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Offline plu

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zinc hydroxide complex
« on: February 05, 2006, 07:40:03 PM »
Zinc hydroxide is not very soluble in water, but in basic solution, it may dissolve as the tetrahydroxozinc complex ion: Zn(OH)42-.
The solubility product constant of zinc hydroxide is Ksp = 2.1 x 10-16.
The formation constant of Zn(OH)42- is Kf = 2.8 x 1015.
a) If 150.0 mg of zinc hydroxide crystals are mixed with 250.0 mL of pure water, what mass of crystals will remain undissolved and what will be the pH of the resulting solution?  Justify your answer by showing your calculations.
b) If 150.0 mg of zinc hydroxide crystals are mixed with 250.0 mL of a 0.100 mol/L NaOH solution, what mass of crystals will remain undissolved?  Justify your answer by showing your calculations.

One of the students I'm tutoring came to me with this problem.  I got answers of 0.1499 g / pH 8.87 and 8.87 x 10-13 g for questions a and b respectively when I tried it but since there was no answer provided with the problem, we have no way are checking these numbers.  Would anybody mind trying the problem and comparing answers?  Thanks!

Offline Mitch

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Re:zinc hydroxide complex
« Reply #1 on: February 05, 2006, 08:45:13 PM »
How can you get an answer for part a? It doesn't tell you the Ksp in neutral pH?
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Offline plu

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Re:zinc hydroxide complex
« Reply #2 on: February 06, 2006, 12:14:47 AM »
The solubility product constant of zinc hydroxide is Ksp = 2.1 x 10-16.

I'm assuming they mean in neutral pH

Offline Borek

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Re:zinc hydroxide complex
« Reply #3 on: February 06, 2006, 02:37:24 AM »
I'm assuming they mean in neutral pH

Doesn't matter. Solubility product is a solubility product.
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Offline Mitch

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Re:zinc hydroxide complex
« Reply #4 on: February 06, 2006, 02:42:40 AM »
What does that mean? You mean Ksp values are not pH dependent?
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Offline AWK

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Re:zinc hydroxide complex
« Reply #5 on: February 06, 2006, 04:26:29 AM »
Solubility, but not solubility product is pH dependent. The  solubility product is a thermodynamic function that depends mainly on temperature and ionic strength, (in much less extent on pressure)
« Last Edit: February 06, 2006, 04:27:02 AM by AWK »
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Offline Borek

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Re:zinc hydroxide complex
« Reply #6 on: February 06, 2006, 08:23:40 AM »
I got answers of 0.1499 g / pH 8.87 and 8.87 x 10-13 g for questions a and b respectively

a - OK

b - 4 mg

Assuming there is some solid left, maximum concentration of Zn2+ is given by solubility product - and is 2.1*10-14M. Concentration of complex is given by Kf[Zn2+ ][OH- ]4 = 5.88*10-3 (assuming Zn2+ concentration as given by solubilty product). Thus there is 0.25*5.88*10-3*99.4=0.146 g of Zn in solution (in form of complex, free ions can be neglected). This confirms our assumption that some solid was left and gives mass of the solid as 150-146=4mg (3.9 to use 2 sd).
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Offline plu

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Re:zinc hydroxide complex
« Reply #7 on: February 06, 2006, 06:09:01 PM »
A-ha!  It has been quite some time since I have done these kinds of questions.  Thanks, mates!

All right, to summarize what I have learned here: 1) The amount of free ions from any particular compound allowed to remain in solution at equilibrium must always follow the Ksp expression for that compound; 2) When dealing with ions that complex in solution, it is best to first calculate the amount of free ions that are allowed to exist in the solution from the given Ksp values, then calculate the amount of complex in solutions.  Nice  ;)

Offline Borek

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Re:zinc hydroxide complex
« Reply #8 on: February 06, 2006, 06:38:39 PM »
A-ha!  It has been quite some time since I have done these kinds of questions.

Over 20 years in my case. No joking - I have spent some time last year on pH, but not on solubilities or complexes.

Quote
The amount of free ions from any particular compound allowed to remain in solution at equilibrium must always follow the Ksp expression for that compound;

Can't be greater than - but can be lower!

Quote
When dealing with ions that complex in solution, it is best to first calculate the amount of free ions that are allowed to exist in the solution from the given Ksp values, then calculate the amount of complex in solutions.  Nice  ;)

But later you have to check if there is solid left. If not - Ksp is of no use.
« Last Edit: February 06, 2006, 06:39:12 PM by Borek »
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