Question

The solubility product of ag2cro4 is 32x10-12. what is the concentration of cro-4 ions in that solution?

Answer 1

Answer : The concentration of $CrO_4^{2-}$ ions in that solution is, $7.937\times 10^{-5}moles/liter$

Solution : Given,

Solubility product = $32\times 10^{-12}$

The equation for the reaction will be,

$Ag_2CrO_4\leftrightharpoons 2Ag^++CrO_4^{2-}$

1 mole of $Ag_2CrO_4$ gives 2 moles of $Ag^{+}$ and 1 mole of $CrO_4^{2-}$.

Thus if solubility of $Ag_2CrO_4$ is 's' moles/liter, solubility of  $Ag^{+}$ is '2s' moles/liter and solubility of $CrO_4^{2-}$ is 's' moles/liter  

The formula used :

$K_sp=[Ag^+]^2[CrO_4^{2-}]$

$32\times 10^{-12}=[2s]^2[s]$

$4s^3=32\times 10^{-12}$

$s=7.937\times 10^{-5}moles/liter$

Therefore, the concentration of $CrO_4^{2-}$ ions in that solution is, $7.937\times 10^{-5}moles/liter$