8. A lake in the Swiss Alps has an alkalinity of 20 mg/L as CaCO 3 . How many mo

Question

8. A lake in the Swiss Alps has an alkalinity of 20 mg/L as CaCO 3 . How many moles/L of acid could be neutralized by the water in this lake? 9. A water supply has very hard water with a calcium ion (Ca ^2+ ) concentration of 300 mg/L. With such hard water, what is the highest concentration of dissolved fluoride (in mg/L) that could be provided in the finished water if no softening is performed? The K sp for CaF2 is 3x10^-11 . Note that the beneficial concentration of fluoride is about 1 ppm, and it can cause health problems above 4 ppm.

Explanation / Answer

8)

C032- + H+ ----> HC03-

HC03- + H+ ----> H2C03

so

1 mole of C032- will neutralize 2 moles of acid

[H+] = 2 x [C032-]

given 20 mg/L of CaC03

we know that

molar mass of CaC03 = 100 g

so

conc of CaC03 = conc in g/L / molar mass

= 20 x 10-3 / 100

= 2 x 10-4

so

conc of CaC03 = 2 x 10-4 mol / L

now

[H+]= 2 x [C032-]

= 2 x 2 x 10-4

= 4 x 10-4

[H+] = 4 x 10-4

so 4 x 10-4 mol / L of acid is neutralized

9)

CaF2----> Ca+2 + 2F-

Ksp = [Ca+2] [F-]^2

given

[Ca+2] = 300 mg/L

conc in mol/L = conc in g/L / molar mass

[Ca+2] = 300 x 10-3 / 40

= 0.0075

[Ca+2] = 0.0075 mol/L

now

Ksp = [Ca+2] [F-]^2

3 x 10-11 = 0.0075 x [F-]^2

[F-] = 6.32 x 10-5 mol/L

now

conc if g/L = molar mass x conc in mol/L

= 19 x 6.32 x 10-5

= 1.2 x 10-3 g/L

= 1.2 x 10-3 x 1000 mg/L

= 1.2 mg/L

so

the conc of fluoride is 1.2 mg/L

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