Question

A fluoridated water supply contains 5 mg/L of F-. What is the maximum amount of Ca+, expressed in grams per liter, that can exist in this water supply?

Ksp(CaF2) = 3.9x10^-11

amount of Ca+ = ?Ag/L

Answer 1

Answer:

Given data,

To determine the maximum amount of Ca+

Ksp of CaF2 = 3.9 x 10^-11

Conc. of F- = 5 mg/L

We know that,

Molar Mass of F- = 19 g/mol

Concentration of F- in mol/L = ( Conc. of F- in mg/L ) / ( 1000 × Molar Mass of F-)  

substitute known values

= ( 5 ) / ( 1000 × 19 )  

= 2.632 × 10^-4 mol / L  

consider reaction which is given as follows

CaF2 ----> Ca2+ + 2F-

So , Ksp = [Ca2+] [F-]²   

3.9 × 10^-11 = [Ca2+] ( 2.632 × 10^-4 )²

[Ca2+] = 0.2873 × 10^-3 mol / L  

Molar Mass of Ca2+ = 40 g/mol

now consider,

Conc. of Ca2+ in g/L = Conc. of Ca2+ in mol/L × Molar Mass of Ca2+

= 5.6316 × 10^-4 × 40  

Concentration = 0.0225 g/L