1st year chem question thread

[nerdgasm]

The problem about what you've given there though is that d5 octahedral, if high spin, has only one possible d configuration. In addition, you can have low spin electron configurations in which you can fill the orbitals differently like d4, in which any of the t2g orbitals can accommodate the extra electron.

Of course, things get a bit more complicated when you consider different geometries and introduce ligand field theory

I think I may have stuffed up a bit in my original comment - I didn't mean to imply that the specific complex itself has only one possible electron configuration (as my sloppy writing seems to imply), but rather, if we were to only look at the number of d-electrons it has, then it could have more than one possible configuration. So, in the examples you give, if we know our complex is d5, it could either be high spin with 3 in the t_2g and 2 in the e_g (with 5 unpaired spins), or it could have all 5 in the t_2g (with 1 unpaired spin), and so on.

In any case, thanks very much for your points. As you can probably tell, I don't go to UoM, so I'm not really aware of what gets covered in the first-year chemistry curriculum, and I don't really want to give info that might be confusing/not in the course this close to the exams.

I haven't studied ligand field theory though (it never really got mentioned except as an alternative to crystal field theory in my experience so far), so I wouldn't be qualified to make any comment there  .

To psyxwar, (sorry for hijacking your question thread), if you're still having any doubts about what kinds of answer you're being expected to give, might you be able to post the question and suggested answers here, or by PM?

[uoiea]

Apologies for necroposting, but since this is still at the top of the front page of this dead forum, I'm going to use it--

5. A solution is prepared by adding 0.002 moles of sodium acetate to a 100 mL of 0.03 M acetic
acid.
(Ka (CH3COOH) = 1.8E-5)
a) Assuming that no volume change occurs upon addition of the sodium acetate what is the
pH of the solution?

So from what I know, since no volume change, I guess CH3COONa completely dissociates into CH3COO- + Na*, granting 0.002 moles of each, since salt + water yada yada

I believe the acid equation is something like
CH3COOH + H2O   ><  CH3COO- + H3O*

Since Ka is 1.8E-5 I guess I can "assume" that the initial amount of CH3COOH doesn't change, so the equation should look like

[H3O]^2 == 1.8E-5 * 0.03

My question is, how do I incorporate the CH3COO- gained from the salt dissociation into this equation? I tried a few methods but I keep getting the wrong answer [pH of 4.6]

[lzxnl]

Apologies for necroposting, but since this is still at the top of the front page of this dead forum, I'm going to use it--

5. A solution is prepared by adding 0.002 moles of sodium acetate to a 100 mL of 0.03 M acetic
acid.
(Ka (CH3COOH) = 1.8E-5)
a) Assuming that no volume change occurs upon addition of the sodium acetate what is the
pH of the solution?

So from what I know, since no volume change, I guess CH3COONa completely dissociates into CH3COO- + Na*, granting 0.002 moles of each, since salt + water yada yada

I believe the acid equation is something like
CH3COOH + H2O   ><  CH3COO- + H3O*

Since Ka is 1.8E-5 I guess I can "assume" that the initial amount of CH3COOH doesn't change, so the equation should look like

[H3O]^2 == 1.8E-5 * 0.03

My question is, how do I incorporate the CH3COO- gained from the salt dissociation into this equation? I tried a few methods but I keep getting the wrong answer [pH of 4.6]

Looks to me you just use the HH equation.

pH = pKa + log([Base]/[Acid]) = 4.74 + log(0.02/0.03) = 4.6

where [base]=0.02 M comes from the given n(acetate) and volume of 100 mL.

[uoiea]

Wow, I completely forgot about the HH equation.

Just a few more - do I have to account for any conjugated base already existing in the solution in equilibrium with CH3COOH? And just to confirm, that 0.02M is from the dissociation of CH3COONa into CH3COO-, right?

Thanks a lot.

[lzxnl]

Wow, I completely forgot about the HH equation.

Just a few more - do I have to account for any conjugated base already existing in the solution in equilibrium with CH3COOH? And just to confirm, that 0.02M is from the dissociation of CH3COONa into CH3COO-, right?

Thanks a lot.

No you don't. You can assume that the acid is not dissociated for the purposes of the question.

And yes. 0.02 M from the sodium salt.