# When the solutions of BaCl2 and Na2SO4 mixed, precipitation of BaSO4 was produced. Before the Precipitation, [Ba(2+)] was 5 * 10^-4 mol dm^-3 and the [SO4(2-)] was 5 * 10^-5 mol dm^-3. Find the...

When the solutions of BaCl2 and Na2SO4 mixed, precipitation of BaSO4 was produced. Before the Precipitation, [Ba(2+)] was 5 * 10^-4 mol dm^-3 and the [SO4(2-)] was 5 * 10^-5 mol dm^-3.

**Find the the precentage of SO4(2-) which was not precipitated. **

**Ksp of BaSO4 in that temperature is 1.1 * 10 ^-10 mol^2 dm^-6**.

**Please help me on this**.

### 2 Answers | Add Yours

`BaSO_4 harr Ba^(2+)+SO_4^(2-)`

`K_(sp) = [Ba^(2+)][SO_4^(2-)]`

Let us assume that xM amount of moles has precipitated once they were mixed.

So at the equilibriun mixture;

`[Ba^(2+)] = 5 xx10^(-4)-x`

`[SO_4^(2-)] = 5 xx 10^(-5)-x`

`1.1xx10^(-10) = (5 xx10^(-4)-x)(5 xx 10^(-5)-x)`

`1.1xx10^(-10) = 25xx10^(-9)-(55xx10^(-5))x+x^2`

Since `x` is very small ` x^2 ~=0.`

`1.1xx10^(-10) = 25xx10^(-9)-(55xx10^(-5))x`

`x = (25xx10^(-9)-1.1xx10^(-10))/(55xx10^(-5))`

`x = 4.53xx10^(-5)`

Remaining `[SO_4^(2-)] = 5 xx 10^(-5)-4.53xx10^(-5) = 4.7xx10^(-6)`

Remaining `[SO_4^(2-)]` % `= (4.7xx10^(-6))/(5xx10^-(5))xx100%` = 9.4%

** So the remaining **`[SO_4^(2-)]`

**.**

*is 9.4%**This is a very small amount which even cannot be identified.*

**Sources:**

Thank you :)