A) Precipitation does not occur
B) Silver chromate gets precipitated first
C) Silver chloride gets precipitated first
D) Both silver chromate and silver chloride start precipitating simultaneously
Correct Answer: C
Solution :
For precipitation, ionic product > solubility product \[({{K}_{sp}})\] For \[A{{g}_{2}}Cr{{O}_{4}}\] ionic product \[={{[A{{g}^{+}}]}^{2}}[CrO_{4}^{-}]\] \[={{({{10}^{-4}})}^{2}}({{10}^{-5}})={{10}^{-13}}\] \[{{K}_{sp}}\] of \[A{{g}_{2}}Cr{{O}_{4}}=4\times {{10}^{-12}}\] Here, \[{{K}_{sp}}>IP\] Thus, no precipitate is obtained. For \[AgCl\], ionic product \[=[A{{g}^{+}}][C{{l}^{-}}]\] \[=[{{10}^{-4}}][{{10}^{-5}}]\] \[={{10}^{-9}}\] \[{{K}_{sp}}(AgCl)=1\times {{10}^{-10}}\] Here, \[IP>{{K}_{sp}}\] So, precipitate will form. Thus, silver chloride gets precipitated first.You need to login to perform this action.
You will be redirected in
3 sec