Question

A solution is 5 mM in each of the following ions:

number	ion	Ksp of M(OH)2
1	Mg2+	1.8e-11
2	Cd2+	2.5e-14
3	Co2+	1.6e-15
4	Zn2+	4.4e-17
5	Cu2+	2.2e-20

Indicate which of the metal ions would precipitate (or start to precipitate) at each of the following pH values. Indicate your answer with the number of the ion. Use 0 to indicate no precipitate. If more than one precipitate is expected, list the numbers in increasing order and separate them with commas. For example, 3,4,5 is ok but 5,4,3 is not.

pH = 5.00:

pH = 9.00:

What is the pH to the nearest 0.1 pH unit at which Mg(OH)₂ begins to precipitate?

pH =

Answer 1

All the given ions make M(OH)2 when react with OH- ions

The ions concentration is given as 5mM = 5*10^-3 M

1)

For Mg+2

Ksp = 1.8e-11 = 1.8*10^-11

At precipitation Ksp = [Mg+2][OH-]² = 1.8*10^-11

If Mg+2 = 5*10^-3 M

5*10^-3 * [OH-]² = 1.8*10^-11

[OH-]= sqrt(1.8*10^-11/ 5*10^-3) =  6*10^-5 M

Least molarity of OH- required to precipitate =  6*10^-5 M

So, pOH = -log(OH-) = -log( 6*10^-5) = 4.22

So, pH = 14 - pOH = 14 - 4.22 = 9.78

So, at pH = 9.8 Mg(OH)2 will just start precipitating

2)

For Cd+2

Ksp = 2.5e-14 = 2.5*10^-14

At precipitation Ksp = [Cd+2][OH-]² = 2.5*10^-14

If Cd+2 = 5*10^-3 M

5*10^-3 * [OH-]² = 2.5*10^-14

[OH-]= sqrt(2.5*10^-14/ 5*10^-3) =  2.23*10^-6 M

Least molarity of OH- required to precipitate =   2.23*10^-6 M

3)

For Co+2

Ksp = 1.6e-15 = 1.6*10^-15

At precipitation Ksp = [Co+2][OH-]² = 1.6*10^-15

If Co+2 = 5*10^-3 M

5*10^-3 * [OH-]² = 1.6*10^-15

[OH-]= sqrt(1.6*10^-15/ 5*10^-3) =  5.66*10^-7 M

Least molarity of OH- required to precipitate =   5.66*10^-7 M

4)

For Zn+2

Ksp = 4.4e-17 = 4.4*10^-17

At precipitation Ksp = [Zn+2][OH-]² = 4.4*10^-17

If Zn+2 = 5*10^-3 M

5*10^-3 * [OH-]² = 4.4*10^-17

[OH-]= sqrt(4.4*10^-17/ 5*10^-3) =  9.38*10^-8 M

Least molarity of OH- required to precipitate =   9.38*10^-8 M

5)

For Cu+2

Ksp = 2.2e-20 = 2.2*10^-20

At precipitation Ksp = [Cu+2][OH-]² = 2.2*10^-20

If Cu+2 = 5*10^-3 M

5*10^-3 * [OH-]² = 2.2*10^-20

[OH-]= sqrt(2.2*10^-20/ 5*10^-3) =  2.097*10^-9 M

Least molarity of OH- required to precipitate =  2.097*10^-9 M

a)

For precipitation at pH =5 or pOH = 14 - pH = 14-5 = 9

the least molarity of OH- required is 10^-pOH = 10^-9

If the molarity of OH- required is >10^-9 the precipitation will not occur.

Which is the case in 1,2,3,4

But precipitation will occur in 5

b)

For precipitation at pH =9 or pOH = 14 - pH = 14-9 = 5

the least molarity of OH- required is 10^-pOH = 10^-5

If the molarity of OH- required is >10^-5 the precipitation will not occur.

Which is not the case with any of the ions

So, precipitation will occur in 1,2,3,4,5