Question

Determine the concentrations of Na2CO3, Na , and CO32– in a solution prepared by dissolving 1.82 × 10–4 g Na2CO3 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 CO32–. Ignore any reactions with water.

Answer 1

Given,

Mass of Na₂CO₃ = 1.82 x 10^-4 g

The volume of water = 1.75 L

Firstly calculating the number of moles of Na₂CO₃,

= 1.82 x 10^-4 g x ( 1 mol / 105.99 g)

= 1.72 x 10^-6 mol of Na₂CO₃

We know, the formula for molarity,

Molarity = Moles of solute / L of soluion

Molarity = 1.72 x 10^-6 mol of Na₂CO₃ / 1.75 L

Molarity of Na₂CO₃ = 9.8 x 10^-7 M

Now, Calculating Molarity for Na⁺,

=  1.72 x 10^-6 mol of Na₂CO₃ x ( 2 mol of Na⁺ / 1 mol of Na₂CO₃) x (1 / 1.75 L)

Molarity of Na⁺ = 1.96 x 10^-6 M

Now, Calculating the molarity of CO₃^2-,

= 1.72 x 10^-6 mol of Na₂CO₃ x ( 1 mol of CO₃^2- / 1 mol of Na₂CO₃) x (1 / 1.75 L)

Molarity of CO₃^2- = 9.8 x 10^-7 M

Now, Calculating the concentration of ionic species in parts per thousnd(ppt),

PPT = ( Mass of ion / Mass of solution) x 1000

Mass of solution = Mass of solute + Mass of solvent

Mass of solution =  1.82 x 10^-4 g + [1.75 L x ( 1000 mL /1 L) x ( 1 g / 1mL)]

Mass of solution =  1.82 x 10^-4 g + 1750 g

Mass of solution = 1750.2 g

calculating the concentraton in ppt for Na⁺,

= 1.72 x 10^-6 mol Na₂CO₃ x ( 2 mol Na⁺ / 1 mol Na₂CO₃) x (22.99 g of Na / 1 mol) x ( 1 / 1750.2 g of soluion) x 1000

= 4.51 x 10^-5 ppt of Na⁺

Similarly, calculating the concentration of CO₃^2-,

= 1.72 x 10^-6 mol Na₂CO₃ x( 1 mol CO₃^2- / 1 mol Na₂CO₃) x (59.98 g CO₃^2-/ 1 mol) x ( 1 / 1750.2 g soluion) x 1000

= 5.88 x 10^-5 ppt of CO₃^2-