Net Ionic Equation for Al(OH)4-?

Question

I need help on the Net Ionic equation for Alum. The equation is:

2Al(s) + 2KOH(aq) + 6H2O(l) --> 2Al(OH)4-(aq) + 2K+(aq) + 3H2(g)

I am in AP chem and sometimes the teacher gets the problem wrong and he couldn't explain this one very well, so I decided to come here for help.

If you could also list the steps that would be really helpful.

Thanks,
Albert

I wanted to add these details :

This is for the AP Chem lab Synthesis of Alum.
The question for the lab says :

Write balanced net ionic equations for
a) Aluminum reacts with KOH and water forming potassium ions, [Al(OH)4]- and hydrogen gas.

The formula that I found was from a college site about this lab but I couldn't find the Ionic one. My teacher's ionic equation doesn't look right, thats why I need the steps and the solution, because he has been know to be wrong.

Answer 1

Forget the potassium ions - they don't do anything. The aluminium metal dissolves in the potassium hydroxide solution, forming aluminate ions and hydrogen gas.

The ionic equation is:

2Al + 2OH- + 6H2O ---> 2[Al(OH)4]- + 3H2

To explain how to get this equation, you need the ideas of half equations, and oxidation and reduction.

For example, Al + 4OH- ---> [Al(OH)4]- + 3e-

and 4H2O + 4e- ----> 2H2 + 4OH-

Now combine them, getting the number of electrons equal, and cancelling down some hydroxide ions.

If it's any consolation, this exercise would be beyond most A-level pupils in the UK.

Answer 2

Equation - 2 K+(aq) + 2 [Al(OH)4]-(aq) + H2SO4 → 2 Al(OH)3 + 2 H2O(l) + 2 K+(aq) + SO4-2(aq)
Net - 2 [Al(OH)4]-(aq) + H2SO4 → 2 Al(OH)3 + 2 H2O(l) + SO4-2(aq)

The reason your equation doesn t make sense is because it requires a strong acid to cause the Aluminum tetra-Hydroxide which is a relatively unstable so therefore uncommon ion to shed an ion and become a more stable aluminum hydroxide.

Answer 3

Al Oh 4

Answer 4

i think of you're lacking something on your equation. Sodium metallic, Na(s), plus HCl will effect in a redox reaction, no longer a metathesis reaction: 2Na(s) + 2HCl ----> 2 NaCl + H2(g) that's if the solvent is HCl. If the solvent is water, then you definately gets sodium hydroxide forming too. once you're reacting a sodium salt like NaOH with HCl, a greater attainable occasion to apply for ionic equations, the molecular equation would be: NaOH(aq) + HCl(aq) ----> H2O(l) + NaCl(aq) This element shows all the molecules as though they weren't ionizing in answer. that's certainly customary that they do ionize nevertheless, so a greater precise thank you to depict the reaction is the ionic equation, which shows the species as they exist interior the reaction medium (water, consequently). the completed ionic equation shows all the ions of their solvated, dissociated state, and any insoluble and/or covalently bonded reactants or products that are shaped besides: Na+ + OH- + H+ + Cl- ----> H2O(l) + Na+ + Cl- the internet ionic equation leaves out something that occurs as its solvated self in the two the reactants and the products. consequently, the Cl- and Na+ ions are interior an identical state on the two side of the reaction, so which you would be able to easily... cancel them out. the sole "product" shaped is the hot, non-ionic molecule of water, and the sole ions in touch in its formation are the H+ and the Cl- ions: OH- + H+ -----> H2O(l) you are able to write Na(s) + HCl in an ionic style, regardless of the shown fact that it would not probable describe the reaction very nicely. Ionic equations are inteded to describe metathesis reactions greater suitable than they are meant to describe redox reactions. nonetheless, the completed and internet ionic equations for Na + HCl could be: 2Na(s) + 2H+(aq) + 2Cl-(aq) -----> H2(g) + 2Na+(aq) + 2Cl-(aq) 2Na(s) + 2H+(aq) -----> H2(g) + Na+

Answer 5

The problem is that the equation for Aluminum hydroxide is Al(OH)3 not Al(OH)4

Answer 6

2Al(s) + 6 H2O(l) --> 2 Al3+(aq) + 6 OH-(aq) + 3H2(g)