Question

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0300 M Ag + ( aq ) . What will be the concentration of Ca 2 + ( aq ) when Ag 2 SO 4 ( s ) begins to precipitate? Solubility-product constants, K sp , can be found in the chempendix.

Answer 1

Answer : The concentration of $[Ca^{2+}]$ ion is, 0.00371 M

Explanation :

First we have to calculate the concentration of $[SO_4^{2-}]$ ion.

The solubility equilibrium reaction of $AgSO_4$ will be:

$Ag_2SO_4\rightleftharpoons 2Ag^{+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^{+}]^2[SO_4^{2-}]$

$K_{sp}=1.20\times 10^{-5}$

$1.20\times 10^{-5}=(0.0300)^2\times [SO_4^{2-}]$

$[SO_4^{2-}]=0.0133M$

Now we have to calculate the concentration of $[Ca^{2+}]$ ion.

The solubility equilibrium reaction of $AgSO_4$ will be:

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][SO_4^{2-}]$

$K_{sp}=4.93\times 10^{-5}$

$4.93\times 10^{-5}=[Ca^{2+}]\times (0.0133)$  

$[Ca^{2+}]=0.00371M$

Thus, the concentration of $[Ca^{2+}]$ ion is, 0.00371 M