Hello- I thought I should go ahead and type in all of the equations that I need

Question

Hello- I thought I should go ahead and type in all of the equations that I need help on. I need to add these together in pairs, so I can get relative equations for each AL(OH) concentration (for [Al(3+)],[Al(OH)(2+)],[Al(OH)2(+)] and[Al(OH)4(-)] ).

1. Al(OH)3(s) + 3[H+] = [Al(3+)] + 3[H2O] pK = -8.5

2. Al(OH)3(s) + 2[H+] = [Al(OH)(2+)] + 2[H2O] pK = -3.5

3. Al(OH)3(s) + [H+] = [Al(OH)2(+)]+ [H2O] pK = 0.8

4. Al(OH)3(s) + H2O = [Al(OH)4(-)]+ [H+] pK = 14.5

The example I have uses Cd equations, but they all work out nicely - here is one pair that was added together and the results:

3. [Cd(2+)] + 4[H20] = [Cd(OH)4(2-)] + 4[H+] pK = -47.44

4. [Cd(OH)2(s)] + 2[H+] = [Cd(2+)] + 2[H2O ] pK = +13.65

____________________________________________________

[Cd(OH)2(s)] +2[H2O ] =[Cd(OH)4(2-)] + 2[H+] pK= -33.79

10^-33.79 =[Cd(OH)4(2-)][H+]^2

A: log[Cd(OH)4(2-)] = -33.79 + 2pH

I need to come up with similar results for each Al concentration. Please help!

Explanation / Answer

Take any 2 equations... 1. Al(OH)3(s) + 3[H+] = [Al(3+)] + 3[H2O] pK = -8.5 2. Al(OH)3(s) + 2[H+] = [Al(OH)(2+)] + 2[H2O] pK = -3.5 Now reverse eqn2 3. [Al(OH)(2+)] + 2[H2O] = Al(OH)3(s) + 2[H+] kc is reversed hence pK=3.5 Now add 1,3 [Al(OH)(2+)] +[H+] = [Al(3+)] + 3[H2O] pK= -5 Hence -log [Al(3+)] + log [Al(OH)(2+)] = -5+pH You can obtain [Al(3+)] from 1 ie; log[Al(3+)]= -8.5+ 3pH Similarly all other results can be obtained..

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