Equilibrium constants at 25 ∘C are
Substance Kc Value of Kc
CaCO3 Ksp 4.5×10−9
H2CO3 Ka1 4.3×10−7
Ka2 5.6×10−11
What is the molar solubility of marble (i.e., [Ca2+] in a saturated solution) in acid rainwater, for which pH=4.20?
CaCO3 ==> Ca^2+ + CO3^2- Ksp = (Ca^2+)(CO3^2-)
H2CO3 ==> H^+ + HCO3^- k1 = ....
HCO3^- ==> H^+ + CO3^3- k2 = ....
pH = 4.20; (H^+) = 6.31E-5
Solubility = (Ca^2+) = S
S = (CO3^2-)+(HCO3^-)+(H2CO3)
Find (CO3^2-) in terms of S.
k2 = (H^+)(CO3^2-)/(HCO3^-)
Plug in k2 and H^+ and solve for HCO3^- so that (HCO3^-) = some number x CO3^2-
Then k1 = (H^+)(HCO3^-)/(H2CO3)
Plug in H^+ and HCO3^- and solve for H2CO3 so that
(H2CO3) = some number x (CO3^2-).
Then go back to S = CO3^2- + HCO3^- + H2CO3 and plug all of that in to obtain
S = some number x (CO3^2-)
Then CO3^2-) = S/some number
Into Ksp, plug S in for Ca; plug S/some number in for CO3^2-; substitute for Ksp, solve for S and you have it.
Post your work if you get stuck. By the way, there is a quicker way to do this and it works on this problem but it doesn't work generall on all problems like it.
Thanks! I got it now
So what is the answer?
To calculate the molar solubility of marble in acid rainwater, we need to use the concept of equilibrium constants and the pH of the solution. Here's how you can find the answer:
1. Write down the balanced equation for the dissolution of CaCO3 (marble) in water:
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
2. Use the solubility product constant (Ksp) to set up an expression for the equilibrium constant (Kc) of the dissolution reaction:
Kc = [Ca2+][CO3^2-]
3. Since marble dissolves in acid rainwater, we need to consider the effect of the acidic environment on the equilibrium. Acid rainwater contains carbonic acid (H2CO3), which dissociates in water to release H+ ions. The dissolution equilibrium will be affected by the concentration of H+ ions, which can be related to pH using the ionization constants (Ka1 and Ka2) of carbonic acid:
H2CO3 ⇌ H+(aq) + HCO3^-(aq)
HCO3^- ⇌ H+(aq) + CO3^2-(aq)
4. We can now write the expression for the equilibrium constant of the dissolution reaction in the presence of the acid rainwater:
Kc = ([Ca2+][CO3^2-]) / ([H+][HCO3^-][CO3^2-])
5. Plug in the values given:
Ksp for CaCO3 = 4.5×10^(-9), Ka1 for H2CO3 = 4.3×10^(-7), Ka2 for HCO3^- = 5.6×10^(-11), and pH = 4.20.
6. Use the pH value to calculate the concentration of H+ ions in the solution:
[H+] = 10^(-pH)
7. Substitute the known values into the equilibrium constant expression:
4.5×10^(-9) = ([Ca2+][CO3^2-]) / ((10^(-pH))[HCO3^-][CO3^2-])
8. Simplify the expression by canceling out the common terms:
4.5×10^(-9) = [Ca2+] / (10^(-pH)[HCO3^-])
9. Rearrange the equation to solve for [Ca2+]:
[Ca2+] = 4.5×10^(-9) * 10^(-pH) * [HCO3^-]
10. Calculate [HCO3^-] using the ionization constant Ka1:
[HCO3^-] = √(Ka1 / [H2CO3])
11. Calculate [H2CO3] using the ionization constant Ka2:
[H2CO3] = √(Ka2 / [CO3^2-])
12. Calculate [CO3^2-] using the relation [HCO3^-] = [CO3^2-] + 2[H2CO3].
13. Finally, substitute the values for [HCO3^-], [H2CO3], and [CO3^2-] into the equation for [Ca2+] to find the molar solubility of marble in acid rainwater.
Remember to convert the molar solubility into the desired units, such as grams per liter (g/L), if necessary.