Buffers and Titrations Flashcards
How to make a buffer.
Two ways:
1) Combine a conjugate pair in roughly equal mole portions.
2) Partially titrate a weak acid with roughly half of an equivalent of strong base, or partially titrate a weak base with roughly half of an equivalent strong acid.
Neutralization
This is the mixing of equal mole portions of an acid with a base, regardless of their concentrations and strengths. IT DOES NOT MEAN TO MAKE THE pH OF THE SOLUTION EQUAL TO 7. When the base is stronger than the acid, the neutralized solutions is slightly basic, so the pH is greater than 7.0 (and vice versa for a stronger acid).
pH range for buffers
pH range = pKa +/- 1
If water is added to a buffer solution with pH = 3.96, what happens to the pH?
A. The pH increases lightly.
B. The pH decreases slightly.
C. The pH remains the same.
D. If the pH is greater than 7,then it decreases. If the pH is less than 7, then it increases.
The answer is C. Addition of water to a buffer equally dilutes the concentration of the weak acid and its weak conjugate base. This means that the mole ratio of the weak base to the weak acid does not change upon the addition of water. According to the Henderson-Hasselbalch equation, the pH of the solution does not change because pKa is constant and the fraction has not changed. The result is that the pH of a buffer does not change when it is diluted. This is why the Henderson- Hasselbalch equation can be written as moles A- over moles HA, as well as [A-] over [HA].
What happens in the body when there is:
1) Retention of CO2
2) Loss of CO2
3) Loss of HCO3-
4) Loss of H3O+
CO2 + H2O H2CO3 HCO3- + H+
1) Blood pH drops (respiratory acidosis)
2) Blood pH rises (respiratory alkalosis)
3) Blood pH drops (respiratory acidosis)
4) Blood pH rises (respiratory alkalosis)
Equivalence point of a titration of a strong base with a strong acid.
pH = 7
Equivalence point of a weak acid titrated with a strong base.
pH > 7
What is the pH after 30 mL of 1.00 M NaOH(aq) has been added to 100 mL 0.50 M HOAc(aq)? HOAc has a pKa = 4.74.
A. 3.51 B. 4.56 C. 4.92 D. 5.97
Because the strong base is twice as concentrated as the weak acid, only half the volume of strong base (relative to the weak acid) is required to reach the equivalence point. This means that 50 mL of 1.00 M NaOH(aq) fully neutralizes the 100 mL of 0.50 M HOAc(aq). The halfway point of the titration is reached when exactly 25 mL of 1.00 M NaOH(aq) has been added. At the halfway point, the pH of the solution equals the pKa of the weak acid. The additional strong base beyond the 25 mL makes the pH of the solution slightly greater than the pKa of the acid, 4.74. The best choice is answer C.
What is the pH of a solution made by mixing 10.0 mL 0.10 M HCO2H(aq) with 4.0 mL 0.10 M KOH(aq)? The pKa for HCO2H is 3.64.
A. 1.34 B. 3.46 C. 3.82 D. 9.36
The best way to solve this question is to think in terms of equivalents. The weak acid and titrant strong base are of equal concentration, so 10 mL KOH is one equivalent (the amount needed to reach the equivalence point.) If 5.0 mL are added, then the acid is half-titrated, so pH = pKa. However, less than 5.0 mL has been added, so pH is a little less than pKa. The pKa is 3.64, so the best answer is choice B. Choice A is too much less than the pKa (more than 1.0 is beyond the 10 : 1 ratio, which would be when less than 1.0 mL of KOH had been added.)
Name three common polyprotic acids.
carbonic acid H2CO3, phosphoric acid H3PO4, sulfuric acid H2SO4.
How can the pH at equivalence for the titration of a weak acid by a strong base be approximated?
Average the pKa of the weak acid and the pH of the titrant.
