Given #K_"sp"*AgCl=1.8xx10^(-10)#, if #[Ag^+]=10^(-10)# and #[Cl^-]=10^(-4)#, will silver chloride precipitate?

1 Answer
Dec 11, 2016

Answer:

#Ag^(+) + Cl^(-) rightleftharpoonsAgCl(s)darr#

#K_"sp"=1.8xx10^-10#

Explanation:

We have the #K_"sp"# value, and thus we know that if the ion product #[Ag^+][Cl^-]>K_"sp"#, silver chloride will precipitate until #[Ag^+][Cl^-]-=K_"sp"#.

For #A.# #[Ag^+]=10^-10*mol*L^-1#; #[Cl^-]=10^-4*mol*L^-1#.

The ion product #[Ag^+][Cl^-]=10^-10xx10^-4=10^-14#,

Since this ion-product is LESS than #K_"sp"# silver chloride will not precipitate. I think you are due to do the next problems.

See here for similar treatments.