# How to Calculate Alkalinity From pH

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Alkalinity is an important water characteristic that quantifies the capacity of water to accept hydrogen ions H+. Concentration of H+ expressed as "pH" determines the acidity of a solution. Alkalinity of natural water is mainly due to the presence of 2 forms of the carbonate ions that act as a buffer system. By a definition, alkalinity is a sum of ion concentrations (represented as brackets): Alkalinity = [HCO3-] + 2 x [CO3(-2)] + [OH-] - [H+]. As an example, calculate alkalinity at pH 10 if the total molar concentration of carbonate ions in water is 0.1 M.

Skill level:
Easy

## Instructions

1. 1

Consider 2 equilibriums for carbonate ions in water. CO2 + H2O = HCO3(-)+ H+; K1 = 4.45E-7 (notation "E-7" means "10 in power -7"). HCO3(-) = CO3(-2) + H+; K2 = 4.69E-11. Note: K1 and K2 are equilibrium constants (see Reference) of the corresponding reactions that will be used in Step 3.

2. 2

Calculate concentration of H+ and OH- ions. [H+] = 1E(-pH) [OH-] =1E-14/[H+] In the example at pH=10; [H+]=1E-10, and [OH-]=(1E-14)/(1E-10)=1E-4=0.0001.

3. 3

Calculate fractions of carbonate ions that depends on pH: Fraction (HCO3(-)) = (K1 x [H+]) / (([H+] x [H+]) + (K1 x [H+]) + (K1 x K2)) Fraction (CO3(-2)) = (K1 x K2) / (([H+] x [H+]) + (K1 x [H+]) + (K1 x K2)) Note: K1, K2 are the constants defined in Step 1. In the example: fraction (HCO3(-))=(4.45E-7 x 1E-10)/((1E-10 x 1E-10) + 4.45E-7 x 1E-10 +(4.45E-7 x 4.69E-11)) = 4.45E-17/(1E-20 + 4.45E-17 + 2.087E-17)=4.45E-17/6.537E-17=0.681 fraction (CO3(-2))=(4.45E-7 x 4.69E-11)/((1E-10 x 1E-10) + 4.45E-7 x 1E-10 +(4.45E-7 x 4.69E-11)) = 2.087E-17/(1E-20 + 4.45E-17 + 2.087E-17)= 2.087E-17/6.537E-17 = 0.319.

4. 4

Calculate concentration of carbonate ions [concentration] = [total concentration carbonates] x fraction. In the example, total concentration of carbonates is 0.1 M. [HCO3-] = 0.681 x 0.1 M=0.0681 M [CO3(-2)] = 0.319 x 0.1 M =0.0319 M

5. 5

Calculate Alkalinity Alkalinity = [HCO3-] + 2 x [CO3(-2)] + [OH-]. Note: The concentration of H+ is low (1E-10) and can be omitted from the equation. In the example: Alkalinity = 0.0681 + 2 x 0.0319 + 0.0001=0.132 (eq/L). CO3(-2) ion can accept 2 hydrogen ions or 2 equivalents. This is why, the CO3(-2) concentration is multiplied by 2 and units of alkalinity are equivalents per liter (eq/L).

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