Question

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)?

Answer 1

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.