Saturday, November 9, 2013

Calculate mass of acetic acid needed to prepare 100 ml solution with pH equal to 3.5

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

Rearranging the above equation gives the following:

C = ((2 × [H+] + Ka)2 - Ka÷ (× 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 10-3.5 = 0.000316 mole/ liter

C = ((2 × 0.000316 + (1.75 × 10-5))2 - (1.75 × 10-5)÷ (× (1.75 × 10-5))
C = ((0.00065)2 - (3.0625 × 10-10 )) ÷ (0.00007)
C = ((4.225 × 10-7) - (3.0625 × 10-10 )) ÷ (0.00007)
C= (4.222 × 10-7÷ (0.00007)
C = 0.006 mole/ liter


Mass of acetic acid needed to make 100 ml solution with a concentration of 0.006 mole/ liter is calculated as the following:

Volume of solution in liters = 100 ml × (0.001 liter/ ml) = 0.1 liter
Number of moles = concentration × volume = (0.006 mole/ liter) × 0.1 liter = 0.0006 mole

Mass of acetic acid = number of moles × molar mass =0.0006 mole × (60.05 g/ mole) = 0.036 g

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Calculate pH of 100 ml solution containing 0.05 g acetic acid and 0.06 g sodium acetate.
Calculate pH of 1 liter solution containing 0.5 g benzoic acid.



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