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Choosing buffers based on pKa In many experiments, we need a buffer that maintains the solution at a specific pH. We may need an acidic or a basic buffer, depending on the experiment. The Henderson-Hasselbalch equation can help us choose a buffer that has the pH we want. pH = pKa + log([conj. base]/[conj. acid]) With equal amounts of conjugate acid and base (preferred so buffers can resist base and acid equally), then … pH = pKa + log(1) = pKa + 0 = pKa So choose conjugates with a pKa closest to our target pH.
Example: You need a buffer with pH of 7.80. Which conjugate acid-base pair should you use, and what is the molar ratio of its components?
buffer capacity: the amount of strong acid or strong base that can be added to a buffer without changing its pH by more than 1 unit; essentially the number of moles of strong acid or strong base that uses up all of the buffer’s conjugate base or conjugate acid. Example: What is the capacity of the buffer solution prepared with 0.15 mol lactic acid CH 3 CHOHCOOH (HA, Ka = 1.0 x 10
- 4 ) and 0.20 mol sodium lactate NaCH 3 CHOHCOO (NaA) and enough water to make 1.00 L of solution? (from previous lecture notes)
Review of equivalence point equivalence point: moles of H
= moles of OH
(moles of acid = moles of base, only when the acid has only one acidic proton and the base has only one hydroxide ion). What is the pH at the equivalence point* for a strong acid-strong base titration? HNO 3 + KOH H 2 O(l) + KNO 3 (aq) I 0.100 moles each *(sample H
= OH
in titrant) C End No products are acidic or basic enough to affect the pH of the solution, so the pH is 7 at the equivalence point. Fig. 16.5, p. 589. Note that KNO 3 (aq) = K
(aq) + NO 3
(aq). K
What is the pH at the equivalence point* for a weak base-strong acid titration? NH 3 + HCl H 2 O(l) + NH 4 Cl(aq) I 0.100 moles each *(sample OH
- = H + in titrant) C End Note: NH 4 Cl(aq) = NH 4
(aq) + Cl
(aq) Cl
- H 2 O(l) NH 4
- H 2 O(l) NH 3 + H 3 O
I C Eq Equilibrium process (Chapter 15 calculations) results in pH < 7 (acidic). Fig. 16.7, p. 593.
pH before equivalence point Before the equivalence point in a titration, the stoichiometric reaction neutralizes only a portion of the sample. Example: Strong acid – strong base titration HNO 3 + KOH H 2 O(l) + KNO 3 (aq) I 0.100 0. C End Calculate remaining (unreacted) HNO 3 => H 3 O
. (Chapter 4 and 5 calculations = stoichiometric) M = mol Volume? HNO 3? KOH? L pH = - log[H 3 O
] (Chapter 15 calculation)
Example: 50.0 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH solution. Calculate the pH after 25.0 mL of titrant has been added.
The half-equivalence point is a special case prior to reaching the equivalence point. At the half-equivalence point, the strong base has converted half of the weak acid into its conjugate base. We now have a buffer solution that contains equal amounts of conjugate acid and base. pH = pKa + log([conj. base]/[conj. acid]) pH = pKa Again, the H-H equation is useful when dealing with buffer calculations.
Titrations for weak base-strong acid work similarly, except the strong acid converts the weak base into its conjugate acid. Next time: pH after equivalence point.
