[Mobulinae]

Hello everyone,

In 2 weeks from now I have an re-exam on water quality. Even though I have been studying properly there is just a question which I completly fail to anwser and I dont even know where to start with. So I figured lets just give it a shot on this website and share it. Could anyone please elaborate which step to take?

Question: A calcareous lake is saturated with respect to CaCO₃ and is in equilibrium with atmospheric CO₂. Ca²⁺ is by far the dominant cation in solution. Carbonate is the dominant acid-base system.
If atmospheric CO₂ has a partial pressure of 10^-3.7 atm, then what would be the pH of the lake?

Looking forward to an anwser.

[Borek]

You have to show your attempts at solving the problem to receive help, this is a forum policy.

Can you start by listing all relevant equilibria and all relevant species?

[Mobulinae]

I am sorry about not following the policy Borek, I will start listing. I spend a few hours on it and came to an anwser. The problem is we don't have a clue wether this is correct, but this is what we got from our manual which is needed to bring with since it is an open book exam.

Relevant equilibria with acidity constant:
CO₂(g) + H₂O(l) H₂CO₃                                                                                     K1=10^1.46
H₂CO₃ H⁺(ub][s2+(aq)                                                                                     K2=10^6.35
HCO₃^-(aq) CO<sub>3[/subr]p](aq)+2+p]+(aq)                                                                                    K3=10^10.33

Becoming
CO2(g) + H2O(l) CO3^2- + 2H⁺(aq)                                              K= (K1*K2*K3)= 10^18.14
CaCO3(s) + H2O  Ca²⁺(aq) + CO3^2- (aq)                              KSO (solubility constant) =  10^-8.3

Total relevant equilibria formula
CaCO3(s)* 2H⁺(aq) CO2(g) + H2O(l)+Ca²⁺(aq)                        K^-1*K^SO= (10^18.14)^-1*10^-8.3 = 10^9.84

Maximum solubility of Ca²⁺ calculation, with PCO2 = 10^-3.7atm
[Ca²⁺] = 10^9.84*[H⁺]²*(PCO2)^-1</sub>     10^9.84*(2)²*(10^-3.7)^-1 = 1.384*10^{14 [/sup}[/color]]In this formula I am doubting if the substitution for [H⁺] is correct

Calculate pH
log[Ca²⁺] = log 1.384*10¹⁴ = pH = 14.14

Can anyone confirm or point me toward my errors and eleborate?

[Borek]

Sorry, I am on vacation ATM and I can't solve the problem properly, so just a few remarks.

Sanity check: your final answer suggests the lake is highly alkaline, when it should be (due to the dissolved CO₂) on the slightly acidic side.

Solubility of the CaCO₃ doesn't require full protonation of the acid, the most important reaction being

CaCO₃ + CO₂ + H₂O Ca²⁺ + 2HCO₃^-

which in turn means your attempts to assume reaction uses two H⁺ to proceed are wrong.

You are right about the need for Ksp and two Ka values, can you think of mass and charge balance equations to accompany the equilibrium equations?