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Answer to Question #19313 in Inorganic Chemistry for Christa

Question #19313
A silver nitrate solution was standardized by dissolving 0.54 grams of NaCl into 120 ml of solution using 4.0 drops of 1.0M K2CrO4 for an indicator (1ml=20 drops). Upon titrating with 43.6 ml of AgNO3, the Ag2CrO4 just began to precipitate. At this point in the titration the system reaches equilibrium and therefore Qsp for silver chromate is equal to Ksp of silver chromate and Qsp for silver chloride is equal to Ksp for AgCl.


A.) calculate the equilibrium concentration of chromate ion.
B.) calculate the equilibrium concentration of the silver ion
C.) calculate the equilibrium concentration of the chloride ion
D.) calculate the percent Cl- remaining in solution
E.) calculate the molarity of the AgNO3 solution used
Expert's answer
the end point being signaled by the appearance of redsilver chromate(Ag2CrO4) .
AgCl(s) → Ag++ Cl-
KSP(AgCl) = 1.35 x 10-5
Ag2CrO4 → 2Ag+ + CrO42-
KSP(Ag2CrO4) = 1.2 x 10-12

A) Ksp =[Ag+]2·[CrO42-]
[CrO42-] = Ksp/[Ag+]2
[CrO42-] = 1.2*10-12/1.35*10-5
[CrO42-] = 6x10-3 M

B) 2[CrO42-] = [Ag+]
[Ag+] = 2 * 6x10-3 M = 1.2 *10-2 M

C) [Cl-] = Ksp/[Ag+]
[Cl-] = 1.35 x 10-5 /1.2*10-2 = 1.125*10-3 M

D) NaCl + AgNO3 → AgCl + NaNO3
CM = n / V; n = 0.54 g / 54.5 g/mol = 0.0092 mol
CM(NaCl) = 0.0092 mol/0.12 L = 0.07 M
[Cl-] = 0.07 M
[Cl-] = 0.07 - 1.125*10-3 = 0.068875 M

E) 0.07 *120 = CM*43.6
CM = 0.0254 M

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