Question

(alkalinity) You added 10.6 g of Na2CO3 and 12 g of acetic acid (CH3COOH) to 1 L water. Write all species that are present in water assuming complete dissociation. Calculate the concentration of each species in moles/L. What is the change in alkalinity of the solution? (hints: use alkalinity equation).

Answer 1

The reaction equation is

CH₃COOH + Na₂CO₃ CH3COONa + CO₂ + H₂O

The balanced equation is expressed as

2 CH₃COOH + Na₂CO₃ 2 CH3COONa + CO₂ + H₂O

So, after complete dissociation CH3COONa will only present in water.

Now,

10.6 g of Na2CO3

Molar mass of Na2CO3 = 106 g/mol

So, 106 g of Na2CO3 = 1 mol

1 g of Na2CO3 = (1/106) mol

10.6 g of Na2CO3 = (10.6/106) mol = 0.1 mol

12 g of acetic acid (CH3COOH)

Molar mass of CH3COOH = 60 g/mol

So, 60 g of CH3COOH = 1 mol

1 g of CH3COOH = (1/60) mol

12 g of CH3COOH = (12/60) mol = 0.2 mol

Again,

2 CH₃COOH + Na₂CO₃ 2 CH3COONa + CO₂ + H₂O

In the above reaction equation

1 mole of Na₂CO₃ reacts with 2 moles of CH₃COOH to produce 2 moles of CH3COONa.

So, 0.1 mole of Na₂CO₃ reacts with 0.2 moles of CH₃COOH to produce 0.2 moles of CH3COONa.

Now,

0.2 moles of CH3COONa is in 1 L water.

So, molarity of CH3COONa = 0.2 mol / 1 L = 0.2 M.

CH3COONa + H2O CH3COOH + OH^-

Kb = [CH3COOH] [OH^-] / [CH3COONa]

5.56 x 10^-10 = x² / 0.2

x² = 1.11 x 10^-10

x = 1.11 x 10^-5

So, [OH^-] = 1.11 x 10^-5 M

pOH = - log [OH^-] = - log(1.11 x 10^-5) = 4.95

pH 14 - 4.95 = 9.05

Hence, after complete dissociation the pH of the water will increase by

9.05 - 7 = 2.05