Question

6.) (8) The Ksp for Calcium Fluoride is 1.46 x 10-10. How many milligrams of Ca2+(aq) can be present in 3.75 L of 0.25M NaF s

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Answer #1

NaF = Na+(aq) + F-(aq)

F-(aq) = 0.25 ( NaF is completely dissociable )

CaF2 = Ca2+(aq) + 2 F-(aq)

S. 2S+0.25(from NaF)

But ksp is very small so F- = 0.25

Ksp = (Ca2+)*(F-)^2

= 1.46*10^-10 = S*(0.25)^2

= 2.336*10^-9 = S

Ca2+ = 2.336*10^-9 M

Now

Moles = molarity* volume

2.336*10^-9*3.75

= 8.76*10^-9

Mass = mole * molar mass

= 8.76*10^-9 * 40.08

= 3.511*10^-7 gram

Or 3.511*10^-4 mg

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