Question

The following equilibrium concentration exists when Agbr is in solution AgBr=Ag +Br , Ksp =7. 7*10-13 , calculate solubility of AgBr in g/l in a solution of 0. 0010M NaBr?

Answer 1

I make `S_"AgBr"=1.44xx10^-7*g*L^-1` in millimolar bromide...

Explanation:

We address the equilibrium...

`Ag^+ + Br^(-) rightleftharpoons AgBr(s)darr`

For which `K_"sp"=[Ag^+][Br^-]=7.7xx10^-13`

Now `[Br^-]=0.01*mol*L^-1` from the common ion...and so if `x*mol*L^-1` `AgBr` dissolve....then...

`K_"sp"=7.7xx10^-13=(x)(0.001+x)`.

We maketh the approximation that `0.001">>"x`

So `x=(7.7xx10^-13)/(0.001)=7.70xx10^-10*mol*L^-1`

A gram solubility of ………………

`7.70xx10^-10*mol*L^-1xx187.77*g*mol^-1=1.44xx10^-7*g*L^-1`.

Now in the ABSENCE of extraneous bromide ion, `S_"AgBr"=sqrt(7.7xx10^-13)=8.77xx10^-7*mol*L^-1`, a GRAM solubility of `0.165*mg*L^-1`... less than `"1 ppm"`... this shows the effect of the phenomenon of `"SALTING OUT..."`, i.e. adding a common ion to the solution.