How do you determine the pH of citric acid?

This site reports that #C_6H_8O_7#, a formal triacid, has #pK_(a1)=3.13#, and #pK_(a2)=4.76#.

Given a starting concentration of citric acid, we solve the competing equilibria simultaneously. It would not be straightforward, and I doubt you would be asked to perform in 1st or 2nd year.

The question was probably ill posed/recognised to/by you.

Citric acid is a pure substance made of solid crystals. Pure substances, as sodium hydroxide and pure hydrochloric acid, nonaqueous substances and mixtures and even skin and hair, have no pH.

Water is the only pure substance having a pH.

In fact pH is a property of aqueous solutions . It indicates the negative exponent of the molar concentration of hydrated hydrogen ions in water, expressed as power of ten.

Then, to get a pH (presence of free hydrated hydrogen ions) you must dissolve a certain amount of citric acid in a given volume of water.

In this way you get a water solution of citric acid, whose pH will depend on its concentration.

This pH will be acidic, quite low (between 2-3) if the solution is not highly diluted, because citric acid is a relatively strong acid among the organic acids.

In that case you could measure experimentally the pH of the solution (faster method) or calculate theoretically that pH (quite cumbersome).

Lemon juice has a citric acid concentration around 50 g/L, that is around 0,25 molar. Its experimental pH is around 2-3. But this cannot be calculated exactly, because citric acid is not the only acid in a lemon, although it is the most abundant. Furthermore there are buffering substances and citrates reducing its acidity when the lemon is ripe.

I hope this add clarity in your understanding of this topic.

However, you can calculate the pH of an aqueous solution of citric acid.

Citric acid is a weak triprotic acid.

Ideally, we should do a systematic calculation involving all three acidity constants.

However, we can get a pretty good approximate answer by assuming that only the first ionization is important.

Let's use an ICE table to calculate the pH of 0.300 mol/L citric acid.

#color(white)(mmmmmmm)"H"_3"Cit" + "H"_2"O" ⇌ "H"_2"Cit"^"-" + "H"_3"O"^"+"#
#"I/mol·L"^"-1":color(white)(mm)0.300color(white)(mmmmmmm)0color(white)(mmmm)0#
#"Cmol·L"^"-1":color(white)(mmml)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmm)"+"x#
#"E/mol·L"^"-1":color(white)(ml)"0.300-"xcolor(white)(mmmmmmll)xcolor(white)(mmmm)x#

#K_text(a1) = (["H"_2"Cit"^"-"]["H"_3"O"^"+"])/(["H"_3"Cit"]) = x^2/(0.300-x) = 7.41 × 10^"-4"#

Check for negligibility

#0.300/(7.41 × 10^"-4") = 405 > 400#. ∴ #x ≪0.300#

Then

#x^2/0.300 = 7.41 × 10^"-4"#

#x^2 =0.300 × 7.41× 10^"-4" = 2.22 × 10^"-4"#

#x = 0.0149#

#["H"_3"O"^"+"] = "0.0149 mol/L"#

#"pH" = "-log(0.0149)" = 1.83#