Recently I've had several requests for info on how to calculate the amounts of acid of various types needed to acidify water to a given pH. This post (in 2 parts) gives formulas which will allow the amount of acid required to be calculated. They are as general as I can make them and apply equally well therefore, to strong monoprotic acids like hydrochloric, and weak, polyprotic acids like citric. It is the latter case which leads to a large part of the complexity. If you put these formulas in a spreadsheet (or program them in FORTRAN or even C++) and play around with them assuming the water pH is less than 8.3 and use strong acids you'll see that there are simplifications which can be made.

The amount of acid is assumed to be that required to move a carbonic/bicarbonate/carbonate system from the original pH of the water (symbolized here by pHo) to a target pH (symbolized simply by pH) plus the amount required to establish the lower pH. Note that there are several factors which make it unlikely that the amount of acid you calculate will bring your water sample to exactly the pH for which you made the calculation. Among these are:

- The water alkalinity is not as reported (e.g. there has been a seasonal variation due to heavy snow melt in your area).
- Part of the alkalinity is due to systems other than the carbonic/bicarbonate/carbonate system (such as nitrate and/or phosphate).
- The acid you are using is not exactly at its labeled strength.
- Depending on your equipment you may not be able to measure the weight or volume of the acid terribly accurately.
- The quantities of acid calculated depend upon the pK's of the acids which vary with temperature.

To use these formulas you must know the alkalinity of the water as the alkalinity is the major factor in determining how much acid is required unless, of course, the alkalinity is very low in which case the acid required is just that necessary to supply the H+ ions which establish the pH.

The formulas follow. Example numerical values are given for the acidification of a water sample with an alkalinity of 100 ppm as CaCO3 and a pH of 8.3 to pH 5 using citric or lactic acid.

Step 1. Compute the mole fractions of carbonic (f1o), bicarbonate (f2o) and carbonate(f3o) at the water sample's pH (example: pHo = 8.3)

pHo = 8.3 r1o = 10^(pHo - 6.38) = 83.17638 r2o = 10^(pHo - 10.33) = 0.009332 do = 1 + r1o + r1o*r2o = 84.95262 f1o = 1/do = 0.011771 f2o = r1o/do = 0.97909 f3o = r1o*r2o/do = 0.009137Step 2. Compute the mole fractions at pHb = 4.3 (the pH which defines alkalinity).

pHb = 4.3 r1b = 10^(pHb - 6.38) = 0.0083176 r2b = 10^(pHb - 10.33) = 0.0000009 db = 1 + r1b + r1b*r2b = 1.0083176 f1b = 1/db = 0.9917510 f2b = r1b/db = 0.0082490 f3b = r1b*r2b/db = 0.0000000077Step 3. Convert the sample alkalinity to milliequivalents/L (example: alkalinity = 100 ppm as CaCO3)

alk = (alkalinity in ppm as CaCO3)/50 = 2.00Step 4. Solve

Alk = Ct*(Change in Carbonic + change in carbonate) for Ct Ct = alk/( (f1b - f1o) + (f3o - f3b) ) = 2.0220Note that this will always be very close to alk (in mEq/L) as long as the pH of the sample is less than or equal 8.3

Step 5. Compute mole fractions at desired pH (example: pH 5)

pH = 5 r1c = 10^(pH - 6.38) = 0.04168 r2c = 10^(pH - 10.33) = 0.00000467 dc = 1 + r1c + r1c*r2c = 1.04168 f1c =1/dc = 0.95998 f2c = r1c/dc = 0.04001 f3c = r1c*r2c/dc = 0.00000018Step 6. Use these to compute the milliequivalents acid required per liter (mEq/L)

Acid required = Ct*(Change in Carbonic + change in carbonate) E = Ct*( ( f1c - f1o) + (f3o - f3c) ) + 10^(-pH) - 10^(-pHo) = 1.93576 + .01Note that this is also very close to the alkalinity (expressed as mEq/L). The alkalinity is the amount of acid required to get to pH 4.3. 'E' is the amount of acid required to get to a somewhat more basic pH. The last two terms (.01) give the acid which would be required if no carbonate or bicarbonate were being "neutralized". This is the amount of acid that would be required if distilled water were being acidified.

Step 7. If the acid is labeled in terms of its normality (i.e. 1 N, 0.1N) recognize that a milliter contains the same number of mEq as the normality of the acid e.g. 1 N acid contains 1 mEq/mL, 0.1N contains 0.1 mEq/L. Of the acids typically used only hydrochloric and sulfuric are likely to be labeled in this way. Divide 'E' by the number of mEq/mL to get the number of mL of acid to add to each liter of the water. Thus if 8.75 N acid (approximate strength of hardware store hydrochloric acid) were being used with the example water 1.94/8.75 = 0.216 mL would be required for each liter being acidified.

