The K_ sp of Ag_2 CrO_4, AgCl, AgBr and AgI are respectively, 1.1× 10^-12 , 1.8× 10^-10 , 5.0× 10^-13 , 8.3× 10^-17 . Which one of the following…

For, ${ \text{Ag}}_{\text{2}} { \text{CrO}}_{\text{4}} \rightleftharpoons { \text{2Ag}}^{\text{+}} { \text{+CrO}}_{\text{4}}^{ \text{2-}}$Ag2​CrO4​⇌2Ag++CrO42-​

solubility product

$${\text{K}}_{ \text{sp}} \text{=}{ ( \text{25} )}^{\text{2}} { \text{×5=4s}}^{\text{3}}$$Ksp​=(25)2×5=4s3

$${\text{K}}_{ \text{sp}} =( 1.1\times { 10}^{ -12} )$$Ksp​=(1.1×10−12)

$$\text{S}=3\sqrt{ \frac{ {\text{K}}_{ \text{sp}} }{4} } =0.65\times { 10}^{ -4}$$S=34Ksp​​​=0.65×10−4

For, $\text{AgCl}\rightleftharpoons { \text{Ag}}^{\text{+}} { \text{+Cl}}^{\text{-}}$AgCl⇌Ag++Cl-

${\text{K}}_{ \text{s}\text{p}} =\text{S}\times \text{S}$Ksp​=S×S $( {\text{K}}_{ \text{s}\text{p}} =1.8\times { 10}^{ -10} )$(Ksp​=1.8×10−10)

$$\text{S=}\sqrt{ {\text{K}}_{ \text{sp}} } =1.34\times { 10}^{ -5}$$S=Ksp​​=1.34×10−5

For, $\text{AgBr}\rightleftharpoons { \text{Ag}}^{\text{+}} { \text{+Br}}^{\text{-}}$AgBr⇌Ag++Br-

$${\text{K}}_{ \text{sp}} \text{=S×S} ( {\text{K}}_{ \text{sp}} { \text{=5×10}}^{ \text{-13}} )$$Ksp​=S×S(Ksp​=5×10-13)

$$\text{S=}\sqrt{ {K}_{ sp} } =0.71\times { 10}^{ -6}$$S=Ksp​​=0.71×10−6

For, $\text{AgI}\rightleftharpoons { \text{Ag}}^{\text{+}} { \text{+I}}^{\text{-}}$AgI⇌Ag++I-

$${\text{K}}_{ \text{sp}} \text{=S×S} ( {\text{K}}_{ \text{sp}} \text{=8}{ \text{.3×10}}^{ \text{-17}} )$$Ksp​=S×S(Ksp​=8.3×10-17)

$$\text{S=}\sqrt{ {\text{K}}_{ \text{s}\text{p}} } =0.9\times { 10}^{ -8}$$S=Ksp​​=0.9×10−8

therefore, the solubility of $\text{A}{\text{g}}_{2} \text{C}\text{r}{\text{O}}_{4}$Ag2​CrO4​ is highest so it is precipitate last.