Question

A solid sample formed by barium chloride and potassium nitrate, weighing 600 mg is subjected to reaction with sulfuric acid solution in excess during the addition of the acid, we observed the formation of a precipitate, filtered, the precipitate, which after washed and dried weighed 0.050 g. The solution resulting from filtration was added a few drops of dilute H2SO4 and there was no precipitate formation. The resulting solution was transferred to a 250 mL volumetric flask and completed to volume with distilled water until measurement.
Check the alternative that indicates the approximate value of the concentration of potassium ions (in ppm) in the resulting solution. (Explain your answer)
Data: molar mass in g / mol: Ba = 137; Cℓ = 35.5; S = 32; M = 1; O = 16, K = 39; N = 14
a) 222 ppm
b) 345 ppm
c) 309 ppm
d) 277 ppm
e) 858 ppm

Answer 1

The precipitate that is most likely to be formed is:

BaCl₂ + H₂SO₄ -> BaSO₄ + 2HCl

0.050 g of BaSO4 are produced, so we can get moles produced:

0.050 g * (1mol/233.43g) = 2.142 x 10^-4 moles

So we can say that that amount of moles of BaCl2 were consumed, having a 1 on 1 stoichiometric relation. We then get the grams:

2.142 x 10^-4 moles * (208.23 g / mol) = 0.0446 grams

As no more precipitate was formed, we can say that that was all barium chloride present, which we'll now subtract from initial mass:

0.6 grams - 0.0446 grams = 0.5554 grams of potassium nitrate

We get the moles:

0.5554 grams * (1mol / 101.1032g) = 0.00549 moles of potassium nitrate

Hence, we have 0.00549 moles of potassium ions, which will weigh:

0.00549 moles * (39 g/mol) = 0.2142 grams = 214.2 mg

Concentration in ppm:

214.2 mg / 0.25 L = 857 ppm, letter e)