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

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:

×= 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×× = 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!

Explanation:

The formula for a trapezoid is  .

So once we put in the given data from the image, that gives us .

First, add the numerators.

16.5+6.5= 23

Now, divide the fraction.

23/2= 11.5

Finally, multiply.

11.5×7= 80.5