Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride sol

Question

Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride solution until no precipitate forms. If 45 mL of silver sulfate were added to react completely with the potassium chloride, what was the original concentration of silver sulfate solution? What concentration of sulfate will remain in solution after the reaction is complete if 45 mL of silver sulfate solution were added to the potassium chloride solution? Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride solution until no precipitate forms. If 45 mL of silver sulfate were added to react completely with the potassium chloride, what was the original concentration of silver sulfate solution? What concentration of sulfate will remain in solution after the reaction is complete if 45 mL of silver sulfate solution were added to the potassium chloride solution? What concentration of sulfate will remain in solution after the reaction is complete if 45 mL of silver sulfate solution were added to the potassium chloride solution?

Explanation / Answer

The reaction will be

Ag2SO4 + 2KCl --> 2AgCl (s) + K2SO4

We know that

Molarity = Number of moles / Volume of solution in litres

Molarity of KCl = 0.5

Volume = 25 mL

so moles of KCl = molarity X volume = 0.5 X 25 mL = 12.5 millimoles

As per stoichiometry of reaction 1 mole of Ag2SO4 will react with two moles of KCl to form 2 moles of silver chloride

so moles of AgCl formed = 2 X 12.5 millimoles = 25 millimoles

The moles of Ag2SO4 reacted = 12.5 / 2 millimoles = 6.25 millimoles

Volume used = 45 mL

so molarity of Ag2SO4 = moles / volume = 6.25 / 45 = 0.139 molar (original concentration of silver sulphate)

The sulphate remained will be = moles of Ag2SO4 used = 0.139 molar

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