Experiment 3 Flashcards
(8 cards)
Buffer solution
-a solution that resists changes in pH upon addition of small amounts of acid/base
-Consists of a weak acid and its salt (conjugate base) or a weak base and its salt (conjugate acid)
-Acid neutralizes base, base neutralizes acid
-When a small amount of acid is added to a buffer, it will be consumed (neutralized by the base component of the buffer and vice versa if base is added) and the pH will only slightly be affected
Henderson-Hasselbach equation
-pH = pKa - log ([HA]/[A-])
-used for weak acid/base and strong base/acid
Equivalence point
point at which the reactant being titrated has been exactly consumed by the addition of the titrant
End point
when the indicator changes color, dependent on the indicator chosen
Weak acid/Strong base titration curve
-Titration curve will start low and gradually increase, rising steeply at equivalence point to have slope of infinity (vertical)
Weak base/Strong acid titration curve
-Titration curve will start high and gradually decrease, falling steeply at equivalence point to have slope of infinity (vertical)
pH at different points of weak acid/strong base titration
-Before addition of NaOH: pH can be found by solving the weak acid equilibrium
Ka = [H3O+][A-][HA]
-Buffer region (after addition of base but before eq, point): henderson-hasselbalch equation can be used to find pH
-At equivalence point: all weak acid has been converted to salt and solution only contains conjugate base, A-. [OH-] can be found through:
A- + H2O ⇄ OH- + HA
Kb = [OH-][HA][A-]
-Beyond equivalence point: when excess strong base is added, the [OH-] of solution is dominated by the quantity of NaOH added. pH is calculated for a strong base solution
Obtaining pKa of unknown acid from titration curve
-At halfway to equivalence point, half of the weak acid has reacted with the strong base, creating a ratio of HA to A- of 1:1
-Since HA:A- ratio is 1:1, the pH = pKa - log([HA]/[A-]) will become
-pH = pKa
-As log(1) = 0
-So you can find pKa by finding ½ VE
