Answer:
3.8x10⁻⁵ M
Explanation:
By the reaction given, the stoichiometry between FeCl3 and AgNO3 and AgCl is 1:3:3. The molar mass of AgCl is 143.32 g/mol, so the number of moles that was formed is:
n = mass/molar mass
n = 0.0036g/143.32g/mol
n = 2.51x10⁻⁵ mol
Thus, the number of moles of FeCl3 that had precipitated was:
1 mol of FeCl3 ------------ 3 moles of AgCl
x -------------2.51x10⁻⁵ mol
By a simple direct three rule:
x = 8.37x10⁻⁶ mol
The precipitation is a reversible reaction, and, for the AgCl, is has an equilibrium constant Kps = 1.8x10⁻¹⁰, then, the solubility, S, (the maximum concentration of the ions that will not precipitate) can be calculated:
S² = Kps
S = √1.8x10⁻¹⁰
S = 1.34x10⁻⁵ mol/L
The volume of the sample was 250 mL = 0.250 L, so the number of moles of Cl⁻ that was not precipitated is:
n = 1.34x10⁻⁵ mol/L*0.250 L
n = 3.35x10⁻⁶ mol
Because of the stoichiometry:
1 mol of FeCl3 ------ 3 moles of Cl-
y ------ 3.35x10⁻⁶ mol
By a simple direct three rule:
y = 1.12x10⁻⁶ mol of FeCl3
The total number of moles of FeCl3 is then:
n = x + y
n = 9.5x10⁻⁵ mol
The concentration is the number of moles dived by the volume:
9.5x10⁻⁵ /0.250 = 3.8x10⁻⁵ M
