Alkalinity is the chemical opposite of acidity. Whereas *acidity* shows up as a low pH reading and represents the capacity of a substance to donate a proton, or hydrogen ion (H+), *alkalinity* shows up as a high pH and signifies the capacity of a substance to accept a proton.

A variety of methods to calculate alkalinity exist. The one used here makes use of the dissociation of carbonic acid, H_{2}CO_{3}, and the equation:

[Alk.] = + 2[CO^{2-}_{3}] + [OH-] - [H+],

where the constituent ions are biocarbonate, carbonate, hydroxide and hydrogen respectively.

In such a problem, you get the concentrations of the ions in g/m^{3}.

## Step 1: Convert g/m3 to eq/m3

In this step, divide the raw concentrations of bicarbonate, carbonate, and hydroxide by their EW values, which is derived from their molecular masses. This yields the concentrations of these ions in eq/m^{3}. These values are 61, 30 and 17 respectively. For example, given:

[HCO_{3}-] = 488 g/m^{3}, [CO^{2-}_{3}] = 20 g/m^{3}, and [OH-] = 0.17 g/m^{3},

divide by 61, 30 and 17 to get

8, 0.67, and 0.01 eq/m^{3}.

## Step 2: Find [H+]

This step requires knowing that [OH-][H+] = Kw = a constant equal to 10^{-14}. You must also first divide the calculated value of [OH-] from Step 1 by 1,000 to convert the concentration to units appropriate for this step. In this case, 0.01 ÷ 1,000 = 10^{-5}.

Thus [H+] = 10^{-14} ÷ 10^{-5} = 10^{-9}.

## Step 3: Multiply [H+] by 1,000

This returns the units to eq/m^{3}.

10^{-9} × 1,000 = 10^{-6}.

## Step 4: Solve for Alkalinity

[Alk.]= 8 + 0.67 + 0.01 - 10-6 = 8.68 eq/L

## Bonus Step

To find the alkalinity in terms of mg/L of calcium carbonate, a commonly used measure of alkalinity, multiply by 50,000:

8.68 eq/L × 50,000 mg/eq = 434 mg/L as CaCO_{3}