# Predicting a precipitate

1. Jan 22, 2009

### 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 $$Al(OH)_{3}$$ = $$2x10^{-8}$$

2. Jan 22, 2009

### Staff: Mentor

You have to try by yourself first.

3. Jan 22, 2009

### 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.

4. Jan 22, 2009

### Staff: Mentor

1.5 or 15?

Why do you divide by two?

5. Jan 22, 2009

### ghostanime2001

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

6. Jan 22, 2009

### ghostanime2001

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

7. Jan 22, 2009

### 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

8. Jan 22, 2009

### Staff: Mentor

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.

9. Jan 22, 2009

### ghostanime2001

1.5g

10. Jan 22, 2009

### ghostanime2001

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

11. Feb 21, 2009

### ghostanime2001

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

12. Feb 21, 2009

### ghostanime2001

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

13. Feb 22, 2009

### 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

14. Feb 22, 2009

### Staff: Mentor

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.

15. Feb 22, 2009

### ghostanime2001

Thats the problem i dont know how to do the calculation!! When i say i dont know how i reaaaaaally dont know how to do the problem !!!!!!!!!!

16. Feb 22, 2009

### ghostanime2001

What dont u understand!?!?!?

17. Feb 22, 2009

### 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 ****

18. Feb 22, 2009

### Staff: Mentor

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

19. Feb 22, 2009

### ghostanime2001

Al2(SO4)3(s) $$\Longleftrightarrow$$ 2Al3+(aq) + 3SO42-(aq)

The question says i have 1.5 grams so im gonna 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) im assuming it goes in the beaker so it's concentration becomes:

0.00438 mol/1.125 L = 0.00389 mol/L

20. Feb 22, 2009

### Staff: Mentor

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?