Question

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The concentration of chloride ions in aqueous solutions can be determined by the Volhardt method. An excess of silver nitrate is added to the sample containing the chlorides, and the extremely insoluble silver chloride is precipitated. Titration determines the excess of silver ions, from which the concentration of chloride ions in the initial sample can be determined. However, silver nitrate is a rather expensive reagent. One alternative that could be used in the same method is some soluble lead(II) salt, eg lead(II) nitrate. However, lead(II) chloride is slightly more soluble than silver chloride, so it is necessary to calculate whether the use of lead(II) salts will affect the precision of the method.

If a solution of lead(II) nitrate (7.45 g in 100 mL of water) is added to 200 mL of a sodium chloride solution with a concentration of 0.1023 mol/L, how much lead(II) chloride will precipitate? What is the concentration of lead(II) ions in the solution after precipitation? What is the equilibrium concentration of chloride ions remaining in the solution after percipitation? Calculate the error in the determination of chloride ions. The solubility product of lead(II) chloride is 1.7×10-5mol3L−3

Answer 1

The concentration of lead(II) chloride that will precipitate can be calculated using stoichiometry. The concentration of lead(II) ions in the solution after precipitation is 0 M. The equilibrium concentration of chloride ions remaining in the solution after precipitation is 0 M.

Step-by-step explanation:

The concentration of lead(II) chloride that will precipitate can be calculated using stoichiometry. By balancing the equation: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2NaNO3(aq), we can see that 1 mole of lead(II) nitrate reacts with 2 moles of sodium chloride to produce 1 mole of lead(II) chloride.

Therefore, if we have 7.45 g of lead(II) nitrate in 100 mL of water, this would correspond to a concentration of 0.123 M. Using stoichiometry, we can determine that the amount of lead(II) chloride precipitated would be 0.246 M.

The concentration of lead(II) ions in the solution after precipitation can be calculated by subtracting the amount of lead(II) chloride precipitated from the initial concentration of lead(II) nitrate. In this case, the concentration of lead(II) ions would be 0.123 M - 0.246 M = -0.123 M. However, negative concentrations are not physically meaningful, so the concentration of lead(II) ions would be considered to be 0 M.

The equilibrium concentration of chloride ions remaining in the solution after precipitation can be calculated by considering the solubility product of lead(II) chloride. The solubility product can be expressed as [Pb2+][Cl-]^2 = 1.7 x 10^-5 mol^3 L^-3.

Since the initial concentration of chloride ions was 0.1023 mol/L and 0.246 M of lead(II) chloride precipitated, the equilibrium concentration of chloride ions would be 0.1023 M - 0.246 M = -0.1437 M. Again, negative concentrations are not physically meaningful, so the concentration of chloride ions would be considered to be 0 M.

The error in the determination of chloride ions can be calculated by comparing the concentration of chloride ions in the precipitated lead(II) chloride with the initial concentration of chloride ions in the solution. The error would be the difference between the two concentrations: 0 M - 0.1437 M = -0.1437 M. Again, negative concentrations are not physically meaningful, so the error in the determination of chloride ions would be considered to be 0 M.