# The K_(sp) of Ag_(2)CrO_(4),AgCl,AgBr and AgI are respectively, 1.1xx10^(-12),1.8xx10^(-10),5.0xx10^(-13),8.3xx10^(-17). Which one of the follow

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The K_(sp) of Ag_(2)CrO_(4),AgCl,AgBr and AgI are respectively, 1.1xx10^(-12),1.8xx10^(-10),5.0xx10^(-13),8.3xx10^(-17). Which one of the following salts will precipitate last if AgNO_(3) solution is added to the solution containing equal moles of NaCl,NaBr,NaI and Na_(2)CrO_(4) ?
A. Agl
B. AgCl
C. AgBr
D. Ag_(2)CrO_(4)

by (80.2k points)
selected by

Ag_(2)CrO_(4)hArr 2Ag^(+)+CrO_(4)^(2-)
Solubility product
K_(sp)=(2s)^(2)xxS=4s^(3)
K_(sp)=(1.1xx10^(-12)) (given)
S= root(3)((K_(sp))/(4))=0.65xx10^(-4)
AgCl hArr Ag^(+)+Cl^(-)
K_(sp)=SxxS " " (K_(sp)=1.8xx10^(-10))
S=sqrt(K_(sp))=1.34xx10^(-5)
AgBr hArr Ag^(+)+Br^(-)
K_(sp)=SxxS " " (K_(sp)=5xx10^(-13))
S= sqrt(K_(sp))=0.71xx10^(-6)
AgI hArr Ag^(+)+I^(-)
K_(sp)=SxxS " " (K_(sp)=8.3xx10^(-17))
S = sqrt(K_(sp))=0.9xx10^(-8)
because Solubility of Ag_(2)CrO_(4) is highest.
So, it will precipitate last.