Question

the solubility product of agcl is ksp. than the solubility of agcl in xm kcl is

Answer 1

AgCl(s) disassociates in water in the following equation:

AgCl(s)⇌Ag^+(aq)+Cl^−(aq)

That means,  

$Ksp=[Ag^+][Cl^−]$

$As, [Ag^+] = [Cl^−] = s$

$Ksp=s^2$

$s = (Ksp)1/2$

The addition of common ion such as chloride results in common ion effect which reduces the solubility of the salt.

CaCl_2 (aq) disassociate as:

$CaCl_2(aq)→Ca^2+(aq)+2Cl^−(aq)$

So, x mol of CaCl_2(s) can produce 2x mol of Cl^−(aq).  

Now, on mixing the two solutions the volumes becomes equal to each other. Which halves the concentration so the new concentration of Cl^− is:

$[2x+(Ksp)1/2 ]/2 mol dm^−^3$

The value of K_s_p is quite low and solving the above equation gives the value x mol dm^−^3.

$Hence, [Cl^−] = x mol dm^−^3$

Now,

$Ksp=[Ag^+][Cl^−]$

$Ksp =[Ag^+][x]$

$[Ag^+]= Ksp /[x] mol dm^−^3$[tex][/tex]

Answer 2

Answer: The addition of common ion such as chloride results in common ion effect which reduces the solubility of the salt.

$AgCl(s)⇌Ag^+(aq)+Cl^−(aq)$

$K_sp=\frac{[Ag^+][Cl^-]}{AgCl(s)}$           [AgCl(s)=1]

$K_sp={[Ag^+][Cl^-]$

1 mole of AgCl gives 1 mole of $Ag^+$ and 1 mole of $Cl^-$

Thus if solubility of AgCl is s moles/liter, the solubility of $Ag^+$ and $Cl^-$ is also s moles/liter.

$K_sp={[s][s]=s^2$

$s=\sqrt {K_sp}$

Now , $KCl\rightarrow K^++Cl^-$

1 mole of KCl gives 1 mole of $K^+$ and $Cl^-$. x moles will give x mole of $K^+$ and x mole of $Cl^-$.

Thus now [Ag^+] = s and  [Cl^-] = s+x

$K_sp=[s][s+x]$

as s<<<x, $K_sp=[s][x]$

$s=\frac{K_sp}{x}$