Question

A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added until the pH of the solution is 8.50. Which of the following is true?

Pb(OH)2, Ksp= 1.4 x10^-20

Mn(OH)2, Ksp= 2.0 x 10^-13

Zn(OH)2, Ksp= 2.1 x10^-16

A) No precipitate will form.

B) Only Pb(OH)2 will precipitate.

C) Only Mn(OH)2 wil precipitate.

D) Only Zn(OH)2 and Pb(OH)2 will preipitate.

E) All three hydroxide will preipitate.

Answer 1

ANSWER : D) only Zn(OH)2 and Pb(OH)2 will precipitate.

We have relation, pH + pOH = 14

pOH = 14 - pH = 14 -8.5 = 5.5

We have relation, pOH = - log [ OH - ]

[ OH - ] = 10 -pOH

[ OH - ] = 10 -5.5 = 3.16 10 -06 M

We know that precipitation takes place when Ionic product > solubility product.

To check precipitation , we need to calculate ionic product.

Consider reaction Pb(OH) 2 (s) Pb 2+ (aq) + 2 OH - (aq)

For above reaction , K sp = [ Pb 2+ ] [ OH - ] 2 = 1.4 10 -20

We have, Q sp = [ Pb 2+ ] [ OH - ] 2 = ( 0.010 ) ( 3.16 10 -06 ) 2

Q sp = 9.98 10 -14

Here Ionic product ( 9.98 10 -14 ) > Solubility product ( 1.4 10 -20 ) , hence precipitate will be formed.

Consider reaction Mn(OH) 2 (s) Mn 2+ (aq) + 2 OH - (aq)

For above reaction , K sp = [ Mn 2+ ] [ OH - ] 2 = 2.0 10 -13

We have, Q sp = [ Mn 2+ ] [ OH - ] 2 = ( 0.010 ) ( 3.16 10 -06 ) 2

Q sp = 9.98 10 -14

Here Ionic product ( 9.98 10 -14 ) < Solubility product ( 2.0 10 -13 ) , hence precipitate will not be formed.

Consider reaction Zn(OH) 2 (s) Zn2+ (aq) + 2 OH - (aq)

For above reaction , K sp = [ Zn 2+ ] [ OH - ] 2 =2.1 10 -16

We have, Q sp = [ Zn 2+ ] [ OH - ] 2 = ( 0.010 ) ( 3.16 10 -06 ) 2

Q sp = 9.98 10 -14

Here Ionic product ( 9.98 10 -14 ) > Solubility product ( 2.1 10 -16 ) , hence precipitate will be formed.