What is the millimolar concentration of sulfate ion?

Question

A sample contains an average of 3.98 x 10^3 ppm sodium ion and 271 ppm sulfate ion. What is the millimolar concentration of sulfate ion in this sample if the average density of the solution in the sample is 1.02 g/mL?

Answer 1

271 ppm sulfate ion can be considered to be 271 g sulfate in 10⁶ g of solution or 2.71 x 10^-4 g sulfate for each gram of the solution.

As the solution density is 1.02 g/mL, the volume of one gram will be 1/1.02 = 0.980 mL/gm.

The weight-volume concentration of sulfate is then 2.71 x 10^-4 g SO4 / g solution / 0.980 mL/g solution = 2.77 x 10^-4 g SO4 / mL or 0.277 mg /mL.

The molecular mass expressed in mg is 96.1mg / millimole, so the millimolar concentration is

0.277 mg/mL / 96.1 mg / millimole x 1000 mL/L = 2.88 millimoles SO4^2- / Liter. I indicated which units cancel by a strike.

Always check math.