1. What mass of lead chloride (PbCl2, MW= 278.2; Ksp= 1.8 x 10^-10) is dissolved in 250 mL of a saturated solution?

Question

2. What mass lead chloride (PbCl2, MW=278.2; Ksp=1.8 x 10^-10) will dissolve in 150 mL of 0.050 M Pb(NO3)2?

Answer 1

PbCl2 ==> pb^+2 + 2Cl^-

Ksp + (Pb^+2)(Cl^-)^2
Set up ICE chart with Pb^+2 = x and Cl^- = 2x, substitute into Ksp and solve for x. x will have units of moles/L; therefore, convert moles/L to grams (g = moles x molar mass) and divide by 4 to find the mass PbCl2 in 250 mL.

For #2, you have a common ion of Pb^+2 from Pb(NO3)2.
PbCl2 ==> Pb^+2 + 2Cl^-
Ksp = etc.
Here Pb^+2 comes from two sources; i.e., PbCl2 and Pb(NO3)2. I think you can ignore the Pb^+2 from PbCl2. Substitute M Pb from Pb(NO3)2, solve for Cl^-, convert to PbCl2 (by taking half of the Cl^-), convert to grams PbCl2, (that will be grams/L), then convert to g/150 mL.
Post your work if you get stuck.

Answer 2

Thank you so much for helping, but I'm confused to why I'm diving it by 4 on the first one?

Ksp=1.8 x 10^-10=(2x)^2(x)
1.8 x 10^-10=4x^3
Divide by 4= 4.5 x 10^-11
Cube Root= 3.556893304 x 10^-4 mols/L
Times Molar Mass: 0.0989527717
Divide by 4: 0.0247381929, but why?

Answer 3

Why divide by 4? Your answer of 0.09895 grams PbCl2 (the 3.55 x 10^-4 is M is in moles/L.) gives grams/liter of solution. The problem asks for the solubility of PbCl2 in grams/250 mL and 250 mL is 1/4 L.
By the way, you have carried out the number of decimal places far to many. The Ksp has two significant figures in it; therefore, you must round the answer you have to two places. I would round 0.02474 to 0.025 g/250 mL.

Answer 4

Does any solid pbcl2 form when 3.5 mg of NaCl is dissolved in 0.25L of 0.12M Pb(NO3 )2