Question

Determine the concentrations of Na2CO3, Na , and CO32– in a solution prepared by dissolving 2.61 × 10–4 g Na2CO3 in 2.25 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of CO32–. Ignore any reactions with water.

Answer 1

Answer – We are given, mass of Na₂CO₃ = 2.61*10^-4 g

Volume = 2.25 L

First we need to calculate the moles of Na₂CO₃

We know,

Moles of Na₂CO₃ = 2.61*10^-4 g / 105.98 g.mol^-1

                              = 2.46*10^-6 moles

So, 1 moles of Na₂CO₃ =1 moles of CO₃^2-

So, 2.46*10^-6 moles of Na₂CO₃ = ? moles of CO₃^2-

= 2.46*10^-6 moles of CO₃^2-

1 moles of Na₂CO₃ = 2 moles of Na⁺

So, 2.46*10^-6 moles of Na₂CO₃ = ?

= 4.96*10^-6 moles of Na⁺

So, molarity of Na₂CO₃ = 2.46*10^-6 moles / 2.25 L

                                      = 1.09*10^-6 M

Molarity of Na⁺ = 4.96*10^-6 moles / 2.25 L

                          = 2.19*10^-6 M

molarity of CO₃^2- = 2.46*10^-6 moles / 2.25 L

                               = 1.09*10^-6 M

We know, 1 ppm = 1 mg/ L

So, mass of Na⁺ = 4.96*10^-6 moles * 22.990 g/mol

                           = 0.000113 g

                           = 0.113 mg

So, pppm of Na⁺ = 0.113 mg / 2.25 L

                            = 0.0503 ppm

Mass of CO₃^2- = 2.46*10^-6 moles * 60.08 g/mol

              = 0.000148 g

                = 0.148 mg

So, ppm of CO₃^2- = 0.148 mg / 2.25 L

                            = 0.0658 ppm