A solution is .01 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is a

Question

A solution is .01 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

Options:

A) No precipitate will form.

B) Only Pb(OH)2 will precipitate.

C) Only Mn(OH)2 wil precipitate.

D) Only Zn(OH)2 will preipitate.

E) All three hydroxide will preipitate.

*I tried to answer by multiplying the concentrations together like this (.01)(10^-8.50)^2. I got A for my answer and would like confirmation that I am doing this question correctly or not.

Please show work!! No points/credit without work!!

Explanation / Answer

Pb(NO3)2 (aq) + 2 NaOH (aq) = Pb(OH)2 (s) + 2 NaNO3 (aq)

Mn(NO3)2 (aq) + 2 NaOH (aq) = Mn(OH)2 (s) + 2 NaNO3 (aq)

Zn(NO3)2 (aq) + 2 NaOH (aq) = ZN(OH)2 (s) + 2 NaNO3 (aq)

Pb(OH)2=Pb+2 + 2[OH-] Considering [Pb+2]=0.1 M and [Mn+2]=0.1 M,[Zn+2]=0.1

given pH =14- pOH

       pOH=14-8.50=5.5

[OH-]=10^-pOH=10^-5.5

also since

Q1=[Pb+2 ][OH-]^2=0.01x(10^(-5.5))^2=1X10^-13

Q2=[Mn+2 ][OH-]^2=0.01x(10^(-5.5))^2=1X10^-13

Q3=[Zn+2 ][OH-]^2=0.01x(10^(-5.5))^2=1X10^-13

for Pb(OH)2 given Ksp=1.4 x10^-20 and therefore Q1= 10^-13 here Q1 >Ksp

or Mn(OH)2 given Ksp= 2.0 x 10^-13 and therefore Q2= 10^-13 here Q2<Ksp

or Zn(OH)2 given Ksp= 2.1 x10^-16 and therefore Q3= 10^-13   here Q3>Ksp

If Q> Ksp equilibrium shifts to the right and a precipate will be wrong

NOTE: There are volume values missing in the question please check at repost the question.

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