Step 8. If the acid is not labeled by its normality then you must compute the number of millimoles (mM) required to give the needed number of mEq and then convert that to a weight or volume. This is not necessary if the acid is labeled in terms of its molarity (e.g. 2 M, 0.1M) in which case each milliliter contains the same number of mM as the strength. One mL of 1M acid contains 1 mM. Start by computing the number of mEq of H+ obtained from 1 mM of the acid at the target pH. To do this you will need all the pK's of the acid being used. The following table gives values you can plug into the formulas which follow (you will need the molecular weights later):

Acid pK1 pK2 pK3 Mol. Wt Acetic 4.75 20 20 60.05 Citric 3.14 4.77 6.39 192.13 Hydrochloric -10. 20 20 36.46 Lactic 3.08 20 20 90.08 Phosphoric 2.12 7.20 12.44 98.00 Sulfuric -10. 1.92 20 98.07 Tartaric 2.98 4.34 20 150.09I hope the chemists will appreciate that I know that hydrochloric acid, for example, only has one hydogen ion to give and that the pK for this ion probably isn't - 10. By using -10 for the pK I insures that the math will calculate 1 millimole of H+ from each millimole of HCl whatever the (reasonable) target pH. Similarly the use of +20 for the second and third pKs will result in calculation of insignificant additional amounts of hydrogen ions from the second and third nonexistant dissociations. This artifice allows the same formulas to be used for any of the acids we are likely to encounter.

The "fraction" (in quotes because it may be a number biggert than one) of moles of acid which release a hydrogen ion are found from the following formulas.The pK's for the example numbers are for citric acid:

pK1 = 3.14 pK2 = 4.77 pK3 = 6.39 pH = 5 r1d = 10^(pH - pK1) = 72.4436 r2d = 10^(pH - pK2) = 1.6982 r3d = 10^(pH - pK3) = 0.0407 dd = 1/(1 + r1d + r1d*r2d + r1d*r2d*r3d) = 0.004963 f1d = dd = 0.004963 f2d = r1d*dd = 0.359553 f3d = r1d*r2d*dd = 0.6106087 f4d = r1d*r2d*r3d*dd = 0.0248750 frac = f2d + 2*f3d + 3*f4d = 1.655395It's not necessary to go through all this for hydrochloric acid which will give frac = 1 for any pH or sulfuric which will give frac ~ 2 unless the pH is below about 4 (e.g. if the water is being acidified for yeast washing).

Step 9. Now divide the mEq required by the "fraction". This is the required number of moles of acid. For the citric example:

mM required = E/frac = 1.946/1.655 = 1.176 mM For lactic acid frac = 0.9881 and mM required = E/frac = 1.946/.9881 = 1.969 mMStep 10. Multiply by molecular weight of the acid (192.13 mg/mM for citric)

mg required = mM required*Mol.wt. (mg/mM) = 1.176*192.13 = 225.94 mg For lactic acid mg required = mM required*Mol.wt. (mg/mM) = 1.969*90.08 = 177.37 mgThis is the weight of acid required to treat each liter of water. If the acid is a solid, like citric, it can be now be weighed out. If it is a liquid, like lactic, it will be necessary to determine the volume of liquid which gives the desired weight of acid (see Step 10).

Step 10. Liquids are usually labeled according to the percentage of their weight which is the acid, for example, 88% lactic acid, 25% phosphoric acid and 28% hydrochloric acid are typical labelings. In order to calculate the volume of liquid which contains a given weight it is necessary to know the specific gravity of the liquid. In some cases this is specified on the label (for example 28% HCl is labeled 18 Baume which converts to about 1.142 specific gravity or 1142 mg/mL). In other cases you will have to determine the specific gravity by the use of tables in the CRC handbook (sulfuric) or weigh a small known quantity of the acid. 88% lactic acid, for example, weighs about 1214 mg/mL (and thus has a density of about 1.214 mg/mL). 25% phosphoric acid weighs about 1170 mg/L (specific gravity 1.170). If unable to obtain a specific gravity value you can use 1000 mg/L. The three examples just given indicate that you would incur errors of 14 - 21% by doing that. This may seem like a lot of error but it really isn't especially if you are going to add measured acid gradually until the target pH is reached.

The mg of acid per mL is the product of the mg/mL weight of the acid times the percent acid expressed as a fraction. Thus for 28% HCl it is 0.28*1140=319.8 mg/mL, for 88% lactic acid, .88*1214 = 1068 mg/mL and for 25% phosphoric acid,.25*1170 = 292.5 mg/mL. The final step is to divide the required mg of acid by the mg/L for the particular strength acid. Using 88% lactic acid to bring a liter of the example water to pH 5 would, thus, require 177.37mg/1068mg/mL = 0.166mL of the 88% solution.

With all that there are bound to be some mistakes but, as always, I hope no serious ones.

*A.J. deLange
ajdel@mindspring.com
*