Question

4. A solution contains 0.018 mol each of I–, Br–, and Cl–. Use the Ksp values given below to determine how much AgCl(s) precipitates out when this solution is mixed with 200 mL of 0.24 M AgNO3.
NOTE: This question isn’t as difficult as it looks. It’s really just stoichiometry.
Ksp AgI = 1.5 x 10–16 Ksp AgBr = 5.0 x 10–13 Ksp AgCl = 1.6 x 10–10
A) 5.0 g B) 3.3 g C) 2.6 g D) 0.0 g E) 1.7 g

Answer 1

answer : E ) 1.7 g

AgI , Ksp = 1.5×10^–16

AgBr , Ksp = 5.0×10^–13

AgI , Ksp = 1.6 ×10^–10

AgNO3 (aq) + Cl- (aq) ---------------------> AgCl (s) + NO3- (aq)

Ksp of AgI and AgBr are lesser than AgCl so they could form precipitate first

actual moles of AgNO3 are present = 200 x 0.24 / 1000 = 0.048

moles of each ion present 0.018

AgNO3 moles needed for I- = 0.018

AgNO3 moles needed for Br- = 0.018

remaining AgNO3 moles = 0.048 -(0.018 +0.018)

                                        = 0.012

now AgCl precipitation.

AgNO3 (aq) + Cl- (aq) ---------------------> AgCl (s) + NO3- (aq)

0.012                 0.018                                    0.012

moles of AgCl formed = 0.012

mass of AgCl formed = moles x molar mass

                                    = 0.012 x 143.32

                                     = 1.72 g

AgCl(s) precipitates = 1.72 g