Question

You have an aqueous solution made so that the formal concentration of the soluble salt Na2SO3 is 0.15 M. Ka1 and Ka2 for H2SO3 are 1.2x10^-2 and 6.6x10^-8, respectively.

a.Write chemical equations showing all of the equilibria that occur in the solution.

b.Write the charge balance equation for the system.

c.Write a mass balance equation that relates the sulfurous and sodium species.

Answer 1

a. Write chemical equations showing all of the equilibria that occur in the solution.

Water dissociates into H⁺ ion and OH^- ion.

H₂O (l) -----> H⁺ + OH^-

The sulfite anion will act as a base and react with water to form the bisulfate ion, or

SO₃^2-(aq) +H₂O ----->HSO₃^1-(aq) + OH^-(aq)

The bisulfate ion can also act as a base and react with water to form sulfurous acid,

HSO₃^1-(aq) + H₂O(l) ----> H₂SO₃(aq) +OH^-

Sulfurous acid is doesn't exist in aqueous solution. It dissociates into sulfur dioxide,

When the bisulfate ion reacts with water, it'll form

HSO₃^1-(aq) + H₂O(l) -----> SO₂(aq) + H₂O(l) +OH^-

b. Write the charge balance equation for the system.

The concentrations of all the (+)’ly charged ions = the concentrations of all the (-)’ly charged ions.

H₂O (l) -----> H⁺ + OH^-

[H⁺] +[Na⁺] --> 2[SO₃^2-]+ [OH^-]

c. Write a mass balance equation

Sodium sulfate will dissociate completely in aqueous solution to give sodium cations and sulfite anions

Na₂SO₃ -------> 2Na⁺(aq) + SO₃^2-(aq)

1 mol of sodium sulfite will produce 2 moles of sodium cations and 1 mole of sulfite anions.

[Na₂SO₃] = 0.15 M (given)

[Na⁺]=2*[Na₂SO₃] = 0.15 x 2 =0.30M

[SO₃^2-]= [Na₂SO₃] = 0.15 M

Mass balance between Na⁺ and SO₃^2-

[Na⁺] = 2[SO₃^2-]