[webassignbuddy]

a) Neutral pure water:

CaF₂  Ca²⁺ + 2 F^-

Ksp = [Ca²⁺][F^-]²
Ksp = (x)(2x)²
Ksp = 4x³
(Ksp/4)^1/3 = x
(3.9 x 10^-11/4)^1/3 = x
2.136 x 10^-4 = x

[F^-]= 2x
[F^-]= 2(2.136 x 10^-4)
[F^-] = 4.27 x 10^-4 (CORRECT)

b) "Hard water" where the [Ca²⁺] = 0.074 M



  I don't understand what to do. I tried plugging 0.074 M into Ksp = [Ca²⁺][F^-]² and tried solving for [F^-] but apparently it's not that simple because the answer for b is "supposed to be the same as A"

[Borek]

I tried plugging 0.074 M into Ksp = [Ca²⁺][F^-]² and tried solving for [F^-]

That's the correct approach, no doubt about it.

[webassignbuddy]

I tried plugging 0.074 M into Ksp = [Ca²⁺][F^-]² and tried solving for [F^-]

That's the correct approach, no doubt about it.

My teacher hasn't released an answer key yet but i tried googling the problem and one person with a similar problem said the answer was supposed to be the same as A.

Which doesnt make sense

[webassignbuddy]

In the meantime, how would I go about answering this question:



I tried making a table but it's proven to be a little more difficult than I thought it would be.

Table calculations

• 25 ml x (0.35 mmol/mL) = 8.75 mmol RWW
• Hg molar mass = 200.59 g = 200590 mg
  510 mg x (1 mmol/200590 mg) = 0.00250 mmol Hg

--------- Hg²⁺ ---- + ---- 2 RWW --- ------- Hg(RWW)₂²⁺
i-------(0.0025)------------(8.75)---------------------(0) mmol
Δ--------(-x)-----------------(-x)---------------------(+x) mmol
f------(0.0025-x)---------(8.75-x)--------------------(x) mmol

That doesn't look right....