The Pearson Square is a useful tool for determining the alcohol (% alc/vol), total acidity (TA), or residual sugar (RS) concentrations in a wine made from a blend of two or more wines. These concentrations all exhibit linear relationships in a blend, i.e. each concentration can be calculated as a ratio of the volumes and concentrations of the wines used in the blend.

But all too often I see winemakers using the Pearson Square to calculate the pH of a blended wine. The pH of wine is a logarithmic function and so the Pearson Square cannot be used as it would imply that the logarithm of a sum is equal to the sum of the logarithms—this is mathematically incorrect.

Let’s take a closer look at what this all means and derive a formula to calculate the pH of a blended wine.

The following illustrates how the Pearson Square is used to calculate the concentration of a parameter in a blended wine.

where

A = concentration of the wine to be used
B = concentration of the wine to be “corrected”
C = calculated or desired concentration
D = number of parts of wine to be used and is equal to C-B
E = number of parts of wine to be “corrected” and is equal to A-C

The Pearson Square can be stated as a mathematical equation as:

Now, we know that pH is related to the concentration of hydronium ions, [H3O+], and so we can rewrite the above equation as:

And knowing that pH = –log[H3O+], we can rewrite the above equation as:

This equation can then be rewritten as a function of the pH values of wines A and B, or pHA and pHB, as follows:

For example, if we have equal volumes of two wines, one with a pH of 3.45 and the second with a pH of 3.70, then,

The astute mathematician/winemaker will note that the Pearson Square would yield a pH of 3.58 for the blended wine, which is, for all intents and purposes, close enough to 3.56. But as the difference in pH values and wine proportions increase, the error in Pearson Square results increases exponentially.

Now, if you need to determine the volume of a wine to be added to another wine to achieve a desired pH, the above equation can be reworked to the following:

For example, if a 20-L volume of wine with a pH of 3.70 needs to be adjusted to 3.60 using a wine with a pH of 3.30, then we would need:

If we were to use the Pearson Square, the result would incorrectly suggest that you would need 1 part of the wine with a pH of 3.30 for every 3 parts of the wine with a pH of 3.70, or D = 20/3 = 6.7 L – quite the difference!!

Daniel Pambianchi