Prepare 500 ml acetic acid solution with a pH value equal to 2.7

General information:
Acetic acid is a monoprotic weak acid. It partially ionizes (dissociates) in solution to produce hydronium and acetate ions. It has a molar mass (molecular weight) equal to 60.05 g/ mole. Ka value for acetic acid at 25 ºC is equal to 1.75 × 10^-5.

Molar concentration of acetic acid solution required to produce certain pH value in solution can be calculated from the following equations:

[H₃O⁺] = [H⁺]
C = ((2 × [H⁺] + Ka)² - Ka² ) ÷ (4 × Ka)

Where C is molar concentration of the monoprotic acid, [H⁺] is molar concentration of hydrogen (hydronium) ion, and Ka is the value of the acid ionization constant.


Calculations:

pH = - log [H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-pH = 10^-2.7 = 0.0020 mole/ liter

C = ((2 × 0.0020 + (1.75 × 10^-5))² - (1.75 × 10^-5)² ) ÷ (4 × (1.75 × 10^-5))

C = ((0.004)² - (3.06 × 10^-10 )) ÷ (0.00007)

C = ((1.6 × 10^-5) - (3.06 × 10^-10 )) ÷ (0.00007)

C= (1.6 × 10^-5) ÷ (0.00007)

C = 0.229 mole/ liter

Use online pH calculator to check that 0.229 mole/ liter acetic acid produces solution pH = 2.7 

Mass of acetic acid needed to make 500 ml solution with a concentration of 0.229 mole/ liter is calculated as the following:

Volume of solution in liters = 500 ml × (0.001 liter/ ml) = 0.5 liter
Number of moles = concentration × volume = (0.229 mole/ liter) × 0.5 liter = 0.115 mole

Mass of acetic acid = number of moles × molar mass =0.115 mole × (60.05 g/ mole) = 6.91 g

To make the preparation, dissolve 6.91 g acetic acid in freshly distilled water to make 500 ml solution (using a volumetric flask).

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