0.10 M Cl^- + 0.10 M CrO4^2- + 0.10 M AgNO3

0.10 M Cl^- + 0.10 M CrO4^2- + 0.10 M AgNO3

 

 

You have a solution consisting of 0.10 M Cl^- and 0.10 M CrO4^2-.

You add 0.10 M silver nitrate dropwise to this solution.

Given that the Ksp for Ag2CrO4 is 9.0×10^(-12),

and that for AgCl is 1.6×10^(-10),

which of the following will precipitate first?

a) silver chloride

b) silver chromate

c) silver nitrate

d) cannot be determined by the information given

e) none of these

 

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AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq)

Ksp = [Ag^+] [Cl^-]

 

 

AgCl 침전에 필요한 [Ag^+]를 계산하면,

[Ag^+] = Ksp / [Cl^-]

= (1.6×10^(-10)) / 0.10

= 1.6×10^(-9) M

 

 

 

 

Ag2CrO4(s) ⇌ 2Ag^+(aq) + CrO4^2-(aq)

Ksp = [Ag^+]^2 [CrO4^2-]

 

 

Ag2CrO4 침전에 필요한 [Ag^+]를 계산하면,

[Ag^+]^2 = Ksp / [CrO4^2-]

 

[Ag^+] = (Ksp / [CrO4^2-])^(1/2)

= ((9.0×10^(-12)) / 0.10)^(1/2)

= 9.5×10^(-6) M

 

 

 

 

AgCl이 먼저 침전될 것이다. 왜냐하면,

AgCl 침전에 필요한 [Ag^+]가 더 작은 값이기 때문이다.

( 더 낮은 농도 값 )

 

 

 

답: a) silver chloride

 

 

 

 

[키워드] AgCl Ag2CrO4 침전 기준문서, Ag2CrO4 AgCl 침전 기준문서