Question

Determine the concentrations of Na_2CO_3, Na^+ and CO_3^2- in a solution prepared by dissolving 2.26 times 10^-4 g Na_2CO_3 in 1.75 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per thousand (ppt). Note: Determine the formal concentration of CO_3^2-. Ignore any reactions with water. [Na_2 CO_3] = Number M [Na^+] = Number M = Number ppt [CO_3^2-] = Number M = Number ppt

Answer 1

Ans. Balanced reaction: Na₂CO_3(s) ------> Na⁺_(aq) + CO₃^2-_(aq)

Stoichiometry: 1 mol Na₂CO₃, when dissolved in water, produces 1 mol each of Na⁺ and CO₃^2-.

Given,

            Mass of Na₂CO₃ = 2.26 x 10^-4 g

            Volume of solvent = 1.75 L

It’s assume that after adding the solute to water, volume of solution = volume of solvent.

Now,

            Moles of Na₂CO₃ = Mas/ Molar mass=

= (2.26 x 10^-4 g) / (105.988 g mol^-1)

= 2.13 x 10^-6 mol

Molarity of Na₂CO₃ in solution = Mole of Na₂CO₃ / Volume (in L) of solution

                                                            = (2.13 x 10^-6 mol) / 1.75 L

                                                            = 1.218 X 10^-6 mol/L

                                                            = 1.218 X 10^-6 M

Therefore, molarity in terms of Na₂CO₃ = 1.218 X 10^-6 M

#1. Na⁺

Since 1 mol Na₂CO₃ fives 1 mol Na⁺, hence

            Molarity of Na⁺ = 1.218 X 10^-6 M

Mass of Na⁺ ion is solution = Moles of Na⁺ x Atomic mass of Na

                                                = 2.13 x 10^-6 mol x (23.0 g mol^-1)

                                                = 2.80 x 10^-5 g

PPT of Na⁺ = 1 part Na⁺ per 1000-part solution.

                        Or, g of Na⁺ / per kg solution.                    ; [1 kg = 1000 parts x 1g ]

It’s assumed that the solution to have the density of water (1.0 g /mL). So 1.218 X 10^-6 M solution has 1.218 x 10^-6 mol Na⁺ = 2.80 x 10^-5 g Na⁺ in 1.0 L= 1.0 kg solution.

So,       PPT, Na⁺ = 2.80 x 10^-5 g Na⁺ / kg solution

                                    = 2.80 x 10^-5 PPT

#2. CO₃^2-

            Since 1 mol Na₂CO₃ fives 1 mol CO₃^2- hence

            Molarity of CO₃^2- = 1.218 X 10^-6 M

Mass of CO₃^2- ion is solution = Moles of CO₃^2- x Molar mass of CO₃^2-

                                                = 2.13 x 10^-6 mol x (60 g mol^-1)

                                                = 1.278 x 10^-4 g

PPT of CO₃^2-= g of CO₃^2- / kg of solution

                        = 1.278 x 10^-4 g / 1 kg

                        = 1.278 x 10^-4