sulfate determination

Question

 0.1 g of the mixture of Na2SO4 and K2SO4 is taken and 100 mL of solution of it is prepared. A portion of 10 ml of this prepared solution is placed in a beaker and some distilled water is added. A mass of 15.5 mg is obtained by precipitation with BaCl2 at pH = 5, then filtering and bringing to a constant weight at 800 oC. Calculate the percentages of Na2SO4 and K2SO4 in the mixture accordingly

Answer 1

You are forming BaSO₄ ppt. The equations are the same, the molar masses are different: You can do the stoichiometry on the two constituents. Let f be the fraction of the sodium in the original mixture. I'm going to call sodium sulfate SS and potassium sulfate KS

0.1g( f ) SS ( mole/ g SS) (1 BaSO₄/1 SS)( g/mole BaSO₄) = g of BaSO₄ from SS. (this will have a *f)

0.1g (1-f) KS (mole/g KS) (1 BaSO₄/1KS) (g/mole BaSO₄) = g of BaSO₄ from KS (this will have a *(1-f))

You add the two results up and equate to .0155 grams of BaSO₄ and solve for f

Now take the f and find the fraction of Na₂SO₄ in the product by dividing the answer to the first equation (with f substituted in) divided by .155 grams. The other fraction is 1- the fraction of Na₂SO₄ You may need to change to percentages)

Challenging problem.