1)Mercury (II) hydroxide is quite insoluble in water (K sp of Hg(OH) 2 = 3.0×10

Question

1)Mercury (II) hydroxide is quite insoluble in water (Ksp of Hg(OH)2 = 3.0×10-26 at 25°C.In which one of the following solutions would you expect Hg(OH)2 to be even less soluble than it is in pure water? Why?

(a) 0.1 M NaCl                   (b) 0.1 M HCl                     (c) 0.1 M NaOH                 (d) 0.1 M NaCN

2)How will the solubility of calcium fluoride differ in acidic water compared to neutral water? Why?

(a) It will be more soluble in acidic water

(b) It will be less soluble in acidic water

(c) The solubility in acidic water and neutral water will be the same

Calculate the Ksp for Mg3(PO4)2, given that its molar solubility is 3.57·10-6 mol/L.

Explanation / Answer

1 ) answer : (c) 0.1 M NaOH

explanation:

Hg (OH)2 <-----------------> Hg+2 + 2OH-

NaOH ---------------------> Na+    + OH-

due to common ion the equilibrium shifts to the left so solubility of Hg(OH)2 decreases

2) answer : (a) It will be more soluble in acidic water

CaF2 <----------------> Ca+2   + 2F-

HCl <---------------->   H + Cl

H + F- --------------> HF

so F- ions are removed by acid H+ . so that will be decrease in concentraion of F- in acid .

3)

Mg3(PO4)2 ----------------> 3 Mg+2   + 2 PO4-3

                                         3S               2S

Ksp = (3S)^3 (4S)^2 = 108 S^5

3.57 x 10^-6 = 108 S^5

S = 0.032 M

solubility   = 0.032 M

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