The maximum allowable concentration of Pb2+ ions indrinking water is 0.05 ppm (that is, 0.05 g of Pb2+ in 1million g of water). Is this guideline exceeded if an undergroundwater supply is at equilibrium wiht the mineral anglesite,PbSO4 (Ksp=1.6 x10-8)?

this aint it chief

...........PbSO4 ==> Pb^2+ + SO4^2-

I.........solid.......0........0
C.........solid.......x........x
E.........solid.......x........x

Ksp = 1.6E-6 = (Pb^2+)(SO4^2-)
Substitute the E line into the Ksp expression and solve for x = (PbSO4) in mols/L. Convert to grams Pb/L which will essentially be grams Pb in 1000 g. Compare with grams/10^6 g H2O

To determine if the guideline for the maximum allowable concentration of Pb2+ ions is exceeded, we need to calculate the concentration of Pb2+ ions in the water when it is at equilibrium with anglesite, PbSO4.

The equilibrium constant expression for the dissolution of anglesite in water is as follows:

PbSO4 ⇌ Pb2+ + SO42-

The equilibrium constant (Ksp) for this reaction is given as 1.6 x 10-8.

From the balanced chemical equation, we can see that 1 mole of anglesite (PbSO4) produces 1 mole of Pb2+ ions.

Now, let's assume the concentration of Pb2+ ions in the water is represented by [Pb2+].

According to the solubility product expression, we have:

Ksp = [Pb2+] [SO42-]

Since the concentration of Pb2+ ions is equal to the concentration of SO42- ions in the equilibrium state, we can write:

Ksp = [Pb2+]^2

Substituting the given Ksp value into the equation:

1.6 x 10-8 = [Pb2+]^2

Taking the square root of both sides:

[Pb2+] = sqrt(1.6 x 10-8)

[Pb2+] ≈ 1.26 x 10-4 M

To convert this concentration to parts per million (ppm), we multiply by the molar mass of Pb2+ (207.2 g/mol) and divide by the density of water (1 g/mL) to obtain:

[Pb2+] ≈ (1.26 x 10-4 M) * (207.2 g/mol) / (1 g/mL) ≈ 0.026 ppm

Comparing this value with the guideline of 0.05 ppm, we can see that the equilibrium concentration of Pb2+ ions when at equilibrium with anglesite does not exceed the guideline for the maximum allowable concentration of Pb2+ ions in drinking water.

To determine if the maximum allowable concentration of Pb2+ ions in drinking water is exceeded when in equilibrium with anglesite (PbSO4), we need to compare the concentration of Pb2+ ions with the given Ksp value.

To start, let's calculate the concentration of Pb2+ ions from the Ksp value of anglesite.

The balanced equation for the dissociation of anglesite is:
PbSO4(s) ⇌ Pb2+(aq) + SO4^2-(aq)

The Ksp expression for anglesite is:
Ksp = [Pb2+][SO4^2-]

Given that Ksp = 1.6 x 10^(-8), we know that [Pb2+] = x (where x is the concentration of Pb2+ ions).

Since anglesite is at equilibrium with the underground water supply, the concentration of Pb2+ ions will be equal to x. Therefore, we can substitute x for [Pb2+] in the Ksp expression:

1.6 x 10^(-8) = x * [SO4^2-]

To solve for x, we need to know the concentration of SO4^2- in the water supply. Without that information, we can't determine if the guideline is exceeded.

If you have access to the concentration of SO4^2- in the underground water supply, you can substitute it into the equation to find the concentration of Pb2+. Then, compare that concentration with the given maximum allowable concentration of 0.05 ppm (parts per million) to determine if the guideline is exceeded.