A solution is 5 mM in each of the following ions: My N number lon Ksp of M(OH)2

Question

A solution is 5 mM in each of the following ions: My N number lon Ksp of M(OH)2 1Mg21.8e-11 2 Cd22.5e-14 3 Co2+ 1.6e-15 4 Zn2+ 4.4e-17 Cu 2.2e-20 Indicate which of the ion. Use 0 to indicate example, 3,4,5 is ok but 5,4,3 is not. metal ions would precipitate (or start to precipitate) at each of the following pH values. Indicate your answer with the number of the no precipitate. If more than one precipitate is expected, list the numbers in increasing order and separate them with commas. For pH 7.00: PH- 8.00: What is the pH to the nearest 0.1 pH unit at which Co(OH)2 begins to precipitate? pH -

Explanation / Answer

1)

Answer

pH = 7 4,5

pH= 8 3,4,5

Explanation

a) pH =7

pOH = 14-7 = 7

pOH = - log[OH-]

-log[OH-] = 7

[OH-] = 1×10-7

if Qsp> Ksp, then precipitation will form

i)Mg2+

Qsp = [Mg2+][OH-]2

Qsp = 0.005M ×(1×10-7M)2

Qsp = 5.0×10-17

Qsp < Ksp

So, precipitation will not form

ii)Cd2+

Qsp = [Cd2+] [OH-]2

Qsp= 0.005M × (1×10-7M)2

Qsp = 5.0×10-17

Qsp < Ksp

Precipitation will not form

iii)Co2+

Qsp <Ksp

precipitation will not form

iv) Zn2+

Qsp > Ksp

Precipitation will form

v) Cu2+

Qsp > Ksp

precipitation will form

b) pH = 8

pOH = 14 - 8

pOH = 6

[OH-] = 1×10-6

Qsp = 0.005M ×(1×10-6M)2

Qsp = 5.0×10-15

i) Mg2+

Qsp < Ksp

precipitation will not form

ii) Cd2+

Qsp < Ksp

precipitation will not form

iii) Co2+

Qsp > Ksp

precipitation will form

iv) Zn2+

Qsp > Ksp

precipitation will form

v) Cu2+

Qsp > Ksp

precipitation will form

2)

Answer

at pH 7.8, Co(OH)2 begins to precipitate

Explanation

Solubility equillibrium of Co(OH)2 is

Co(OH)2 <--------> Co2+(aq) + 2OH-(aq)

Ksp =[Co2+][OH-]2 = 1.6×10-15M3

[Co2+][OH-]2 = 1.6×10-15M3

substituting the value of [Co2+]

0.005M × [OH-]2 = 1.6×10-15M3

[OH-]2 = 3.20×10-13M2

   [OH-] = 5.66×10-7M

pOH = - log(5.66×10-7) = 6.25

pH = 14 - 6.25

pH = 7.75

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