The concentration of Ag+ ions is 1x10^-6 M in AgBr solution. Find minimum concentration of Br- ions necessary to cause precipitation of AgBr. ( Ksp of AgBr= 4x10^-13 M^2)?

1 Answer
May 7, 2017

Answer:

We interrogate the equilibrium reaction:

#Ag^(+) + Br^(-) rarr AgBr(s)darr#

Explanation:

We have been given the thermodynamic equilibrium constant for this reaction, i.e.:

#K_"sp"=[Ag^+][Br^-]=4xx10^(-13)#.

#K_"sp"# is the so-called #"solubility product"#, which have been measured for a host of inorganic salts. And so we simply substitute in #[Ag^+]#, and solve for #[Br^-]#............

#[Br^-]=K_"sp"/([Ag^+])=(4xx10^-13)/(1xx10^-6)=4.0xx10^-7*mol*L^-1#

And thus, should #[Ag^+]>4.0xx10^-7*mol*L^-1#, a precipitate of creamy yellow #AgBr# salt will be observed.