Question

In order to have some sense about how carbon dioxide affects concentration of carbonate ion in the water, let us assume pure water instead of seawater is used as the solvent for simplification.

The composition of the total salt components of the sea water shows 0.42 g Ca²⁺/1000 kg seawater. Given the density of sea water 1.02 g/mL, calculate the molar concentration of Ca²⁺.

I got 1.07x10^-5 M Ca²⁺ when I solved it but it doesn't seem correct.

Answer 1

Consider 1000 kg of Sea water

Mass of Calcium ions in it = 0.42 g

Molar Mass of Calcium = 40.078 g/mol

Moles of Calcium = Mass/Molar Mass = (0.42 g) / (40.078 g/mol) = 0.010479

Density of water = 1.02 g/mL = (1.02 g/mL) * (1000 ml/L) * (0.001 kg/g) =

Converting to kg/L, we have,

1.02 g/mL = (1.02 g/mL) * (1000 ml/L) * (0.001 kg/g) = 1.02 kg/L

Volume of water = Mass / Density = (1000 kg) / (1.02 kg/L) = 980.392 L

Molarity = Moles of Calcium / Volume of water = (0.010479) / (980.392 L) = 1.07 x 10^-5 M

Thus, your answer is correct.