Tap water often contains many ions, such as calcium and iron. Such water is call

Question

Tap water often contains many ions, such as calcium and iron. Such water is called “hard” water. In addition, fluoride ions are often deliberately added to the water supply in order to prevent tooth decay. Typically, the concentration of Ca2+ ions in tap water is 8.00 ppb (by mass), and the concentration of F- ions is 1.00 ppb (by mass). Using these concentrations, calculate Q (the reaction quotient) of the reaction where solid CaF2 dissolves to form Ca2+ and F- ions. Based on this value of Q, would any solid CaF2 precipitate from tap water?

Explanation / Answer

CaF2 <--< Ca+2 + 2F-

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

[Ca+2] = 8 microgram per liter = 8*10^-6 g / L

molarity of Ca+2 = mass/MW = (8*10^-6)/40.078 =1.9961*10^-7 M

[Ca+2] = 1.9961*10^-7 M

for F_

[F-] = 1 microgram per liter = 10^-6 g / L

MW of F = (10^-6)/18.998403 = 5.26360*10^-8 M

now

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

Q = (1.9961*10^-7 )(5.26360*10^-8)^2 = 5.53*10^-22

KSp = CaF2 3.45×10–11

since Q << KSp, no precipitate forms

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.