[escapeartistq]

Could someone please show me how to solve this problem?

To 20 cm³  of a aqueous solution of Al³⁺, concentration 1.0 * 10^-2 mol/dm^-3 are added sodium hydroxide in order to maintain the pH of the solution at 6.0

a) justify the formation of a precipitate.

b) calculate the concentrations of the ions in the solution and the mass of the obtained precipitate.

Ks (Al(HO)₃) = 2.0 * 10 ^-32

Thanks in advance. I could solve part a) but I am having trouble solving b)

[Grumples]

I'm sorry, but for me, that's totally illegible.  And for the record, please don't say something like 1.0 e-2 M : that looks very much like you are talking about a one molar solution of electrons, which is impossible.  Instead, please say 1.0 * 10^-2.  I know it takes more time to write, but it takes much less time to read and understand.  To put in the superscript and subscript, look above the text box where it says "add UBBC tags".

[escapeartistq]

Sorry. It was my first post and English is not my mother tongue :-(
I edited the post so, hopefully, it looks better now.
Thanks.

[Borek]

a) justify the formation of a precipitate.

What is there to justify? It just precipitates:) Perhaps it is worth to add that Al is amphoteric so the precipitate will get dissolved at higher and lower pH.

calculate the concentrations of the ions in the solution and the mass of the obtained precipitate.

Write down the equation for the Kso, calculate pOH from known pH. Ask youself what is known and what is unknown. Rest should be a breeze.

[Donaldson Tan]

grumples: i agree escapeartistq shouldn't use 1e-2. However, the correct notation should be 1E-2  because using the small e would suggest the exponential function. 1E-2 is the standard shorthand for 1 X 10^-2

[escapeartistq]

You can show quantitatively that it precipitates by calculating Q = |Al³⁺| |HO^-|³ = (1.0 * 10 ^-2) * (1.0 * 10 ^-8)³ = 1.0 * 10 ^-26 > Ks

but from here on I am lost.

The equations involved in the equilibrium are:

Al(H₂O)₆³⁺(aq) <=> Al(HO)(H₂O)₅²⁺(aq) + H⁺(aq) (1)

Al(HO)₃(s) <=> Al³⁺(aq) + 3HO^-(aq) (2)

NaHO(s) -> Na⁺(aq) + HO^-(aq) (3)

Al(HO)₃(H₂O)₃(s) + HO^-(aq) <=> Al(HO)₄(H₂O)^-(aq) + H₂O(l) (4)

Equation (2) shows that the addition of NAHO moves the equilibrium to <- and equation (4) shows that the added HO^- and the existing Al³⁺ will be consumed moving the equilibrium to ->

Am I right?  How do I proceed with the calculations, because I am getting wrong values.


The answer to the problem says that:
|Al³⁺| = 2.0 * 10 ^-8 M
|HO^-| = |Na⁺| = 1.0 * 10 ^-8 M
precipitate mass = 1.6 * 10 ^-2 g

[Borek]

However, the correct notation should be 1E-2  because using the small e would suggest the exponential function. 1E-2 is the standard shorthand for 1 X 10^-2

Define what do you mean by 'correct notation'. C (as well as FORTRAN) program output can use e or E depending on the decision of the programmer - none of the outputs is referred to as 'correct' or 'better' in the books and manuals I have here. Pascal uses capital E. OpenOffice uses capital E. So probably it is only a matter of convention, although E seems to be used more often.

[Borek]

Am I right?  How do I proceed with the calculations, because I am getting wrong values.

Don't try to overdo the problem. Forget about all equilibriums other than the one defined by the Kso. Do what I told you to do previously and you will get the same concentrations for OH- and Al(3+). No idea what the mass of precipitate is, but you may safely assume all Al precipitated so it is trivial to calculate.

The only thing that is obviously wrong in the answers provided is the assumption that [OH-] = [Na+]. Concentration of Na+ must be much higher, as some of the OH- ions are now part of the precipitate.

[GCT]

calculate the concentrations of the ions in the solution and the mass of the obtained precipitate.

Ks (Al(HO)₃) = 2.0 * 10 ^-32

Thanks in advance. I could solve part a) but I am having trouble solving b)

I'll try giving you a general idea for solving this problem...

-you are given the equilibrium pH, this indicates the equilibrium concentration [0H-] for the Ksp equation and thus you can simply plug this in (you'll have to do some work first).

-You are given the initial aluminum concentration, since you know the Ksp value and [0H-] value, you can deduce the solvated aluminum concentration at equilibrium.

-Subtract this concentration from the initial aluminum concentration and deduce the moles of aluminum precipitated and next the compound as a whole.

[escapeartistq]

Thank you for the hint GCT! It was easier than I expected.

[GCT]

glad I could be of help