Discussion Forum

An aqueous solution contain unkonwn concentration of Ba^2+. When 50 mL of a 1M solution of Na₂SO₄ is added , BaSO₄ just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO₄ is 1×10^-10. What is the original concentration of Ba²⁺?

Answer:

Explanation:

It is given that the final volume is 500 mL and this final volume has arrived when 50 mL of 1M Na₂SO₄ was added to unknown Ba²⁺solution. 

So, we can interpret the volume of unknown Ba²⁺ solution as 450 mL i.e.

450 mL Ba²⁺  Solution   +     50mL  Na₂SO₄  Solution →   500mL BaSO₄ Solution

From this, we can calculate the

the concentration of SO₄^2- ion in the solution

via

 M₁V₁=M₂V₂

1×50 =M₂ × 500     (as 1 M Na₂SO₄ is taken into consideration)

M₂= $\frac{1}{10}=0.1 M$

Now for just precipitation,

 Ionic product= Solubility product  ( K_SP)

i.e.

     $[Ba^{2+}][SO_{4}^{2-}]=K_{sp}  $ of BaSO₄

  Given K_sp of BaSO₄  - 1× 10^-10

So,    [Ba²⁺] [01]= 1 × 10^-10

      [Ba²⁺] =1 × 10^-9 M

Reminder  This is the concentration of  Ba²⁺ ions in the final solution. Hence, for calculating  the [Ba²⁺] in the original solution

we have to use

M₁V₁=M₂V₂

as M₁× 450 = 10^-9 × 500

So,

          M₁=1.1 × 10^-9 M