Citric acid polyprotic acid equilibria

[i_love_science]

Homework Statement

Citric acid (H3C6H5O7) is a triprotic acid with Ka1 = 8.4*10^-4, Ka2 = 1.8*10^-5, and Ka3 = 4.0*10^-6. Calculate the pH of 0.15 M citric acid.

Relevant Equations

polyprotic acid equations (below)

Using the first equilibrium, H₃A <-> H₂A^- + H⁺, the [H⁺] is solved to be 0.01165 M ~ 1.1 * 10^-2 M.

Using the second equilibrium, H₂A^- <-> HA^2- + H⁺, the additional change in H⁺ is found to be 1.8 * 10^-5 M. The solution says that this second equilibrium is negligible, because 1.1 * 10^-2 M + 1.8 * 10^-5 M ~ 1.1 * 10^-2.

The question asks to find the pH though, not specifically the concentration [H⁺]. I don't understand why it would not be better to add 0.01165 M + 1.8 * 10^-5 M = 0.011668 M, and continue with this type of calculation with the third equilibrium as well? Would it make the answer more precise? I got pH 1.93 using this method, but the solution gives 1.96.

Could anyone explain why, or whether my reasoning is right? Thanks.

 

[Borek]

It is always a matter of deciding "which answer is already good enough".

For most practical applications if the difference between two ways of calculating pH is below 0.1 pH unit it doesn't matter, as there are other sources of error, typically larger (or at least comparable). But it is not like it is an approach set in stone, there is a lot of personal choice in whether the difference of 0.05, 0.1 or 0.2 can be considered to be negligible.

I have no idea how you calculated the value of 0.01165, assuming a monoprotic acid with Ka=8.4e-4 I got pH of 1.97 ([H⁺ = 0.01081M). I would understand the value of 0.01122 as that's what you can get from approximated calculations, but 0.01165 seems like some kind of error to me.

Using pH calculator and your set of Ka values (they are often listed as slightly different, that's one of these error sources I mentioned earlier) I got an 'exact' pH of 1.97 using all three Ka for calculations, and NO difference between pH values (slight difference in calculated H⁺ concentration was too low) no matter whether I did calculations using one, two or three Ka values/dissociation steps.