Question

The maximum allowable concentration of Pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of Pb2+ in 1 million grams of water). (a) What is the Pb2+ concentration of an underground water supply at equilibrium with the mineral cerussite (PbCO3) (Ksp = 3.3 × 10−14) ? × 10 g / LEnter your answer in scientific notation. (b) Does this concentration exceed the allowable concentration guideline? Yes No

Answer 1

Answer:

a. 0,049g of Pb²⁺ in 1 million grams of water.

b. No

Explanation:

The equilibrium of PbCO₃ in water is:

PbCO₃ ⇄ Pb²⁺ + CO₃²⁻

Where ksp is defined as:

ksp = [Pb²⁺] [CO₃²⁻] (1)

The initial concentration of PbCO₃ in molarity is:

$\frac{10g}{1L}$×$\frac{1mol}{267,21g}$= 0,0374M

a. The concentrations in equilibrium are:

[PbCO₃} = 0,0374M-x

[Pb²⁺] = x

[CO₃²⁻] = x

Replacing in (1)

3,3x10⁻¹⁴ = x²

x = 1,82x10⁻⁷M

Thus, [Pb²⁺] = 1,82x10⁻⁷M, this value in g of Pb²⁺ per million grams of water is:

1,82x10⁻⁷M×$\frac{267,21g}{1mol}$×$\frac{1x10^6g}{1000g}$ = 0,049 g of Pb²⁺ in 1 million grams of water

As this value is lower than the maximum allowable concentration of Pb²⁺, the concentration doesn't exceed this maximum value

I hope it helps!