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
It is not important not to mix R-22 and R-401A refrigerants in the same container because:
d. they use different oil.
The Freon in both systems are essentially the same but the oils in each system are not compatible which will render the mixture of the two systems useless.
Answer:
The answer should be 40.3044 g/mol
B).Proton acceptor did that help at all
