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`350`  mL of 3.2 M Pb(NO3)2 and `200`  mL of 0.020 M NaCl are added together. Ksp for...

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roshan-rox | Valedictorian

Posted August 13, 2013 at 6:39 PM via web

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`350`  mL of 3.2 M Pb(NO3)2 and `200`  mL of 0.020 M NaCl are added together. Ksp for the lead chloride is `1.6xx 10^(-5).`

 Will precipitation of PbCl2 occur?  

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

Posted August 13, 2013 at 7:29 PM (Answer #1)

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`Pb(NO_3)_2rarr Pb^(2+)+2NO_3^-`

Amount of `Pb^(2+)` moles = `(3.2/1000)xx350` = `1.12 mols`

`NaCl rarr Na^+Cl^-`

Amount of `Cl^-` `= (0.02/1000)x200 = 0.004 mols`

When we mix these together volume will become (350+200) = 550. So the concentration of `Pb^(2+)` and `Cl^-` will change.

`[Pb^(2+)] = (1.12/550)xx1000 = 2.036M`

`[Cl^-] = (0.004/550)xx1000 = 0.0073M`

`PbCl2 harr Pb^(2+)+2Cl^-`

`Ksp = [Pb^(2+)][Cl^-]^2`

For the current solution;

`[Pb^(2+)][Cl^-]^2 = (2.036)(0.0073)^2 = 1xx10^-4`

but Ksp of `PbCl_2` is `1.6xx10^-5` .

`[Pb^(2+)][Cl^-]^2>Ksp`

So PbCl2 will precipitate.

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