Question

What is the molar solubility of ag2co3 (ksp= 4 × 10–13) in 0.1 m na2co3 solution ?

Answer 1

molar solubility:-
ag3co4 (ksp=5×11-14) in 0.1 m na2 cog3 solution

Answer 2

Answer: $20\times 10^{-13}moles/liter$

Explanation: The equation for the reaction will be as follows:

$Ag_2CO_3\leftrightharpoons 2Ag^++CO_3^{2-}$

1 mole of $Ag_2CO_3$ gives 2 moles of $Ag^{+}$ and 1 mole of $CO_3^{2-}$.

Thus if solubility of $Ag_2CO_3$ is s moles/liter, solubility of  $Ag^{+}$ is 2s moles\liter and solubility of $CO_3^{2-}$ is s moles/liter

Therefore,  

$K_{sp}=[Ag^+]^2[CO_3^{2-}]$

$K_{sp}=[2s]^2[s]$

Now , $Na_2CO_3\rightarrow 2Na^++CO_3^{2-}$

1 mole of $Ag_2CO_3$ gives 2 moles of $Ag^+$ and 1mole of $CO_3^{2-}$

0.1 moles of  $Ag_2CO_3$ gives 0.2 moles of $Ag^+$ and 0.1 mole of $CO_3^{2-}$

Thus now $[Ag^+]$= 2s M and $[CO_3^{2-}$= s+0.1 M

$K_{sp}=[2s]^2[s+0.1]$

as s<<<0.1, $K_{sp}=[2s][0.1]$

$4\times 10^{-13}=[2s][0.1]$

$s=20\times 10^{-13}M$