250 mg of CaF2 added to 1L of water, [Ca++] = 2.1 x 10^-4M what Ksp?
17. 250 mg of CaF2 added to 1L of water, [Ca++] = 2.1 x 10^-4M what Ksp?
In the first place, the amount (250mg) of CaF2 does not matter excepting that there be some solid left at equilibrium. This is a simple problem.
The amount of CaF2 dissolved is s = 2.1 x 10^-4M and from the stoichiometry the concentration of F^- is 2 x [Ca++] = 2s.
Ksp = (s)(2s)^2 = 3.7 x 10^{-11} M^3.
Lets use the tabular technique to solve this problem.
Lets add an addtional part to the question. Knowing s we can compute the amount of calcium fluroide that is left. In the liter of solution the number of moles of Ca^++ = 2.1 x 10^-4M x 1L and with a molecular weight of 78., 2.1 x 10^-4M * 78 = 16.4 mg has gone into solution. The amount of ppt left is 250 - 16.4 = 233.6 mg.
37. [Pb^++] = 0.0012M to which is added enough chloride so that [Cl^-] = 0.010M. Q = (0.0012)(0.010)^2 = 1.2 x 10^{-7} << Ksp = 1.7 x 10^{-5}M^3. The solubility product has not been exceeded and so no lead chloride will ppt.
55. A solution is 0.10M in both Ba^++ and Ca^++.
The Ksp(BaF2) = 1.7 x 10^{-6}, and Ksp(CaF2) = 3.9 x 10^{-11}.
(a) What amt [F^-] will ppt max amount of CaF2 without pptn BaF2?
[F^-] = 1.7 x 10^{-6}/0.1 = 1.7 x 10^{-5}M Just slightly below this concentration the solubility product will not be exceeded for BaF2. At [F^-] = 1.7 x 10^{-5}M the Ca^++ concentration will be:
(b) [Ca^++] = (3.9 x 10^{-11})/(1.7 x 10^{-5}M)^2 = 1.7 x 10^{-7}M.
63. Keq = ? for 
.
We have the solubility products:
Subtract the second solubility product (equilibrium) from the first to obtain
the reaction with
K = Ksp(Zn(OH)_2)/Ksp(Zn(CN)_2) = 5.6 x 10^{-6}. This small value for the equilibrium
constant means that [CN^-] >> [OH^-].
It is not practical to form Zn(CN)_2 by adding CN^- to Zn(OH)_2.
70. Aragonite is more soluble than Calcite because it's Ksp is about 1.6 Ksp(Calcite).
73.
(a) [Ba++] = 0.015 M what is the minimum F^- concentration to ppt Ba++?
From (a) problem 53 Ksp = 1.7 x 10^{-6} M^3 therefore, 
(b) 100 mL of 0.015M Ba^++ How many mg of NaF needed to just begin ppt? From part (a) we need [F^-] = 0.0107M. In 100 mL this corresponds to 0.00107 moles. The Mw(NaF) = 42 g/mol so 42 g/mol x 0.00107 moles = 44.9 mg
(c) After adding 500mg of NaF to 100 mL 0.015M Ba^++ what is the [Ba^++]?
This is 0.500g/42g/mol = 0.0119 mole of F^-. Assume all the barium ion is taken down
by this excess F^-. That will leave 0.0119 - 0.0015 = 0.0104 moles F^- in the 100 mL.
Therefore, [F^-] = 0.0104moles/0.100L = 0.104 M. This is in equilibrium with the solid
barium fluoride and so [Ba^++] = Ksp/(0.104)^2 = 0.00016M.
80. 
where pH = 10.49 what Ksp?
pOH = 14 - pH = 3.51; [OH^-] = 3.1 x 10^{-4}M. [Mg^++] = s, [OH^-] = 2s = 3.1 x 10^{-4}M
Ksp = s(2s)^2 = 4s^3 = 1.48 x 10^{-11}