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When the solutions of BaCl2 and Na2SO4 mixed, precipitation of BaSO4 was produced....

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shihan | Student, Undergraduate | Salutatorian

Posted July 25, 2013 at 4:52 AM via web

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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.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 25, 2013 at 7:05 AM (Answer #1)

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`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:

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shihan | Student, Undergraduate | Salutatorian

Posted July 25, 2013 at 7:38 AM (Answer #2)

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Thank you :)

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