It's nice to see someone come back, especially with a relevant experimental result. Especially one that corresponds exactly to prediction, though as Borek says there may be an element of fluke in its coming out so exact. That's why I said at the start - measure it.
Main influences I know why measured value would deviate from calculated are ionic strength of the solution, and temperature. You'd have to go back to the original work that is referenced in the tables of pK's to see how closely the conditions where they were measured resemble yours. For some other factors such as the difference between concentration and activity, well they could well have been present in in the original measurements, and therefore the 'errors' cancel out. Anyway although I would not be much troubled by some discrepancy between predicted and measured the prediction ought to be quite better than ballpark.
Your equation is an approximation valid at high concentrations. Getting on top of this subject around pH's which generates such a large fraction of the homework questions involves approximations. That is I want to say: it's not like there is The Theory and then, oh, here you can use a convenient approximation and there another one. Rather the approximations are an essential part of the (limited) subject the student has to master. In fact there are several different kinds of approximation, of varying goodness and generality used, and you need to have an understanding of them and how good they are, any limitations.
For that of your equation you don't have to guess, you can easily see at what point it breaks down. I'll just treat it like a monobasic acid, which for present purposes we have seen is good enough. When the total concentration of your acid is equal to K
a then from the equilibrium equation.
$$ K_a = \frac{[H^+][A^-]}{[HA]} $$
It follows that if ##[A^-] = [HA]## then ##[H^+] = K_a ## or ##pH = pK_a## and vice versa.
So in your case when K
a = 7.4×10
-4, [H
+] = [A
-] = 7.4×10
-4 so total citrate molarity is about 1.5×10
-3. At that concentration then half of it is A
- contrary to the assumptions giving your equation which are that the acid is concentrated enough that nearly 100% of it is HA, so approximately equal to the total acid molarity, C, used in your formula.
So you expect the formula to be noticeably failing around 1.5×10
-3 M. But if you increase the citrate concentration tenfold you expect A
- to be only about a tenth of the total acid, and your equation to hold moderately well. By 5×10
-2M it should be good enough for all practical purposes. But if on the other hand you go to concentrations 10 times lower than K
a the acid will be nearly all dissociated and [H
+] = [A
-] ≅ C, so then plot of pH against - log C will be approximately linear with slope 1. This slope will thus be: close to at ½ high concentrations as predicted by your equation changing to 1 at lower, changeover happening around pH = pK
a ≈ - log 2C.
This is all illustrated in an overall equation
$$ [H^+] = \frac{K_a + \sqrt{K_a(K_a + 4C)}}{2} $$
plotted here:
Red pH (ordinate) against - log C (abcissa) for monobasicn acid K
a = 7.4×10
-4
Blue pH = (3.13 + C)/2
Green pH = C
Even that equation is not the complete answer, obviously the linearity cannot continue forever - it has to level off by pH 7 (you cannot make a solution and alkaline by diluting acid!). In other words we have to take the dissociation of the water molecules into consideration at high dilutions. And all this is just for a monobasic acid; with citric when you get much below 10
-4 M its second dissociation is significant (you could use your original formula though with the second pK
a because then the first dissociation is approximately 100%.) It must be a rare that such calculations are useful however.
(Sorry this answer has been delayed; this was due to difficulties doing the plot which turned out to be nothing but silly mistakes in inputting the formula. And it was unnecessary to solve the equation for [H
+] either to plot the curve, or to see the different linear approximations at high and low concentrations - I could have just used
$$C = \frac{[H^+]^2}{K_a} + [H^+] $$
)