A 100 mL solution contains 0.0010 mol/L of calcium chloride and 0.0010 mol/L of strontium chloride....

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A 100 mL solution contains 0.0010 mol/L of calcium chloride and 0.0010 mol/L of strontium chloride. 50 ml of 0.1000 mol/L sodium sulfate is added. Determine the percentage by mass of calcium sulfate in the product. The solubility products value of calcium sulfate is 9.1 x 10 and for strontium sulfate it is 3.2 x 107. [6 marks] 50 mL of an aqueous solution of 0.1 mol/L toluene was extracted into 10 ml of octanol. If Log K is 3.6, calculate the concentration of toluene left in the aqueous phase. b. [2 marks] A weak acid A with K, = 1.82 x 10 has a partition coefficient K between water and an organic solvent of 2.5. A weak base B with a K = 3.45 x 10 has a K of 1.2 with the same organic solvent is mixed with A. Is it possible to obtain an extract that is 90% pure, or greater, in either A or B? Explain. [2 marks] A 100 mL solution contains 0.0010 mol/L of calcium chloride and 0.0010 mol/L of strontium chloride. 50 ml of 0.1000 mol/L sodium sulfate is added. Determine the percentage by mass of calcium sulfate in the product. The solubility products value of calcium sulfate is 9.1 x 10 and for strontium sulfate it is 3.2 x 107. [6 marks] 50 mL of an aqueous solution of 0.1 mol/L toluene was extracted into 10 ml of octanol. If Log K is 3.6, calculate the concentration of toluene left in the aqueous phase. b. [2 marks] A weak acid A with K, = 1.82 x 10 has a partition coefficient K between water and an organic solvent of 2.5. A weak base B with a K = 3.45 x 10 has a K of 1.2 with the same organic solvent is mixed with A. Is it possible to obtain an extract that is 90% pure, or greater, in either A or B? Explain. [2 marks]

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a SO42 in final solution 01000 M 50 mL 150 mL 00333 M Ca2 Sr2 in solution after addition 00010 M 50 mL 150 mL 0000667 M Since SrSO4 is less soluble th... View the full answer