Question

Determine the concentrations of Na2CO3, Na , and CO32– in a solution prepared by dissolving 1.45 × 10–4 g Na2CO3 in 2.00 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. [Na2CO3] = __________ M [Na+] =___________ M = _________ ppt [CO3^2-]=___________M = ___________ppt

Answer 1

Solution :-

Lets first calculate the moles of the Na2CO3

Moles of Na2CO3 = 1.45*10^-4 g / 105.988 g per mol = 1.368*10^-6 mol

Molarity of the Na2CO3 = 1.368*10^-6 mol / 2.00 L = 6.84*10^-7 M

Mole ratio of the Na2CO3 to Na+ is 1 : 2

So

Molarity of Na+ = 6.84*10^-7 M * 2 = 1.37*10^-6 M

Mole ratio of the Na2CO3 to CO3^2- = 1 : 1

So the concentration of CO3^2- = 6.84*10^-7 M

Now lets calculate the concetration in paprts per thousand

Parts per thousand = 1 part per thousand

So

Concentration of the Na+ =( (1.37*10^-6 mol*2) * 23 g )(1 ppt / 2 L ) = 3.10*10^-5 ppt

Concentration of CO3^2- = (1.37*10^-6 mol * 60 g / 1 mol )( 1ppt / 2.00 L) = 4.11*10^-5 ppt