[ghostanime2001]

This was one of the questions in my grade 12 exam and it goes like this:

If 1.5 grams of Al₂(SO₄)_3(s) is added to a beaker containing 1125 mL (1.125 L) of 0.015M NaOH solution will a precipitate occur? K_sp of Al(OH)_3(s)= 2.0 x 10^-8

First step i know to do is:
1.5 g Al₂(SO₄)_3(s) / 342.14 g/mol Al₂(SO₄)_3(s) = 0.00438 mol

then:
0.00438 mol / 1.125 L = 0.00389 M

The balanced equation i believe is this (please check the states and see if i labelled them correctly):

Al₂(SO₄)_3(s) + 6NaOH_(aq) --> 2Al(OH)_3(s) + 3Na₂SO_4(aq)

A question: should Al₂(SO₄)_3(s) be Al₂(SO₄)_3(aq) since its being put into a beaker containing NaOH solution and the solvent is largely water ?

I also know that: 2Al(OH)_3(s) is the precipitate formed but how can i determine if it REALLY will precipitate out when i add the 1.5 grams of Al₂(SO₄)_3(s) into the NaOH_(aq) solution ??

One more question: Is this a common ion + precipitate type of problem just makng sure cuz there are tutorials in the internet this problem just seems it has something extra built-in.

[Astrokel]

A question: should Al2(SO4)3(s) be Al2(SO4)3(aq) since its being put into a beaker containing NaOH solution and the solvent is largely water ?

Follow the question, (s).

I also know that: 2Al(OH)3(s) is the precipitate formed but how can i determine if it REALLY will precipitate out when i add the 1.5 grams of Al2(SO4)3(s) into the NaOH(aq) solution ??

Ionic product vs. solubility product.

One more question: Is this a common ion + precipitate type of problem just makng sure cuz there are tutorials in the internet this problem just seems it has something extra built-in.

No common ion.

http://www.science.uwaterloo.ca/~cchieh/cact/c123/ksp.html
http://www.ncl.ac.uk/dental/oralbiol/oralenv/tutorials/ksp.htm

[ghostanime2001]

I am not sure but i think im very close. My steps are as follows:

Al₂(SO₄)_3(s) + 6NaOH_(aq)  3Na₂SO_4(aq) + 2Al(OH)_3(s) 
1.5g                                                        0.00876 mol
342.14 g/mol
0.00438 mol
0.00389 M

2Al(OH)_3(s)  2Al³⁺_(aq) + 6OH^-_(aq)                its just being simplified down to:
Al(OH)_3(s)            Al³⁺_(aq)     +     3OH^-_(aq)
0.00438 mol          0.00438 mol       0.01314 mol
0.00389 M            0.00389 M          0.01168 M

Q_sp = [Al³⁺][OH^-]³
      = [0.00389 M][0.01168 M]³
      = 6.19 x 10^-9

Q_sp < K_sp
 :delta: No ppt

Please verify i did my work right. Im still not very sure i did this correct. If there are mistakes please help me with it. Ive been solving this problem for a long while. THanks.

[Astrokel]

You don't really need an ICE table and your values are wrong. Calculate [Al³⁺] and [OH^-] in the solution and put it in your ionic product formula.

P.S: I realised this has a common ion effect because of the limiting reagent. But it won't affect how you calculate the answer, still the same way. My bad.

[Borek]

Al₂(SO₄)_3(s) + 6NaOH_(aq) --> 2Al(OH)_3(s) + 3Na₂SO_4(aq)

Note: according to IUPAC rules it should be

Al₂(SO₄)₃(s) + 6NaOH(aq) --> 2Al(OH)₃(s) + 3Na₂SO₄(aq)

(no subscript for states). Most teachers are not aware, so it probably doesn't matter.

[ghostanime2001]

I dont understand what you meant here when u said my values are wrong i followed the same stoichiometry i learned in grade 11 so whats wrong with my values ?

You don't really need an ICE table and your values are wrong. Calculate [Al3+] and [OH-] in the solution and put it in your ionic product formula.

[Al3+] = 0.00389 M (If u think this value is wrong then please explain why)
[OH-] = 3(0.01314 mol) because theres 3 molecules of hydroxide ions = 0.03942 mol / 1.125 L = 0.03504 M (If u think this value is also wrong then please explain why)

[ghostanime2001]

Will anyone out there help me out ?

[Astrokel]

How many mole of Al³⁺ in 1 mole of Al₂(SO₄)₃?
How many mole of OH^- in 1.125L 0.015M NaOH?

[Borek]

because theres 3 molecules of hydroxide ions = 0.03942 mol

You are mixing concentration and stoichiometry coefficients, these are separate things.

[ghostanime2001]

you know what f&#$ everything just tell me what i should do i have no clue right now about anything

[Astrokel]

you know what? http://www.chemicalforums.com/index.php?topic=58.0

[ghostanime2001]

WHAT DO U WANT ME TO DO IF I DONT UNDERSTAND ANY OF IT I REALLY DONT

[ghostanime2001]

are you gonna help or not !

[ghostanime2001]

The COMPLETE BALANCED CHEMICAL EQUATION
Al₂(SO₄)_3(s)     +     6NaOH_(aq)            3Na₂SO_4(aq)      + 2Al(OH)_3(s)       <---- precipitate

The NEW SOLUTION
Al₂(SO₄)_3(s)    2Al³⁺_(aq) + 3SO₄^2-_(aq)
0.00438 mol      0.00876 mol
                      0.00779M

NaOH_(aq)   Na⁺_(aq) + OH^-_(aq)
0.015M                       0.0169 mol
0.0169 mol                  0.015M

A precipitate can only occur when Q>K_sp

Q CALCULATION
Al(OH)3(s)    Al3+(aq) + 3OH-(aq)
Q = [Al3+(aq)][OH-(aq)]³
Q = [0.00779][0.015]³     
Q = 2.629 x 10^-8

Q (2.629 x 10^-8) > K_sp (2.0 x 10^-8)

 :delta: Al(OH)_3(s) precipitate does form

NOW this got to be right !!!!!!

[Borek]

Looks OK.

Next time try to be more polite and less demanding, it will both increase your chances of getting help and decrease your chances of getting banned.