Question

How much Na2SO4 solution (in L) must be added to initiate the precipitation of CaSO4?

Answer 1

Answer:

$V=0.48L$

Explanation:

Hello,

In this case, the rest of statement is missing, therefore, it was found to be:

We add 100 ml of 0.01 M Na₂SO₄-solution to 100 ml of 0.02 M CaCl₂-solution. A precipitate of CaSO₄ is formed. What is the (minimal) volume of 0.01 M Na₂SO₄-solution that needs to be added to the mix for the CaSO₄ precipitate to be just dissolved completely? Consider CaSO₄ solubility constant to be 2.5x10⁻⁵

In this case, Ca⁺², SO₄⁻² and H₂O are out interest substance, thus, if we introduce $V$ as the additional volume of Na₂SO₄-solution, we've got:

$V_T= V+0.2L$

$n_{SO_4^{-2}}=(V+0.1 L)(0.01 \frac{mol}{L})$

$n_{Ca^{+2}}=(0.1 L)(0.02\frac{mol}{L} )$

Hence, we have:

$Ksp=[Ca^{+2}][SO_4^{-2}]=\frac{n_{Ca^{+2}}*n_{SO_4^{-2}}}{V_T} =2.5x10^{-5}$

Replacing the previous values:

$2.5x10^{-5}=\frac{(0.1L)(0.02\frac{mol}{L} )}{V+0.2L} \frac{(V+0.1L)(0.01\frac{mol}{L} )}{V+0.2L}$

Solving for $V$, we obtain:

$V=0.48L$

Best regards.