A sample containing 0.175g of Cl- is dissolved in 50 mL of deionized H2O.
The solution is titrated to the equivalence point with a AgNO3 solution. The final volume of the titrated sample is 75 mL.
a. What is the concentration of the AgNO3 solution?
b. How many grams of AgCl are precipitated?
The sample of the compound containing Cl- has 0.175 g of Cl- ions. It is dissolved in 50 mL of deionized water and titrated with AgNO3. The volume of the final solution is 75 mL. This gives the volume of the AgNO3 solution as 25 mL.
AgNO3 reacts with the compound containing Cl- ions to yield AgCl. The mass of Cl- is 35.453 g/mole. 0.175 g of the sample contains 4.936*10^-3 mole of Cl-. As this reacts with AgNO3 in a 25 mL solution, the concentration of AgNO3 is 4.936/25 = 0.197 M.
The mass of AgCl precipitated is the mass of 4.936*10^-3 moles of AgCl. As the mass of AgCl is 143.32 g/mole, the mass of AgCl precipitated is 4.936*10^-3*143.32 = 0.707 g
The concentration of the AgNO3 solution is 0.197 M
The mass of AgCl precipitated is 0.707 g