# PH problem

1. Feb 12, 2016

### terryds

1. The problem statement, all variables and given/known data

6.84 gram aluminium sulfate (Al2(SO4)3) is added into water so the volume of the solution becomes 2 liters.
If we know that the atomic mass Al = 27, S = 32, and O = 16, determine the pH of the solution!
(Kb Al(OH)3 = 5*10^-9)

2. Relevant equations
$[H+] = \sqrt{\frac{Kw}{Kb}M}$

3. The attempt at a solution

n= 6.84/342 = 2*10^-2 moles
Al2(SO4)3 have 2*10^-2 moles

$Al_2(SO_4)_3+6H_2O\rightarrow2Al(OH)_3+3H_2SO_4$

Moles of Al(OH)_3 produced are 2 * 2* 10^-2 = 4 * 10 ^-2 moles
Molarity = 4*10^-2 / 2 = 2 * 10^-2

$[H+] = \sqrt{\frac{10^{-14}}{5\times 10^{-9}}*2*10^{-2}} = 2 \times 10^{-4}$
pH = 4 - log 2

That's the pH I get from the Al(OH)3 solution
My question is..
Does the H2SO4 that's also produced affect the pH?
If it does, how to calculate the pH? By summing the hydrogen concentrations up from H2SO4 and Al(OH)3 then directly using - log[H+] ??

2. Feb 12, 2016

### epenguin

I would like to know where this problem comes from as it seems odd, and have a more complete quotation. Were you given the relevant equations or are they what you thought is relevant?

The chemical reaction you show is not relevant here, in fact I don't think it's really right. Al(OH)3 is pretty insoluble, Aluminium sulphate is quite soluble. To get the hydroxide you would have to make the solution alkaline.

Where does the weak acidity of this salt come from? In aqueous solution Al3+ forms the complex with water Al(H2O)63+. This has a slight tendency to lose a proton. maybe this is helpful

So it must have a pKa, of course it must have a pKb and as you have quoted is correct - it is just surprising and unusual to see it quoted for an acid. Maybe it is an academic excercise.

3. Feb 13, 2016

### terryds

That's my homework..
The equation is just what I think relevant...
So, this means that Al(OH)3 doesn't affect pH?
What about H2SO4? Is it produced?
How to calculate the pH then?

4. Feb 13, 2016

### epenguin

You may find it helpful to revise about acids, bases and salts. Aluminium sulphate is a salt. It could be produced by the reaction you quote, but with the arrow reversed.
Asking what has happened to the sulphuric acid is like asking the same question about sodium sulphate. You can think of it as having been neutralised by the base Al(OH)3.

To a first approximation salts are neutral. However to a better approximation, It is not quite like sodium sulphate because that is the salt of a strong acid and strong base, whilst aluminium sulphate is salt of a strong acid and weak base. The salt of a strong acid and weak base will be slightly acid. This is because - just take the example of ammonium sulphate - the conjugate base SO42- of the strong acid has little tendency to be protonated, while the conjugate NH4+ of the weak base NH3 has some tendency to 'lose' the proton (i.e. transfer it to water). So there is a greater concentration of protons than in pure water. Maybe it hasn't been very apparent but there is an exact analogy:
[Al.(H2O)6]3+ plays a role like NH4+, and [Al.(H2O)5OH]2+ like NH3

I would still like to know where this Kb you quote comes from. If this is the Kb for the equilibrium between the last mentioned forms, it fits reasonably with the pKa I found here http://chemwiki.ucdavis.edu/Core/In...iod_3_Elements/Chlorides_of_Period_3_Elements of 3.3 to 3.6. But Al(OH)3 is something different.

As to how you calculate the pH, this is pretty much as you have done it. However I should do it again. It seems to me that the total molarity [Al] is twice your figure, taking into account the two atoms/molecule in the starting formula.

Also I should say that [Al.(H2O)6]3+ is not all that weak an acid. It's more in the category 'moderately weak' or 'moderately strong' - base. likewise its conjugate a moderately weak - or moderately strong base. In fact I get pH around 3.5, getting to be a moderately acid solution, and the last-mentioned species about half deprotonated.

Last edited: Feb 14, 2016