Slaked lime absorbs CO2 out of the air and slowly converts to carbonate. Does this need to be taken into account when calculating the amount of lime to add?
Yes and no. If you use Hubert Hanghoffer's method which relies only on pH readings then you don't calculate or measure anything but pH. It goes without saying, however, that you will be well served by having an idea as to how much lime is required before you begin.
Let's say I have a 5 year old bag of Mrs Wage's' Pickling Lime that I bought for making pickles. It's been stored in the original (opened) paper bag in a dry place. How much is still lime and how much is chalk?
Impossible to guess but, fortunately it is pretty simple to determine at least approximately by measuring the alkalinities (note plural) of a 'solution'. Start by putting some of the powder in a test tube or on a plate and drop some acid on it (vinegar will do). If it fizzes then there is appreciable carbonate and you will want to know how much. The presence of the carbonate is by no means detrimental. In fact it is beneficial as a source of nucleation sites.
Measure out 100 mg of the 'lime' you have at hand and dissolve it in cold DI water (RO will do if your TDS is low). Make up to 100 mL. Shake or stir thoroughly. As the solubility of lime is 1730 mg/L at 20 °C the powder should all dissolve eventually, though it may take a lot of stirring, if it is all Ca(OH)2. If it does that is a good indication that there is no CaCO3 (we are assuming no other significant levels of impurities here) and you need go no further.
If undissolved powder remains measure the P alkalinity. This done by adding a bit of phenolpthalein to the beaker or flask and then adding 0.1 N acid until the color disappears. The number of mL added acid is the P alkalinity in mEq/L. What has happened at this point is described by:
xCa(OH)2 +yCaCO3 + (2x+y)H+ ---> (x + y)Ca++ + 2xH2O + yHCO3-
From this it is apparent that
P = 2x + y
where x is the number of millimoles of lime in the powder and y the number of millimoles of calcium carbonate. In words, most of the OH- ions have been protonated to form water and most of the carbonate ions have been protonated to form bicarbonate ions. The 'most of' phrases explain why the answer you get will be approximate. The pH at this point is about 8. Of course, if you have a pH meter you can use it to titrate to pH 8. If you are color blind, as I and 1 in 12 males are (in my whole life I've only met one color blind female and, believe me, that was the least of her problems) the pH meter isn't really an option.
If there was undissolved powder, i.e. carbonate, this will take a looong time to dissolve so you must be patient. The phenolpthalein will clear or the meter will indicate pH 8 but upon standing or stirring the pH will start to creep back up. A colorless but turbid solution indicates that dissolution of the powder is not complete even though the pH is 8 or less. If this happens, start over again with 50 mg of powder or even less. You must have a perfectly clear solution stable at pH 8.
Now add Methyl orange (or whatever indicator your test kit uses for M alkalinity determination or use the meter) and continue titration to the M end point (pH 4.5). The total number of mL of acid added (from the beginning - not from the P end point) is the M alkalinity in mEq/L.
In this step
yH+ + yHCO3- ---> yH2O + yCO2
so the M alkalinity is
M = 2(x+y)
Just knowing y answers your question. As the molecular weight of calcium carbonate is 100 the weight of calcium carbonate in the powder you started with is 100y and dividing this by the mass of the powder you started with gives you the percentage by weight of chalk in your powder but this requires that you have means for measuring weight and volume accurately. If you don't you can still do the test if you don't. In this case you put a 'pinch' of powder in a flask and dissolve with DI (or RO) water. Now add drops of acid of reasonable strength and measure P and M in units of drops.
Now M - P = y and M + P = 4x + 3y from which it is clear that
x = [(M + P) - 3(M - P)]/4 = P - M/2
The weights in mg/L of, respectively, lime and chalk in the powder are thus, respectively
mL = wL(P-M/2)f and mC = wC(M-P)f
where wL is the molecular weight of lime (74.1), wC the molecular weight of chalk (100) and f a constant which converts the measured units of alkalinity to mEq/L. It depends on the strength of the acid and the volume in which the powder is dissolved and the weight of the powder dissolved. If the strength of the acid is 0.1 N, the weight 100 mg and the volume 100 mL then f = 1 and M and P are in units of mL acid.
Then the ratio of lime to chalk in the powder is
rLC = wL(P-M/2)f/wC(M-P)f = (wL/wC)(P-M/2)/(M-P)
and the percentage Ca(OH)2 in the powder is then
% = 100r/(1+r)
Do keep in mind that P and M can be measured in any units such as drops of lactic acid. Diluted hydrochloric acid from the hardware store can also be used (be careful with it!). Vinegar is not a terribly good choice as it is not quite acidic as we would like but will do in a pinch. Adjust the amounts of powder tested, water volume and acid dilution so that M and P are reasonably large numbers. M = 2 and P = 1 drops would not be good alkalinities. Test 10 times more solution or dilute the acid 10 times. Also keep in mind that you do not need to know the weight of powder, volume of water or strength of the acid. Just the drop counts to the M and P pH's.
Going back to the case where there is excess powder for a minute: You can also take the approach that while the undissolved powder may contain some undissolved lime that the solution over it is saturated, look up the strength of saturated solutions of lime at various temperatures and assume that this is the concentration of Ca(OH)2 in the liquid.