Ba^2+(aq) + SO4^-2(aq) -------> BaSO4 (aq) Calculate the concentration of sulfat

Question

Ba^2+(aq) + SO4^-2(aq) -------> BaSO4(aq)

Calculate the concentration of sulfate in seawater in unites of mmol/L and mg/L, assuming that all of the S is present in seawater as sulfate ion, SO4^-2. (The concentration of sulfur is 8.85 x 10^2). Based on this estimate, calculate the volume of seawater required to produce 100 mg of barium sulfate, BaSO4, and the volume of 0.3 M BaCl2 solution that would be required to precipitate all of the sulfate ion (+20% extra Ba^2+). Please show calculations.

Explanation / Answer

I assume conc of S = 8.85 x10^2 uM

hence sulphate conc = S conc = 8.85 x10^2 x10^-3 mM =0. 885 mmol/L

sulphate conc = 0.885 x96 = 84.960 mg/L

milli moles of BaSO4 = 100 / 233.43 = 0.4284

Ba2+ and SO42- react in 1:1 ratio

so water containing SO42- required = ( 0.4284/0.885) = 0.484 liters = 484 ml

milli moles of BaCl2 reqiured = milli moles od BaSO4 = 0.4284

including 20 % extraa mmol of BaCl2 = ( 0.4284 x 120/100) = 0.51408

0.3 x V = 0.51408

V = 1.7 ml

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