What is the molar solubility of marble (i. e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express your answer to?

Missing in your question:

Ksp of (CaCO3) = 4.5 x 10 - 9

Ka1 for (H2CO3) = 4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9

CaCO3 (s) → Ca + 2 (aq) + CO3-2 (aq)

2) equation 2 for Ka1 = 4.7 x 10^-7

H2CO3 + H2O → HCO3 - + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

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

so, form equation 1& 2&3 we can get the overall equation:

CaCO3 (s) + H + (aq) → Ca2 + (aq) + HCO3 - (aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:

when the inverse of equation 3 is:

CO3-2 (aq) + H3O + (aq) ↔ HCO3 - (aq) + H2O (l) Ka2^-1 = 1.79 x 10^10

when we add it to equation 1

CaCO3 (s) ↔ Ca2 + (aq) + CO3-2 (aq) Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:

when H3O + = H+

CaCO3 (s) + H + (aq) ↔ Ca2 + (aq) + HCO3 - (aq) K = 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]

5.6 = - ㏒[H]

∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142