Question

150 mL of 0.0008 M ammonium sulphate is mixed with 50 mL of 0.04 M calcium nitrate.the ionic product of CaSO4 will be : (Ksp = 2.4 × 10^-5 for CaSO4)​

Answer 1

Answer : The ionic product of $CaSO_4$ will be, $3.2\times 10^{-5}$

Explanation :

The dissociation of ammonium sulfate will be,

                          $(NH_4)_2SO_4\rightarrow 2NH_4^++SO_4^{2-}$

Concentration   0.0008 M   $(2\times 0.0008M)$  $0.0008M$

The dissociation of calcium nitrate will be,

                          $Ca(NO_3)_2\rightarrow Ca^{2+}+2NO_3^{-1}$

Concentration   0.04 M      $0.04M$     $(2\times 0.04M)$

The dissociation of calcium sulfate will be,

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

The expression for ionic product of calcium sulfate will be,

$\text{Ionic product}=[Ca^{2+}][SO_4^{2-}]$

Now put all the values in this expression, we get the ionic product of calcium sulfate.

$\text{Ionic product}=0.04\times 0.0008=3.2\times 10^{-5}$

Therefore, the ionic product of $CaSO_4$ will be, $3.2\times 10^{-5}$