Question

In modern construction, walls and ceilings are constructed of "drywall", which consists of plaster sandwiched between sheets of heavy paper. Plaster is composed of calcium sulfate, CaSO4·2H2O. Suppose that you had a leak in a water pipe that was dripping water on a drywall ceiling 0.36 inches thick at a rate of 1.17 L per day. Use the Ksp of calcium sulfate to estimate how many days it would take to dissolve a hole 1.20 cm in diameter. Assume that the density of the plaster is 0.97 g cm-3. days

Answer 1

Plaster is composed of calcium sulfate, CaSO₄·2H₂O

The volume to be dissolved is obtained from

dissolve a hole 1.20 cm in diameter so radius is 0.6 cm

0.36 inches thick is 0.914 cm thick

so we need to dissolve a cylinder of CaSO₄·2H₂O

volume of a cylinder is got from the formula

V=r²h

V= 3.14 x (0.6)² x 0.914

V = 1.03 cm³

density of plaster is 0.97 g /cm³

so weight of CaSO₄.2H₂O that has to be dissolved is 0.97 g/cm³ x 1.03 cm³ = 1 g

This means (1/172.1) x 136.1 = 0.79 g of CaSO₄. (MW of CaSO4 2H2O is 172.1 and that of CaSO₄ is 136.1)

Ksp of CaSO₄ is reported to be 2.4 x 10^-5g/L

The dissolution of calcium sulfate forms equal amounts of calcium ions and sulfate ions according to the following equation.

CaSO₄ Ca²⁺(aq) + SO₄^2-(aq)

If we let [Ca²⁺] = [SO₄^2-] = x , then we can solve for the Molarity of calcium sulfate that would be soluble

Ksp = [Ca²⁺][SO₄^2-]

2.4 x 10^-5 = (x)(x)

solving for x we get [Ca²⁺]= [SO₄^2-] = 4.90 x 10^-3 M

We can assume that 4.90 x 10^-3 moles of CaSO₄ will dissolve.

MW of CaSO₄ = 136.1 g/mol,

Which means

(4.90 x 10^-3 mol/L) x (136.1 g/mol) = 0.666 g/L

So 1 L of water will be able to dissolve 0.666 g/L.

The amount of water that drip from the pipe is 1.17 L per day. This will dissolve

0.666 g/L x 1.17 L/day = 0.78 g/day

The total CaSO4 that has to be dissolved to make a hole of 1.20 cm in diameter is 0.79 g that would take

0.79/0.78 = 1.01 days to accomplish.