Question

Determine the concentrators of Na_2CO_3 Na^+. and a solution prepared by of water Express alt three concentrations molanty express the concentrations of the ionic species in parts per million (ppm) Note Determine the formal concentration of CO_1^2+ Ignore any reactions with water
General chemistry 2. Help please.

Answer 1

Answer – We are given, mass of Na₂CO₃ = 2.06*10^-4 g

Volume = 2.25 L

First we need to calculate the moles of Na₂CO₃

We know,

Moles of Na₂CO₃ = 2.06*10^-4 g / 105.98 g.mol^-1

                              = 1.94*10^-6 moles

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

So, 1.94*10^-6 moles of Na₂CO₃ = ? moles of CO₃^2-

= 1.94*10^-6 moles of CO₃^2-

1 moles of Na₂CO₃ = 2 moles of Na⁺

So, 1.94*10^-6 moles of Na₂CO₃ = ?

= 3.88*10^-6 moles of Na⁺

So, molarity of Na₂CO₃ = 1.94*10^-6 moles / 2.25 L

                                      = 8.64*10^-7 M

Molarity of Na⁺ = 3.88*10^-6 moles / 2.25 L

                          = 1.73*10^-6 M

molarity of CO₃^2- = 1.94*10^-6 moles / 2.25 L

                           = 8.64*10^-7 M

We know, 1 ppm = 1 mg/ L

So, mass of Na⁺ = 3.88*10^-6 moles * 22.990 g/mol

                           = 8.93*10^-5 g

                           = 0.0893 mg

So, pppm of Na⁺ = 0.0893 mg / 2.25 L

                            = 0.0397 ppm

Mass of CO₃^2- = 1.94*10^-6 moles * 60.08 g/mol

              = 0.000117 g

                = 0.117 mg

So, ppm of CO₃^2- = 0.117 mg / 2.25 L

                            = 0.0519 ppm