Lecture 13

Question 1

How to get Ka from [HA], [A-] and [H+]

Answer 1

[H+] = Ka ([HA]/[A-])
–> Ka = [H+][A-]/[HA]

Question 2

What is the Henderson-Hasselbalch equation and what is it used for

Answer 2

pH = pKa + log ([A-]/[HA]) –> BASE OVER ACID
–> used to calculate pH when [A-]/[HA] is known
–> pKa = -log(Ka)

NOTE: this equation already does the assumption

Question 3

What is buffering capacity and between two buffers (weak acid/base + common ion -salt), which one has a higher BC

Answer 3

• Capacity to resist pH change (how much)
• Amount of H+/OH- buffer can absorbs without significant change in pH
• Buffer that is more concentrated has higher BC
  –> higher concentration = higher BC

Question 4

When you add an H+ or an OH-, which compound can they not coexist with

Answer 4

H+ –> can’t coexist with weak base
OH- –> can’t coexist with weak acid

Question 5

When given Kb, but asked to find pH (with Ka), what are the two methods

Answer 5

1. Use ICE table and solve for [OH-] (using Kb)
   Convert to pH
2. Convert Kb to Ka (Kw = Ka x Kb)
   Use Henderson-Hasselbalch eq to get pH
   –> pH = pKa + log ([A-]/[HA])
   –> [base] and [acid] must be known

Question 6

When is a buffer more resistant to pH change (also when is it more effective)

Answer 6

When ([A-]/[HA]) = 1
–> higher BC = base and acid have the same [ ]

Most effective when pH = pKa

Question 7

How to figure out which weak acid/base will be best for a certain buffer at a certain pH?

Answer 7

find -log of each Ka given
–> the one closest to the pH of the buffer will have the best BC

Question 8

How to do titration problems (STRONG ACID-STRONG BASE)

Answer 8

1. Identify species in soln
2. Find initial pH (before titration)
3. Add strong acid/strong base (VOLUME INCREASES)
   –> identify new species by adding H+ or OH-
4. FIGHT TO THE DEATH (between H+ and OH-)
   –> in moles NOT [ ] so do V x Initial [ ]!!
   BR
   AF
   Final [ ] (VOLUME GOES UPPP)
   –> either OH- or H+ goes to 0
5. find the pH

Question 9

When do we get to the equivalence point in titration problems

Answer 9

When there is the same amount of acid and base
–> after the fight, both go to 0
–> solution is neutral

On graph: when tangent line begins to turn

Question 10

What happens after the equivalence point is reached and you add more

Answer 10

If it was an acid problem, becomes base problem
If it was a base problem, becomes acid problem
–> NOT FOR STRONG ACID-STRONG BASE PROBLEMS

Question 11

At what pH is the equivalence point in strong acid-strong base titrations?

Answer 11

pH = 7

NOTES: if not strong acid-strong base, pH varies

Question 12

How to do the assumption

Answer 12

(x/M) x 100 < 5%