1. The solubility product of PbBr 2 is 8.9 ? 10 -6 . Determine the molar solubil

Question

1. The solubility product of PbBr2 is 8.9 ? 10-6. Determine the molar solubility in the following.

(a) pure water

(b) 0.12 M KBr solution

(c) 0.15 M Pb(NO3)2 solution

2. As a result of mineral erosion and biological activity, phosphate ion is common in natural waters, where it often precipitates as insoluble salts, such as Ca3(PO4)2. Calculate Qsp when [Ca2+]init = [PO43-]init = 4.609 10-6M in a given river.
Will Ca3(PO4)2 precipitate?  Ksp of Ca3(PO4)2 = 1.2 ? 10-29. 1. The solubility product of PbBr2 is 8.9 ? 10^-6. Determine the molar solubility in the following. (a) pure water (b) 0.12 M KBr solution (c) 0.15 M Pb(NO3)2 solution 2. As a result of mineral erosion and biological activity, phosphate ion is common in natural waters, where it often precipitates as insoluble salts, such as Ca3(PO4)2. Calculate Qsp when [Ca2+]init = [PO43-]init = 4.609 x 10^-6M in a given river. Will Ca3(PO4)2 precipitate? Ksp of Ca3(PO4)2 = 1.2 ? 10^-29.

Explanation / Answer

1. The solubility product of PbBr2 is 8.9*10-6.

Molar Solubility S
(a) pure water
PbBr2 = Pb(2+) + 2Br-
Ksp = S*(2*S)^2 = 4*S^3 = 8.9*10-6
S = 1.31*10^-2 M

(b) 0.12 M KBr solution
PbBr2 = Pb(2+) + 2Br-
Ksp = S*(2*S+0.12)^2
S << 0.12. So, S*(2*S+0.12)^2 approximately S*(0.12)^2 = 0.0144S = 8.9*10-6
S = 0.06*10^-2 M

(c) 0.15 M Pb(NO3)2 solution
PbBr2 = Pb(2+) + 2Br-
Ksp= (S+0.15)*(2*S)^2
S << 0.15. So, (S+0.15)*(2*S)^2= 0.15*4*S^2 = 8.9*10-6
S = 0.385*10^-2 M

2.

(i)
Calculate Qsp when
[Ca2+]init = [PO43-]init = 4.609*10-6M in a given river.
Ca3(PO4)2 = 3Ca(2+) + 2PO4(3-)

Qsp = ([Ca2+]init)^3*([PO43-]init)^2 = ([Ca2+]init)^3*([PO43-]init)^2 =
(4.609*10-6)^5 = 2079*10^-30

(ii)
Ksp of Ca3(PO4)2 = 1.2*10-29
Qsp > Ksp
So Ca3(PO4)2 will precipitate.

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