Question
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?
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?
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?
Answers
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.
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.
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?
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?
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.
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.
Does any solid pbcl2 form when 3.5 mg of NaCl is dissolved in 0.25L of 0.12M Pb(NO3 )2
Related Questions
Zn + PbCl2 --> ZnCl2 + Pb
The initial mass of a zinc strip is 2.684. After being left overnight...
In an experiment to determine the solubility of lead chloride (PbCl2), 5.6g of (NH4)2SO4 was added t...
What mass of lead (v) trioxonitrate (v) pb (Na3)2 would be required to yield a gram of lead (v) chlo...