Question

Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?

Answer 1

[Br⁻¹] = [NaBr] = 0.1

Ksp = [Ag⁺] x [Br⁻]
5 x 10⁻¹³ = [Ag⁺] x 0.1
[Ag⁺] = 5 x 10⁻¹²

Answer 2

Answer : The maximum concentration of silver ion is
`5* 10^(-12)m`

Solution : Given,

`K_(sp)` for AgBr =
`5* 10^(-13)`

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

`NaBr(aq)\rightleftharpoons Na^++Br^-`

The concentration of NaBr solution is 0.1 m that means,

`[Na^+]=[Br^-]=0.1m`

The equilibrium reaction for AgBr is,

`AgBr\rightleftharpoons Ag^++Br^-`

At equilibrium s s

The expression for solubility product constant for AgBr is,

`K_(sp)=[Ag^+][Br^-]`

The concentration of
`Ag^+` = s

The concentration of
`Br^-` = 0.1 + s

Now put all the given values in
`K_(sp)` expression, we get

`5* 10^(-13)=(s)(0.1+s)`

By rearranging the terms, we get the value of 's'

`s=5* 10^(-12)m`

Therefore, the maximum concentration of silver ion is
`5* 10^(-12)m`.