Question

"Hard" water contains alkaline earth cations such as Ca^2+, which reacts with Co3^2- to form insoluble deposits of CaCO3. Will a precipitate of CaCO3 form if a 250 mL sample of hard water having [Ca^2+] = 8.0 x 10^-4M is treated with the following: (a) 0.10 mL of 2.0 x 10^-3 M Na2CO3 (b) 10 mg of solid Na2CO3.

Answers the following questions (fill in the blanks):

For part (a) IP = ________ (answer in 2 significant figures).    A precipitate forms (yes or no) ________?

For part (b) IP = _________ (answer in 2 significant figures).    A precipitate forms (yes or no)________?

Answer 1

Given data:

[Ca²⁺] = 8*10^-4 M

The volume of hard water = 250 mL

Therefore, the no. of mmol of Ca²⁺ = 8*10^-4 M * 250 mL, i.e. 2*10^-1 mmol

(a) The no. of mmol of CO₃^2- = 0.1 mL * 2.0 x 10^-3 M, i.e. 2*10^-4 mmol (which is added to hard water)

Now, the concentration of Ca²⁺ in the hard water = 2*10^-1 mmol / 250.1 mL, i.e. 7.997*10^-4 M

And the concentration of CO₃^2- in the hard water = 2*10^-4 mmol / 250.1 mL, i.e. 7.997*10^-7 M

Therefore, the ionic product of CaCO₃ in the hard water (IP) = [Ca²⁺] * [CO₃^2-]

= 7.997*10^-4 M * 7.997*10^-7 M

i.e. IP = 6.39*10^-12 M²

The solubility product of CaCO₃ (Ksp) = 3.3*10^-9 M²

Here, IP < Ksp

Hence, a precipiate doesn't form (no).

(b) The no. of mmol of CO₃^2- = 10 mg /106 g mol^-1, i.e. 9.43*10^-2 mmol

Now, the concentration of Ca²⁺ in the hard water = 2*10^-1 mmol / 250 mL, i.e. 8*10^-4 M

And the concentration of CO₃^2- in the hard water = 9.43*10^-2 mmol / 250 mL, i.e. 3.77*10^-4 M

Therefore, the ionic product of CaCO₃ in the hard water (IP) = [Ca²⁺] * [CO₃^2-]

= 8*10^-4 M * 3.77*10^-4 M

i.e. IP = 3.02*10^-7 M²

The solubility product of CaCO₃ (Ksp) = 3.3*10^-9 M²

Here, IP > Ksp

Hence, a precipiate forms (yes).