Question

What mass of Cu(OH)2 (MM= 97.56 g/mol) will dissolve in ,500 L of a pH solution (Ksp=2.2*10 ^=20). In which solution is Cu(OH)2 most soluble and least soluble: A. .01 M NaCl, B. .01M CuCl2, C. .01 HCL, or D. .01NaOH.

Answer 1

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

the solubility product constant is given by

Ksp = [Cu+2] [OH-]^2

consider the solubility of Cu(OH)2 be s

then

[Cu+2] = s

[OH-] = 2s

so

Ksp = [s] [2s]^2

Ksp = 4s^3

4s^3 = 2.2 x 10-20

s = 1.765 x 10-7 mol/L

so

the solubility of Cu(OH)2 is 1.765 x 10-7 mol/L

now

given

volume = 500 L

so

moles of Cu(OH)2 soluble = (solubility in mol/ L ) ( volume in L)

moles of Cu(OH)2 soluble = 1.765 x 10-7 x 500

moles of Cu(OH)2 soluble = 8.82587 x 10-5 mol

now

we know that

mass = moles x molar mass

so

mass of Cu(OH)2 soluble = 8.82587 x 10-5 x 97.56 g

mass of Cu(OH)2 soluble = 8.61 x 10-3 g

mass of Cu(OH)2 soluble = 8.61 mg

so

8.61 milligrams of Cu(OH)2 will dissolve in 500 L of the solution

2)

Cu(OH)2 is the most soluble in C) 0.01 M HCl

Cu(OH)2 is the leat soluble in D) 0.01 M NaOH

this is is because

H+ from HCl will remove OH- ions from the solution

so according to Lechatlier principle

Cu(OH)2 will be more soluble in the solution

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

according to Le Chatiler the equilibrium will shift in a direction to counter the change

here

as OH- ions are being remove , the system will try to increase OH- ions

as as result it will dissolve to give more OH- ions

now

OH- from NaOH will increase the OH- ions in the solution

so according ot Le Chatlier Principle

less Cu(OH)2 will be soluble in the solution