Calculating pH of Limestone Quarry Water | H2CO3 pk1=6.37, pk2=10.32

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The discussion centers on calculating the pH of water in a limestone quarry, using the provided dissociation constants for carbonic acid (H2CO3). The initial attempt at a solution involves the hydrolysis of carbonate ions (CO3 2-) and the equilibrium expression for bicarbonate (HCO3-). It is noted that the pH calculation is complicated by the carbonate concentration and the solubility product of calcium carbonate. A suggestion to use the average of the two pKa values for an amphiprotic salt is deemed inappropriate, as the system starts with fully neutralized carbonate. The complexity of the solution is acknowledged, emphasizing the influence of carbonate hydrolysis on pH determination.
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Homework Statement


Find the pH of water in limestone (CaCO3) quarry.
H2CO3 pk1=6.37, pk2=10.32

Homework Equations


pH=pKa + log [conjugate base]/[acid]
Kw=10^-14

The Attempt at a Solution


CO3 2- + H2O -> HCO3- + OH-
Ka= [HCO3-][OH-]/[CO3 2-]
10^-10.32=x^2/[10^-6.37]
[OH-]=x=.01059
-log[OH-]=1.98
14-1.98=12.02
is this correct?
 
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since its a salt, I think you can do pH= 1/2 (pKa1 + pKa2)
 
wallace13 said:
since its a salt, I think you can do pH= 1/2 (pKa1 + pKa2)

No, that would work for solution of amphiprotic salt, here you start with fully neutralized carbonate.

A lot depends on the initial cncentration of carbonates - and this depends on the calcium carbonate solubility product. Besides, solubility will increase due to carbonates hydrolyzis, so finding an exact answer can be tricky. No idea how deep you are expected to get into it.

coookiemonste said:
Ka= [HCO3-][OH-]/[CO3 2-]
10^-10.32=x^2/[10^-6.37]

Why do you think [CO32-] = Ka1?

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