Asked by Mika
Assuming the sample is pure sodium sulfate, how many mL of 5% BaCl2.2H20 solution would be required to precipitate the sulfate. Density of barium shloride solution of 1.00 g/mL.
Answers
Answered by
DrBob222
That all depends upon how much of the Na2SO4 is involved. moles Na2SO4 = grams/molar mass
Determine the M of the BaCl2.2H2O soln. 5% means you have 5.0 g BaCl2.2H2O/100 mL. moles = grams/molar mass; then M BaCl2.2H2O = moles BaCl2.2H2O/0.1L = ?M.
Write the equation and convert moles Na2SO4 to moles BaCl2.2H2O. Then
M BaCl2.2H2O = moles Ba^2+ required/L soln and solve for L.
Determine the M of the BaCl2.2H2O soln. 5% means you have 5.0 g BaCl2.2H2O/100 mL. moles = grams/molar mass; then M BaCl2.2H2O = moles BaCl2.2H2O/0.1L = ?M.
Write the equation and convert moles Na2SO4 to moles BaCl2.2H2O. Then
M BaCl2.2H2O = moles Ba^2+ required/L soln and solve for L.
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