A solution contains 1.0 x 10-4 M Cu^+ and 2.0 times 10^-3 M Pb^2-. If a source o

Question

A solution contains 1.0 x 10-4 M Cu^+ and 2.0 times 10^-3 M Pb^2-. If a source of I is added to this solution, will PbI2 (Ksp = 1.4 times 10-8) or Cul (Ksp = 5.3 times 10-12) precipitate first? Specify the concentration of 1- necessary to begin precipitation of each compound. Determine the K sp of calcium fluoride (CaF2), given that its molar solubility is 2.14 times 10^-4 moles per liter. For Cu(OH)2, Ksp = 1.6 times 10^-19 What is the molar solubility of Cu(OH)2? 3.4 times 10^-7 M 6.4 times l0^-7'M 2.7 times 10^-11M 5.1 x 10^-10M 1.7 times 10^-10M The solubility of silver sulfate in water at 100 degree C is approximately 1.4 g per 100 ml.. What is the solubility product of this salt at 100 degree C? 5.7 times 10^-8 3.5 times 10^-7 8.3 times 10^-6 4.1 times 10^-5 3.6 x 10^-6 A solution has initial [Cu+2] 0.010 M and initial [NH3] = 1,000M, The complex ion [Cu(NH3)4]^+2 forms in solution with Kf = 5.0 times 10^13. Calculate the [Cu^+2] when equilibrium is achieved. What is pH of NaOH solution (v=20 ml, 0.07 M) If 13 mL of 0.09 M HC1 added to the above mixture what is the pH of the solution?

Explanation / Answer

Answer:

1).

The salt which posses lower Ksp value will precipitate first.

Among the given Ksp value of CuI (5.3*10-12) is lower than Ksp value of PbI (1.4*10-8), thus CuI will precipitate first.

2).

CaF2 Ca+2 +2F-

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

= [2.14*10-4][2*2.14*10-4]2

= 39.2

3).

Cu(OH)2 Cu+2 + 2OH-

Ksp = 1.6*10-19

1.6*10-19 = [S][2S]2

1.6*10-19 = 3S3

S3 = (16*10-18)/3

S3 = 5.33

S = 1.7467

4).

Quantity of Silver sulphate dissolved in 100 mL = 1.4 grams

Quantity of Silver sulphate dissolved in 1000 mL = 0.14grams

Converting grams of Silver sulphate into moles/lit.

0.14/311 = 4.5*10-4 mol/Lit

2Ag(SO4)2 [2Ag+]2 + [SO4-2]

Ksp = [2S]2 [S]

=[2*4.5*10-4]2[4.5*10-4]

Ksp=3.65*10-10

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