Buffers and Titrations Review Week of 10/22/18-10/26/18 Geoff Geberth Important acid and base eqns. and defs. 2 Buffers Definition: A solution that contains substantial amounts of a compound in both its protonated and deprotonated forms that is resistant to pH changes as a result Perfect/Ideal Buffer:
pKa = desired pH Contains equal amounts of acid and base Buffer capacity: The total amount of acid or base that can be added to the buffer before neutralizing all of one of the forms of the compound For equal volume solutions, more conc = higher capacity For equal conc., greater volume = higher capacity 2 main ways to make a buffer: 1. Mix the conjugate pair 2. Partial neutralization 3 Which of the following has the greater buffer capacity?
50 mL mixture that is .2 M NH4Br and .3 M NH3 50 mL mixture that is .5 M NH4Br and .6 M NH3 There is more stuff to react away, thus it can resist the change for longer. Remember: if the volumes are the same, the higher concentration will have the higher capacity Capacity measures how well it can resist a change 4 Making buffers 1. Mix the conjugate pairs Get HA and the salt NaA Get B and the salt BHCl 2. Partial neutralization of HA or B (whats usually done in a lab)
Use the stock to make its own conjugate Add strong acid/base in a limiting amount Treat this like a limiting reaction problem 5 How a buffer works There is some mixture of an acid or base and its conjugate Remember: pH is a measure of the H+ concentration If we add a strong acid, the added proton is neutralized by the weak base, and is converted to weak acid If we add a strong base, the weak acid neutralizes the strong base and converts to the weak base We basically turn strong acids and bases to weak ones, where some quantity is protonated, thus minimizing the change to the overall H+ concentration
6 Pictures always help Weak acid (HA) Conj. Base (A-) Add strong base
Add strong acid Weak acid (HA) Conj. Base (A-) Weak acid
(HA) Conj. Base (A-) ++ 7 Henderson-Hasselbalch (H-H) Eqn Relates concentrations of protonated and deprotonated forms Predicts the pH of a buffer Also works for bases and Kb: Which goes in the numerator? Remember: Products over reactants!
8 H-H Eqn practice Calculate the pH of a buffer solution that is .45 M in NH4Cl and .15 M in NH3 ( Note we dont care about the volume! From Kb, we can calculate for NH3 9 H-H practice (cont) How would you prepare an NH4Cl-NH3 buffer ( with a pH of 9? For every 1 mole of NH4Cl salt, we need .56 moles of NH3 10
Titration Definition: procedure in which a strong acid/base of accurate concentration is added stepwise in small amounts (aliquots) to incrementally neutralize a solution Performed to either determine K or measure the concentration of an unknown Other definitions Analyte- the unknown solution for which you would like to know the concentration (the stuff in the beaker) Titrant- the known solution which you are using to determine the unknown (the stuff in the burette) Indicator- compound that changes the color of the solution used to determine when the pH has undergone changes 11
Titration (cont.) Equivalence point- point at which the number of moles of added titrant are equal to the moles of acid or base in the analyte solution Half equivalence point- point at which exactly half of the original analyte has been neutralized (pH = pKa at this point) End point- point at which an indicator changes color, signaling the stop of the titration Dependent on the indicator chosen Proper selection of indicator matches end point with equivalence point 12 Indicators Special types of weak acids or bases that have different colors when protonated or deprotonated
Only requires 1-2 drops to be visible Doesnt impact pH or overall concentration because quantity is so minute Approximate range is 1 off the pKa Center of range is pH = pKa 13 Protonated state If you find a molecule in solution, does it have a particular proton attached, or is the proton free in solution? 14
Polyprotic acids Acids with more than one acidic proton Each proton has its own equilibrium (multiple Kas) One proton comes off completely before the next H3PO4 is a great example Fraction of species diagram for diprotic Be able to do this for up to 6 protons 15 Titration curve for a polyprotic XS Base 3rd Proton
pH 2nd Proton 1st Proton mL NaOH 16 A long titration problem Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH. Calculate the pH after the addition of each of the following volumes of base:
10 mL 25 mL 50 mL 75 mL 100 mL 17 A long titration problem (10 mL) Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH. after neutralizing =6.67x10-2 M =1.67x10-2 M
=1.74 18 A long titration problem (25 mL) Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH. This is halfway to the first equivalence point + 0 1-
19 A long titration problem (50 mL) Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH. At the first equivalence point we get: This can be derived by looking at the definitions of K, but you havent yet discussed this in class 20 A long titration problem (75 mL) Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH.
This is halfway between the first and second equivalence points + 0 1- 21 A long titration problem (100 mL) Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH. At the second equivalence point, only the basic salt, NaA, is in solution
22 A long titration problem (100 mL) Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH. R A- (aq) + H2O (l) <-> HA (aq) + OH- (aq)
I (M) 0.033 - 0 ~0 C (M) -x -
+x +x E (M) 0.033 - x - x x 23
A long titration problem (100 mL) Consider the titration of 50.0 mL of a .1 M solution of the protonated form of the amino acid analine ( With .1 M NaOH. R A- (aq) + H2O (l) <-> HA (aq) + OH- (aq) I (M)
0.033 - 0 ~0 C (M) -x - +x
+x E (M) 0.033 - x - x x 24
