What is the maximum solubility of AgCl (ksp=1.8x10^-10) in an HCl solution of pH=2.8 ?
The maximum solubility of AgCl in pure water is 1.3x10^-5
1 Answer | Add Yours
AgCl is a solid.But if we put that inside water small amount of solid AgCl will dissolve in water. The reaction is as follows.
`K_(sp) = [Ag^+][Cl^-]`
So the amount of solubility in both ions is equal since
Now imagine the situation AgCl in HCl solution.
As in the above condition AgCl dissolve in HCl also.
If we assume x moles of each `Ag^+` and` Cl^-` is dissolved in 1L of HCl solution;
So the solubility of AgCl in HCl is x mol/l
In the solution we have Ag^+ from AgCl and Cl^- from AgCl as well as HCl.
We know that;
PH = -log[H^+]
[H^+] = 10^(-PH)
In the HCl solution we have 10^(-2.8) `Cl^-` .
`K_sp = [Ag^+][Cl^-]`
In the HCl solution
[`Ag^+` ] = x
[`Cl^-` ] = x+10^(-2.8)
It is given that in water AgCl solubility is 1.3x10^-5. Then as a approximation we can take x=1.3x10^-5
Since 10^(-2.8)>>>1.3x10^-5 we can write;
(1.3x10^-5+10^(-2.8))` ` `~~` 10^(-2.8)
`K_sp` = x*10^(-2.8)
Maximum solubility of AgCl in HCl is `1.136xx10^-7` mol/l.
same temperature conditions prevail in both AgCl in water solution and AgCl in HCl solution.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes