[Ineedhelp1993]

CaCO₃(s) + H₃O⁺(aq)  Ca²⁺(aq) + HCO_3- (aq) + H₂O (l)

First equilibrium reaction involves the solubility equilibrium of CaCO₃ in pure water:

CaCO₃ (s)  Ca²⁺ (aq) + CO₃^2- (aq)

Second equilibrium reaction:

H₃O⁺ (aq) + CO₃^2- (aq)  H₂O (l) +HCO₃^-

Equilibrium constant expression for the net reaction:
K_c= [Ca²⁺} [HCO₃^-] / [H₃O⁺]

The expression that gives the equilibrium constant for the net reaction:
Ksp/Ka₂

What is the numerical value of the equilibrium constant?

I really do not know how to figure this out. please help. thank you.

[Ineedhelp1993]

I know the Ksp of CaCO₃ is 4.96 x 10^-9

The Ka₂ of H₂CO₃ is 5.6 x 10 ^-11

I divided these two numbers and got 89 but that cant be the answer since its not one of the choices.

[Borek]

The expression that gives the equilibrium constant for the net reaction:
Ksp/Ka₂

That would be my approach.

What are the choices given?

[Big-Daddy]

Your second reaction is the reverse of carbonic acid's second dissociation so the equilibrium constant will be Ka2^(-1). Your first reaction is obviously a Ksp. What you are doing is adding the first and second reactions to reach your desired one, so Kc(overall)=Ksp*Ka2^(-1) as you say = 88.6 which is roughly 89 (M). I am confident your answer is correct.

However maybe something to do with the presence of the solid means that we cannot add the reactions as normal for equilibrium?

[Ineedhelp1993]

The choices are:
a) 160
b) 0.021
c) 4.9 x 10^-19
d) 3.7 x 10 ^-15
e) 1.6 x 10 ^-21

[Ineedhelp1993]

However maybe something to do with the presence of the solid means that we cannot add the reactions as normal for equilibrium?

Well, the first equation I wrote is the deterioration of marble from acid rain and it's a net reaction. I don't know if that's what you're trying to say? but thanks for replying.   

[Borek]

The choices are:

Seems to me like none of the answers given is a correct one.

[Ineedhelp1993]

._. Do you know how I would find the Ksp and Ka2 to answer the question?

[Big-Daddy]

The choices are:

Seems to me like none of the answers given is a correct one.

Is it something to do with the fact that the solution might not have reached saturation? You once told me that if this is the case we drop Ksp from the list of equations. So then of course I have no clue how to solve the problem.

[Borek]

._. Do you know how I would find the Ksp and Ka2 to answer the question?

You take them from tables. Values that you are using are reasonably close to the ones I checked.

Is it something to do with the fact that the solution might not have reached saturation? You once told me that if this is the case we drop Ksp from the list of equations. So then of course I have no clue how to solve the problem.

No, question asks about the equilibrium constant, not about concentrations in some solution.

[Ineedhelp1993]

._. Do you know how I would find the Ksp and Ka2 to answer the question?

You take them from tables. Values that you are using are reasonably close to the ones I checked.


I think my professor gave the wrong expression, I think I have to multiply Ksp and Ka2 because two equations are put together to make one and in the book it said if that is done then I should multiply but my professor said I should divide. /:
When I do multiply the values above I get 2.8 x 10^-19 which is close to C
._. would that be the answer?

[Borek]

Don't guess, DERIVE the result.

[tex]K_{sp} = [Ca^{2+}][CO_3^{2-}][/tex]

[tex]K_{a2} = \frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}[/tex]

Divide K_sp/K_a2, compare with the reaction quotient for the CaCO₃+H⁺. If they are identical, K_sp/K_a2 must be the right answer, no matter what others tell you.

[Ineedhelp1993]

ty
turns out I was right and the answers on the paper were wrong
so it was 88 or 89
thx for your help