# The solubility product constant of Ag_(2)CrO_(4) and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarit

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The solubility product constant of Ag_(2)CrO_(4) and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarities of their saturated solutions.

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Ag_(2)CrO_(4) dissolves in water and the equilibrium in the saturated solution is
Ag_(2)CrO_(4)(s)= 2Ag^(o+)(aq)+CrO_(4)^(2-)(aq)
Let the solubolty of AgBr be S_(1)
[Ag^(o+)(aq)]=2S_(1)
and [CrO_(4)^(2-)(aq)=S_(1)]
K_(sp)=[Ag^(o+)(aq)]^(2)[CrO_(4)^(2-)(aq)]=(2S_(1))^(2)xxS=4S_(1)^(3)
or S_(1)=(K_(sp)/4)^(1//3)=((1.1xx10^(-12))/4)^(1//3)=0.65xx10^(-14) M
Similarly AgBr=Ag^(o+)(aq)+Br^(Θ)(aq)
Let the solubility of AgBr be S_(2)
[Ag^(o+)(aq)]=S_(2) and Br^(Θ)(aq)]=S_(2)
K_(sp)=[Ag^(o+)(aq)][Br^(Θ)(aq)]
S_(2)xxS_(2)=S_(2)^(2)
or S_(2)=(K_(sp))^(1//2)=(5.0xx10^(-13))^(1//2)
=(0.5xx10^(-12))^(1//2)=0.707xx10^(-6) mol L^(-1)
Ratio of the molarities of silver chromate to silver bromide comes to =(0.65xx10^(-4))/(0.707xx10^(-6))=91.9 silver chromate is more soluble.