Predicting Precipitate Formation with Al2(SO4)3 and NaOH Solutions

[ghostanime2001]

If 1.5g of Al_{2}(SO_{4})_{3)_{(s)} is added to a beaker containing 1125 mL of 0.015M NaOH solution will a precipitate occur ? Ksp of Al(OH)_{3} = 2x10^{-8}

 

[Borek]

You have to try by yourself first.

 

[ghostanime2001]

15g/342g/mol = 0.00438 mol/1.125 L = 0.00389 M

Al_{2}(SO_{4})_{3)_{(s)} --> 2Al^{3+} + 3SO_{4}^{-2}
[Al^{3+}]=0.00389 mol/l divided by 2 moles of 2Al^{3+} produced=0.001945M
NaOH_{(aq)} --> Na^{+}_{(aq)} + OH^{-}_{(aq)}
[OH] = 0.015 M 1:1 mole ratio

Q=(0.001945)(0.015)^3
Q=6.56x10^-9
Q<Ksp
therefore precipitate forms.

 

[Borek]

1.5 or 15?

Why do you divide by two?

 

[ghostanime2001]

Because 2 moles of Al3+ are produced for every Al2(SO4)3 that dissolved

 

[ghostanime2001]

and Al(OH)3 produces 1 mole of Al3+ ions

 

[ghostanime2001]

This problem is a little different instead of having 2 solutions being mixed together and findting the concentration of the diluted participants its rather putting a mass of something in this case Al2(SO4)3(s) into a beaker full of 1125ml = 1.125 L

 

[Borek]

Because 2 moles of Al3+ are produced for every Al2(SO4)3 that dissolved

Once again - if two moles of Al³⁺ are produced per each mole of salt dissolved, why do you divide by two?

Once again: was it 1.5 g, or 15 g, as you started with one number, but you used th eother in your calculations.

 

[ghostanime2001]

1.5g

 

[ghostanime2001]

i just did my exam and surprisingly this question was on it. Let me set up the problem properly

 

[ghostanime2001]

so is anyone going to help me out on this or not ?

 

[ghostanime2001]

Ive trie and tried and i tried what the ****

 

[ghostanime2001]

If 1.5 grams of Al2(SO4)3(s) is added to a beaker containing 1125 mL of 0.015M NaOH solution will a precipitate occur? Ksp of Al(OH)3(s)=2.0x10^-8

 

[Borek]

Show your calculations. You have already tried once, but you were wrong - and I have pointed to two problems in your calculations. You have ignored my post so far.

 

[ghostanime2001]

Thats the problem i don't know how to do the calculation! When i say i don't know how i reaaaaaally don't know how to do the problem !

 

[ghostanime2001]

What don't u understand??!?

 

[ghostanime2001]

it looks like u ignored my my POST so far when u asked me if it was 15 or 1.5 grams I ALREADY TOLD YOU its 1.5 grams man ****

 

[Borek]

You have correctly started solving the question by calculating concentration of Al³⁺. However, you calculated it wrong.

 

[ghostanime2001]

Al₂(SO₄)_3(s) \Longleftrightarrow 2Al³⁺_(aq) + 3SO₄^2-_(aq)

The question says i have 1.5 grams so I am going to take 1.5g/342.14g/mol molar mass = 0.00438 mol

Then since the Al₂(SO₄)_3(s) is being submerged into the NaOH solution that has volume 1125 mL (1.125 L) I am assuming it goes in the beaker so it's concentration becomes:

0.00438 mol/1.125 L = 0.00389 mol/L

 

[Borek]

This is concentration of aluminum sulfate, but not of Al³⁺. How many moles of Al³⁺ are introduced into the solution per each mole of dissolved sulfate?

 

[ghostanime2001]

0.00438 mol x 2 mol Al³⁺ = 0.00876 mol Al³⁺

 

[ghostanime2001]

2 moles of Al^3+ ions in solution per each mole of aluminum sulfate dissolved

 

[ghostanime2001]

So is anybody going to decide if my answer is right or not ?