TBR, Gchem, Buffers, Passage 5, #30, 31, 33 -Got them right but need help understanding explaination

[Sammy1024]

Sorry for my many questions. I got the following questions correct by logically thinking it out, but I thought I should understand how to calculate the answers?

30. By roughly how much do the two equivalents points in the first graph differ?
A. Fewer than 2.0 pH units
B. Fewer than 4.0 pH units, but more than 2.0 pH units
C. Fewer than 8.0 pH units, but more than 4.0 pH units
D. More than 8.0 pH units

I looked at the graph and I roughly estimated the numbers and got the correct answer, but I don't understand how they went about calculating the problem.

How do they figure out that it should be 0.05 M HoAc and 0.05 M OAc-. And then how do they assume that pKa is roughly 5.0?

31. The pH at equivalence is GREATEST for which of the following titrations?

A. The titration of 0.10 M H3CCO2H by NaOH
B. The titration of 0.10 M H3CCO2Na by HCl
C. The titration o f 0.10 M CH3NH3CI by NaOH
D. The titration of 0.10 M CH3NH2 by HCl

I eliminated B and D because the pH wouldn't be highest with HCl being titrated. Then I just looked at the graphs and the equivalence point for CH3NH3Cl looked higher so I picked that, but i wanted to understand their reasoning to be safe.

I don't understand why they start talking about the conjugate base and all that after the "(pH at equivalence is less than 7.0)."

33. How does the pH at point a in Figure 1 compare to the pH at point d in Figure 1?

A . The pH at point a is more than 1.0 pH unit greater than the pH at point d.
B. The pH at point a is greater than the pH at point d, but the difference is less than one pH unit.
C. The pH at point a is less than the pH at point d, but the difference is less than one pH unit.
D. The pH at point a is more than 1.0 pH unit lower than the pH at point d.

I looked at the graph and a and d seemed very close, so it had to be within 1 pH unit.

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[chenzt]

How do they figure out that it should be 0.05 M HoAc and 0.05 M OAc-. And then how do they assume that pKa is roughly 5.0?

You started out with 0.1M 25 ml of acetic acid, and titrated it with 25 ml of 0.1M base. Double the total volume while keeping the acetate amount the same, the concentration will decrease in half to 0.05M

pKa is roughly 5 because of henderson hasselbach equation. pH = pKa + log (A-/HA)

If 25ml is equivalence point, then 12.5ml is the halfway point (where A- = HA)
HH equation becomes pH = pKa + 0 (whatever A- and HA it is, A-/HA will equal to 1, so log of it will equal to 0).
pH = pKa; at halfway point you ballpark pH to be ~5 (or if you know that pka of acetic acid is 4.76, then you can skip that entire step)

I don't understand why they start talking about the conjugate base and all that after the "(pH at equivalence is less than 7.0)."

[H+] + [A-] + [OH-] --> [A-] + H2O (because of le chatelier, HA will keep on dissociating to H+ and A- to react with OH- until none is left)
A- + H2O --> HA + OH-

When in a weak acid, it titrates with a base and leaves only a conjugate base. The conjugate base have some reactivity to it, react with water and reform some OH-. This is why weak acid at equivalence point have a pH > 7. Weaker the acid, the greater the pH at equivalence point (because weaker acid = stronger conjugate base). Same thing is true for weak base, but opposite. Weaker the base, the lower the pH is at equivalence.

I looked at the graph and a and d seemed very close, so it had to be within 1 pH unit.

Based on Q 30 you should've noticed that both are acetic acid, but point D has twice the volume as point A because of the additional acid you added. Same moles of acetic acid but at twice the volume = half the concentration. so higher pH

 

[Sammy1024]

Thank you so much! I really appreciate your wisdom and knowledge!!!! :]