[chrisf]

10.0mL of 0.10M Pb(NO3)₂ is mixed with 10.0mL of 0.10M NaCl. After equilibrium is attained, it is determined that the concentration of Cl^- in solution is 0.0218M. Calculate the K_sp of PbCl₂.

I've written out the reaction.

Pb(NO₃)₂ + 2NaCl -> PbCl₂ + 2NaNO₃

Moles of NaCl = 0.10mol/1L * 0.010L = 0.001mol NaCl or 0.001mol Na⁺ and 0.001mol Cl^-

Moles of Cl^- at equilibrium = 0.0218mol/1L * 0.020L = 0.000436mol Cl^-

Moles of Cl^- reacted = 0.001mol Cl^- - 0.000436mol Cl^- = 0.000564mol Cl^-

Moles of Pb reacted = 0.000436mol Pb⁺/2 = 0.000218mol Cl^-

Moles of Pb left in solution = 0.001mol Pb⁺ - 0.000218mol = 0.000782mol Pb⁺

K_sp = [Pb⁺][Cl^-]²

When I substitute all this back in, I get K_sp = 3.72x10^-11 which is nowhere near my expected answer (I looked it up) of 1.7x10^-5.

Could someone please enlighten me where I'm going wrong?

Thanks!

[Borek]

Show what you have multiplied, most likely you forgot to convert from moles to concentrations.

[chrisf]

Ah hah! That was probably it...

[Cl^-] = 0.000436mol/0.020L = 0.0218M
[Pb²⁺] = 0.00782mol/0.020L = 0.0391M

K_sp = (0.0391)(0.0218)² = 1.86x10^-5

That's pretty close... hehe.

[Borek]

Moles of Cl^- reacted = 0.001mol Cl^- - 0.000436mol Cl^- = 0.000564mol Cl^-

Moles of Pb reacted = 0.000436mol Pb⁺/2 = 0.000218mol Cl^-

Think it over.

[chrisf]

Moles of Cl^- reacted = 0.001mol Cl^- - 0.000436mol Cl^- = 0.000564mol Cl^-
Moles of Pb reacted = 0.000564mol Pb⁺/2 = 0.000282mol Cl^-

 

[vhpk]

He means that you must use the " charge reservation " rule , I suppose, that is:
[Cl^-].1 + [NO₃^-].1 = [Na⁺].1 + 2[Pb²⁺]
Solve the equation to find out the concentration of Pb2+ left. Hence find the Ksp

[Borek]

No need for playing with charge.

Chris: you have already corrected number when quoting my post, they are OK now. Calculate how much Pb²⁺ was left and finish your calculations.

[chrisf]

If I do it that way, I come out with a K_sp that's nothing near the "accepted" K_sp. Hmm..

[Borek]

Show your math.

[chrisf]

Show your math.

Nevermind. 

Moles of Pb left in solution = 0.001mol Pb+ - 0.000282mol = 0.000718mol Pb+
[Pb2+] = 0.000718mol/0.020L = 0.0359M

Ksp = (0.0359)(0.0218)2
Ksp = 1.70x10-5