Question

A 1.546g sample containing a mixture of only k2So4 and kno3 was dissolved in water an treated with BaCl2, precipitating the So42- as Baso4(pksp=9.96) the resulting precipitate was isolated and yielded 863.5mg of baso4. What is % in mass of K2So4 and Kno3

Answer 1

Mass of precipitate is 863.5 mg BaSO4

Moles of BaSO4 = 0.863.5/233.43 = 0.0037 mol BaSO4

pKsp = -log(Ksp)

10^-9.96 = -log(Ksp)

Ksp = 1.09*10^-10

This indicates it is very much insoluble, so we can assume approximately 100% of BaSO4 precipitates

Since there are 0.0037 mol BaSO4, there must have been 0.0037 mol SO4 in the original sample, indicating 0.0037 mol of K2SO4. 0.0037 mol K2SO4 * 174.26 g/mol K2SO4 = 0.6645 grams

0.6645 grams K2SO4 / 1.546 grams total * 100% = 43%

K2SO4 = 43%

KNO3 = 57%