Calculating pH and concentration using pKa values?
Task:
Carbonic acid has pKa1 = 6.35 and pKa2 = 10.33. (In this exercise, assume that all substances remain dissolved!)
(a) Calculate the pH value obtained when 10 g of sodium carbonate is dissolved in 1 L of pure water.
(b) Calculate the concentration of carbonic acid under these conditions.
Problem/Approach:
For (a), I would proceed as follows: Calculate the amount of Na2CO3, or rather the concentration, calculate the pKb value, and then enter the formula for the pOH value for weak bases. This then yields the pH value. However, according to research, this is incorrect.
For (b) I would fiddle around with the MWG.
Thank you very much for your help!
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We gave a γ=10 g/l solution of Na2CO3, which corresponds to c=γ/M=0.094 mol/l. I have assumed that you have used water-free Na2CO3 for crystal soda Na2CO3 ⋅10 H2O it would be 0.035 mol/l). The pKa values of carbonic acid are 6.35 and 10.33. The solution reacts basicly because
CO32 + H2O ⟶ HCO3 ̄ + OH ̄
First we want the pH. The formula for weak bases
pH=14-1⁄2(14-pK2-lg(c)=11.65
and this is not so bad at all, with a better formula only a marginally different value 11.64 comes out.
Next, you want to know how much free CO2 is in the solution. It is possible to do so differently; a fast and sloppy method is based on the generalization of carbonic acid, which is determined by the acid constant Ktot=K1⋅K2:
CO2 + 2 H2O⟶ CO32 ̄ + 2 H3O+
Ktot=K1K2 = c2(H3O+)⋅c(CO32 ̄) / c(CO2)
c(CO2) = c2(H3O+) ⋅ c(CO32 ̄) / Ktot
Since we know the pH, c(H3O+)=10 ̄pH can now be used and the recently low concentration of 2.35⋅10 ̄8 mol/l is obtained.
This calculation, however, contains the approximation that we have assumed that the equilibrium concentration of carbonate corresponds to the level concentration of 10 g/l = 0.094 mol/l. But she does not really; we have already calculated pH=11.64 and therefore c(OH ̄)=0.0044 mol/l, so the Balanceconcentration of carbonate only 0.090 mol/l (because 0.0044 mol/l of carbonate reacts to HCO3 ̄+OH and is therefore absent).
If we now use the correct value c(CO32 ̄)=0.090 mol/l in the above equation, we get the correct result c(CO2)=2.24⋅10 ̄8 mol/l.