium with excess undissolved solid AgCl(s).


Hint: calculate K3 first and use the equilibration constant expression to find the concentration of AgCl(aq).

Homework Equations


Ksp = [AgCl] I thinK?



The Attempt at a Solution




So I think I'm on the right track but missing something

To get k3 is just combining k1 and k2, Since they easily cancel out to make AgCl(s) -> AgCl (aq), I think k3 is just 200 + 1.9E-10 = 200

k3 = 200? That seems odd but then even if that is true I'm not sure which is next as with solubility I've always had problems where it is Ksp=[A], but this one is just Ksp=[A] as solids wouldn't count right?
 
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Maharg said:

Homework Equations


Ksp = [AgCl] I thinK?

No. First, try to write corect formulas for both Kso and complex formation. Just as if there were no other reactions.
 
Oh I was thinking about it wrong.

Is this correct?

Ag+ + Cl- -> AgCl(aq) k1 = [AgCl]/[Ag+][Cl-]= 2E2
AgCl(s) -> Ag+ + Cl- k2 = [Ag+][Cl-] = 1.9E-10

K3 = k1 k2 = [AgCl(aq)] = 3.8E-8

AgCl(aq) = 3.8E-8 mol/L ?

The last part is only thing I'm not sure about.
 
Last edited:
Kso*Kf, looks OK to me.
 
Yes thank you it was right. I appreciate the help.
 

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