Question

The solubility of agcl (s) with solubility product 1.6x 10-10 in 0.1m nacl solution would be

Answer 1

The use of common ion effect is used here

Answer 2

Answer: $1.6\times 10^{-9}M$

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

$AgCl\rightarrow Ag^++Cl^-$

$K_{sp}=[Ag^+][Cl^-]$

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.

$K_{sp}=[s][s]$

$s=\sqrt K_{sp}$

Now , $NaCl\rightarrow Na^++Cl^-$

1 mole of NaCl gives 1 mole of $Na^+$ and $Cl^-/tex]</p><p>0.1 moles will give 0.1 mole of  [tex]Na^+$ and 0.1 mole of $Cl^-$

Thus now $[Ag^+]$=sM and $[Cl^-]$=(s+0.1)M

$K_sp=[s][s+0.1]$

as s<<<0.1, $K_sp=[s][0.1]$

$s=\frac{K_sp}{0.1}$

$s=\frac{1.6\times 10^{-10}}{0.1}=1.6\times 10^{-9}M$