Question

Samples containing magnisium sulfate weighing 0.3546, 0.3412, and 0.3553 were dissovled in dilute HCl. Barium chloride solution is added in excess to precipitate barium sulfate. Approximatley how many mililiters of 5% BaCl₂ solution would be required to precipitate all the sulfate if we assume that the samples are pure magnesium sulfate? Assume that the density of the barium chloride solution is 1.00 g/mL.

Answer 1

total MgSO4 added = 0.3546+0.3412+0.3553 = 1.0511 g

MgSO4 moles = mass / Molar mass of MgSO4

           = 1.0511 /120.366 = 0.0087325

SO42- moles = 0.0087325  

Now reaction between sulphate ions and Ba2+ ions is

Ba2+ (aq) + SO4^2-(aq) --> BaSO4 (s)

hence Ba2+ moles needed = 0.0087325 = BaCl2 moles ( since 1 BaCl2 gives 1Ba2+)

BaCl2 mass needed = moles x Molar mass of BaCl2

                = 0.0087325 x 208.23 g/mol

          = 1.82 g

we had 5 % BaCl2 solution which means 5 g BaCl2 per 100 g solution

solution mss needed for 1.82 g BaCl2 = ( 1.82) x ( 100 /5) = 36.37 g

density is given 1g/ml

volume = mass/ density

volume of solution needed = 36.37 g / ( 1g/ml) = 36.37 ml