Solubility of lead(II) chloride from Ksp = 1.2×10⁻⁵

[aisha]

use you accepted value for the solubility product of lead 2 chloride to determine the solubility of lead 2 chloride in mol/L and g/L

the accepted value is ksp=1.2*10^-5

Ksp=[Pb] [Cl]^2

1.2*10^-5 = (x) (2x)^2

x=0.014 mol/L

0.014 mol/L * 276.1 g/mol
=3.98 g/L

I don't know if I did this right can someone please tell me ASAP

 

[Unco]

What did you represent with x? [Pb2+], right?

The solubility of PbCl2 is the concentration of Pb2+ at equilibrium (Cl- has a higher concentration); the molar mass of Pb2+ is 207g/mol.

 

[Renge Ishyo]

The result in the initial post seems accurate to me. The only problem I see is that solubility is traditionally -x (not just x), because it represents the amount of PbCl_2 *dissolved* per liter. Your instructor may or may not care about this technicality. Also, your value used for the molar mass was 276.2. Using a periodic table I found online I got a value of 278.14, and so I would have reported the answer as:

- .014 mol/L * 278.14 g/mol = - 3.9 g/L

I think your answer for the bottom problem had a typo in it either way because when you multiply .014 * 276.1 you get 3.86 and not 3.98 like you are reporting.

Cheers.