Question

A solution contains 0.020 mole each of I”, Br, and Cl. When the solution is mixed with 200. mL of 0.28 M AgNO3, what mass of AgCl(s) precipitates out, and what is Ag+)? Assume no volume change AgI: Ksp = 1.5 x 10-16 AgBr : Ksp = 5.0 x 10-18 AgCl : Ksp = 1.6 x 10-10 00 Mass of AgCl(s) = [18+] - M

Answer 1

Here is the answer of the given probelm (see below):

Kp = 5.0 * 10-13 Q. A solution contains 0.020 mole of each of I, Br, and Cl. When the solution is mixed with 200 mL of 0.28 M AgNO3, what mass of AgCl (s) precipitates out, and what (Ag ]? Assume 110 volume change. AgI (5) K = 1.5 * 10-16 AgBr (s) AgCi (3) K=1.6 10-10 Answer: Reaction: T + Br + Cl + AgNO3 - Agi + AgBr + AgCi From the given solubility product it clear that AgI will be precipitated first then AgBr and finally AgCl. Now, moles of AgNO3 solution = 0.2 0.28 = 0.056 Precipitation occurs in equal molarity of (Ag) and [X] where, X=1, Br", CT Given: concentrations of [I'] = 0.020 mole, [Br] = 0.020 mole and [CI] = 0.020 mole As AgI will precipitate first then remaining concentration of [Ag)-(0.056-0.020) = 0.036 mole So, once AgBr will precipitated then remaining concentration of [Ag 1 - (0.036-0.020) = 0.016 mole Now we have 0.016 moles of Ag and 0.02 moles of C1- Thus, precipitate of AgCl will be = 0.010 moles Molar mass of AgC1 = 143.32 g/mol Mass of AgCl ppt = 0.010 x 143.32 g = 1.432 g Concentration of Ag* = 0.01 M Mass of AgC1 (5) 1.432 g [Ag] 0.01 M