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What is the maximum solubility of AgCl (ksp=1.8x10^-10) in an HCl solution of pH=2.8...

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sanashafi | Student, Undergraduate | eNotes Newbie

Posted July 21, 2012 at 1:33 AM via web

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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

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 29, 2012 at 3:01 AM (Answer #1)

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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.

`AgCl_((S))-----gt Ag_((aq))^++Cl_((aq))^-`

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

 

So the amount of solubility in both ions is equal since

`Ag^+:Cl^(-)=1:1`

 

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.

`HCl-------gt H^++Cl^-`

 

We know that;

     PH = -log[H^+]

[H^+] = 10^(-PH)

          = 10^(-2.8)

 

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

`K_sp =(x)(1.3xx10^-5+10^(-2.8))`


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)

      x =1.8x10^-10/10^(-2.8)

         = `1.136xx10^-7`

 

Maximum solubility of AgCl in HCl is `1.136xx10^-7` mol/l.

 

Assumption:

same temperature conditions prevail in both AgCl in water solution and AgCl in HCl solution.

 

Sources:

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