One way lead of preventing lead from dissolving into water is by the addition of

Question

One way lead of preventing lead from dissolving into water is by the addition of phosphate inhibitors that prevent the lead from dissolving. Lead (II) phosphate (K_sp = 1.6 middot 10^-32) is one mineral that can be precipitated out of solution to help suppress the lead from dissolving the pipes. a. What is the molar solubility of Lead (II) phosphate in water? b. What is the lead (II) concentration in pure water in mg/L? c. What is the concentration of lead (II) in mg/L, when the water is treated with a 1.0 millimolar, 1mM, of a phosphate salt? d. If the lead concentration exceeds the action level of 0.015 mg/L, does the water require further treatment for either parts b or c?

Explanation / Answer

Q5.

a)

molar solubility of Lead (II) phoasphate:

Pb3(PO4)2 <-> 3Pb+2 + 2PO4-3

Ksp = [Pb+2]^3 [ PO4-3]^2

assume S = Solubility of 1 mol of Pb3(PO4)2

[Pb+2] = 3S

[PO4-3] = 2S

1.6*10^-32 = (3S)^3(2S)^2

27*s^2* 4*S^2 = 1.6*10^-32

108 * S^5 = 1.6*10^-32

S= ((1.6*10^-32)/108 ) ^(1/5)

S = 1.71450*10^-7 mol of Pb3(PO4)2 per liter

b)

in mg /L:

1 mol of Pb3(PO4)2 --> 811.542722 g

so

1.71450*10^-7 mol of Pb3(PO4)2 per liter * 811.542722 g / mol --> (1.71450*10^-7 )(811.542722) * 1000 = 0.1391389 mg per liter

c)

if we add

PO4-3 --> 1mM = 1*10^-3 M

then:

Ksp = [Pb+2]^3 [ PO4-3]^2

1.6*10^-32 = (3S)^3 * (1*0^-3)^3

27S^3 = (1.6*10^-32) / ((10^-3)^3) = 1.6*10^-23

S = ((1.6*10^-23) / 27 ) ^(1/3)

S = 8.39947*10^-9 M

in mass:

S = (8.39947*10^-9)(1000)*(811.542722 ) = 0.0068165 mg/L

d)

For B... it does exceeds

For C, it does NOT exceeds ( thats the point of adding the salt)

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