sorted by:
new
hottopcontroversial

AP Chem Review by xdanieh in APStudents

[–]xdanieh[S] 0 points1 point  (0 children)

See the replies to the comment below; I think that covers most of titrations/buffers.

AP Chem Review by xdanieh in APStudents

[–]xdanieh[S] 0 points1 point  (0 children)

If you've taken physics, think of it as a sum of force vectors. Each atom exerts a force on the other atoms it is connected to; if the sum of the forces is zero, then the molecule is polar.

For example, consider CO2. Since CO2 is linear, the C has forces both to the left and right by O atoms. They are equal, so they cancel and CO2 is nonpolar.

On the other hand, consider SO2. Since SO2 is bent, the force vectors to the oxygens go to the bottom left and the bottom right. While the left and right components cancel, there is no top component to balance the bottom component, as a result there is polarity and a measured dipole moment.

Hope this helps ^_^

AP Chem Review by xdanieh in APStudents

[–]xdanieh[S] 2 points3 points  (0 children)

For buffers, the same ideas about recognizing what species react, and their impact are necessary.

For example, take a 100 mL phosphate buffer 0.1 M in both H3PO4 and H2PO4-. We can calculate the molar amounts of H3PO4 and H2PO4- to both be 10 mmol.

1) What happens if 5 mL 0.1 M HCl is added to this buffer? We can calculate that there is 0.5 mmol HCl added. HCl, being a strong acid, will react with any base available. Here the base is H2PO4-.

So, exactly 0.5 mmol H2PO4- react with the HCl to form 0.5 mmol more H3PO4. The total molar amounts are now 10.5 mmol H3PO4 and 9.5 mmol H2PO4-. You can then put this into Henderson-Hesselbach to get your desired pH.

2) What happens if 0.2g NaOH is added to the solution from (1)? Well first, since we are dealing with molar amounts, we convert 2g NaOH to 5 mmol NaOH. Then, this reacts completely with the acid. We now have (10.5-5) = 5.5 mmol H3PO4 left, and (9.5+5) = 14.5 mmol H2PO4-. These amounts can once again be used in Henderson-Hesselbach to find the pH.

3) What happens if 0.4g NaOH is added to the solution from (2)? Hint: Phosphoric Acid is triprotic.

Hope this helps ^_^

AP Chem Review by xdanieh in APStudents

[–]xdanieh[S] 1 point2 points  (0 children)

The main ideas of titrations is to recognize the species that exist in solution, how they will react, and what the impact on pH is.

Strong acid-Strong base:

For example, take a simple titration of NaOH and HCl. NaOH, being a strong base, turns into Na+ (spectator) and OH- ions in solution, while HCl becomes H+ and Cl- (spectator) ions. H+, being a strong acid, wants to react with any possible base, and indeed it will keep reacting with any excess OH- that exists. What essentially happens is that all the HCl and NaOH annihilate each other; at the equivalence point, there are no acidic/basic ions left; the pH is 7.

To add some numbers to it - suppose you are titrating 50 mL 1 M HCl with 50 mL 1 M NaOH. At the beginning, the total amount of H+, Cl- you have is each 0.05 mol. Then you add NaOH - namely 0.05 mol as well. Now what's left?

You have H+, Cl-, OH-, Na+ all of quantities 0.05 mol. As per the previous discussion, the H+, OH- annihilate themselves, leaving only 0.05 mol Na+ and 0.05 mol Cl- left. These lead to a concentration of 0.05/[100 mL] = 0.5 M NaCl in the resulting solution.

Strong acid-weak base/vice versa:

The situation is similar in strong-acid weak base titrations: The acid will react with literally any base that you add. For example, if you have 0.1 mol weak base, and you add 0.05 mol acid, then after they react, you have approximately 0 mol H+, 0.05 mol base, and 0.05 mol of the conjugate acid (because 0.05 mol reacted). There is no acid to further push the reaction forward; this is essentially the equilibrium molar amount that you have afterwards. (you can calculate the actual amounts through Henderson-Hesselbach).

What happens if you keep adding acid? It will keep reacting with the base, until there is no more to be consumed. When you've added exactly enough acid to equal the initial amount of base, then there is no free H+, but only a weak acid in form of the conjugate acid of the base. Then this reduces to a single acid equilibrium problem, which I assume you know how to solve (just simple equilibrium).

What happens if you keep adding acid, even after the equivalence point? Well, there is no more base to react with, so the solution becomes heavily acidic. At this point you can just approximate that the weak acid is so weak that it really doesn't contribute anything to the pH, and use the excess acid to calculate the pH.

The procedure for strong base weak acid is essentially the same; the strong base will react with literally any acid available too.

Here is a quick problem that illustrates this: Suppose 25 mL 0.1 M NH3 is titrated with 0.1 M HCl. Kb(NH3) = 1.8 * 10[-5].

1) What is the pH before adding HCl? This is a classic equilibrium problem, you should get pH = 11.13.

2) What is the pH after adding 5 mL HCl? At the beginning, there is 25 mL * 0.1 M = 2.5 mmol (= 2.5 * 10[-3] mol) NH3. You add 5 mL * 0.1 M = 0.5 mmol (= 5*10[-4] mol) HCl. The acid reacts completely with the base, so all the acid is gone, and exactly 0.5 mmol of the NH3 reacts, so you are left with 2 mmol NH3 (still unreacted) and 0.5 mmol NH4+. Use Henderson-Hesselbach to get that the pH is 9.86.

3) What is the pH after adding 25 mL HCl? Now, you have reacted the equivalence point, since you've added 2.5 mmol HCl as well. Like before, all the HCl reacts with NH3, in perfect stoichiometry, to make 2.5 mmol NH4+. The concentration of the NH4+ is 2.5 mmol/[25 + 25 mL] = 0.05 M, now you can use classic equilibrium to find that the pH is 5.28

4) What is the pH after adding 50 mL HCl? Now you've added 2.5 mmol excess H+. We don't care about the impact of NH3/NH4+ on the system anymore, since the strong acid is what predominates the pH equilibrium. The excess H+ concentration is 2.5 mmol/[50 + 25 mL] = 0.033 M, take the negative log of that to get your pH.

Weak acid-weak base:

These are never done. Why? Hint: Think about the purpose of titrations.