Chem 116, Fall 2006

UMass Boston, Chem 116, Fall 2006

There are only 3 ways to prepare a buffer

Goal: to create approximately equal quantities of weak acid CHEM 116 and its conjugate base in the same solution Molecular Structure, Acid Strength 1. Add 1 part HA and 1 part A– to water at the same time. November 16, 2006 Prof. Sevian 2. Add 1 part HA and 0.5 parts OH– to water. The OH– will react and convert 0.5 part of the HA to its conjugate, leaving 0.5 parts HA and 0.5 parts A–.

3. Add 1 part A– and 0.5 parts H+ to water. The H+ will react and convert 0.5 part of the A– to its conjugate, leaving 0.5 parts A– and 0.5 parts HA.

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Agenda Main points about titration

Titration curve for Acetic Acid z A closer look at titration

z Titration endpoint/equivalence point 14 z Buffer region of a titration

z When its ok to use Henderson-Hasselbach equation 12

z Titrating polyprotic acids 10 Equivalence z How molecular structure influences acid strength point 8 Titration

z Lewis theory pH midpoint 6 z Group problem 4

Initial 2 z Exam #3 is on Tuesday. Do NOT forget to bring a Buffer molarity region calculator! of acid 0 0 0.1 0.2 0.3 0.4 0.5 moles of NaOH added z The final exam is scheduled for Monday, December 18, 11:30AM- 2:30PM. Location TBA.

What’s so special about the titration midpoint? Titration midpoint on the titration curve

z It’s in the buffer region Titration curve for Acetic Acid –5 z It is where you have added exactly 0.5 parts of strong base for 1 Ka = 1.8u10 so pKa = 4.74 part of weak acid 14 z So, stoichiometry predicts these initial conditions: 12

– aq l – aq aq CH3COO ( ) + H2O ( ) ' OH ( ) + CH3COOH ( ) 10

0 M 0.150 M 0.300 M 8 Titration

pH midpoint + 0.150 - 0.150 - 0.150 6 pKa Initial 0.150 10-7 0.150 4 conditions 2 z Carrying this further, to the equilibrium point: 0 Change - x + x + x 0 0.1 0.2 0.3 0.4 0.5 moles of NaOH added Equilibrium 0.150–x § 0.150 y 0.150+x § 0.150

Titration of At the titration midpoint a polyprotic acid [CH3COOH][OH ] (e.g., H C O ) Kb [OH ] but ONLY at the titration midpoint 2 2 4 [CH COO ] 3 Starting with 100 mL of 0.100 M What does this mean? oxalic acid Second Kb [OH ] means and pK p First equivalence b OH Adding 0.100 M point so NaOH to it Titration midpoints pK w pKa pK w pH -2 Ka1 = 5.9x10 or pKa1 = 1.23 -5 Ka2 = 6.4x10 pKa pH but only at the titration midpoint! pKa2 = 4.19

6 From Chemistry & Chemical Reactivity 5th edition by Kotz / Treichel. C 2003. Reprinted with permission of Brooks/Cole, a division of Thomson Learning: www.thomsonrights.com. Fax 800-730-2215.

Henderson-Hasslebalch equation is valid in More about polyprotic acids buffer region only! (K b A– (aq) + H O (l) ' OH– (aq) + HA (aq) Example: phosphoric acid is triprotic equation) 2 Intial [A–] 10-7 [HA] + – K –3 o o 1) H3PO4 + H2O ' H3O + H2PO4 a1 = 7.5 u 10 Change - x + x + x – + 2– –8 2) H PO + H O ' H O + HPO Ka2 = 6.2 u 10 2 4 2 3 4 – x – -7 + x = y x Equilibrium [A ]o – § [A ]o 10 [HA]o + § [HA]o 2– + 3– K –13 3) HPO4 + H2O ' H3O + PO4 a3 = 3.6 u 10

Conclusions: Henderson-Hasslebalch equation is just the approximation that x is – z The first proton is the dominant reaction and produces the vast small compared to [A ]o and [HA]o majority of the acidity of a phosphoric acid solution [HA][OH ] [HA]o x [OH ] [HA]o [OH ] z If phosphate were added to water, the first ionization step K b | [A ] [A ]o x [A ]o 3– 2– – K = K / K PO4 + H2O ' HPO4 + OH b3 w a3 Rearranging, would produce the vast majority of the OH– in the resulting basic solution § [A ]o · § [A ]o · 9 [OH ] K ¨ ¸ pOH pK log¨ ¸ b ¨ ¸ b ¨ ¸ ©[HA]o ¹ ©[HA]o ¹

The small x approximation with a buffer Example: pH of a solution of polyprotic acid (Kb ' OH ¯(aq) A¯(aq) + H O (l) HA (aq) Exercise 16.13, pp. 692-693 equation) 2 +

-7 What is the pH of a 0.10 M solution of oxalic acid, H2C2O4? Intial [A¯]o 10 [HA]o + – What are the concentrations of H3O , HC2O4 , and oxalate ion Change - x + x + x 2– K -2 K -5 C2O4 ? a1 = 5.9u10 , a2 = 6.4u10 x -7 + x = y x Equilibrium [A¯]o – § [A¯]o 10 [HA]o + § [HA]o

K z If the equilibrium constant (in this case b) is very small, then very little A¯will hydrolyze, so x will be small

z If x is small, then subtracting or adding it to the A¯and HA concentrations will not affect them very much, so the A¯and HA concentrations at equilibrium will not be very different from what they were initially

z At equilibrium, the OH¯concentration will be some value, call it y

z Since x no longer appears in the equilibrium concentrations, you y 10 have only one variable to solve for ( )

Henderson-Hasslebalch equation/approximation An authentic buffer problem Only works when: You desire to make a buffer with pH 3.50. The buffer needs to - i.e., z Approximately equal quantities of HA and A present ( be able to withstand dilution (adding water) and addition of buffer region) small amounts (0.01 moles) of strong acid or base. The z The small x approximation must be valid chemicals you have available are:

z Acetic acid and sodium acetate

z Citric acid and sodium citrate § [A ]o · pOH pKb log¨ ¸ ¨ ¸ z Oxalic acid, sodium hydrogen oxalate, and sodium oxalate © [HA]o ¹ z Monosodium hydrogen phosphate, disodium hydrogen phosphate, and sodium phosphate § [A ]o · pH pKa log¨ ¸ ¨[HA] ¸ © o ¹ Which two chemicals should you choose, and in what quantities? z Advice: don’t use the H-H equation unless you are absolutely sure you are in the region where it is valid 15

Let’s say you want to make a buffer solution An authentic titration problem

What do you need to consider?

z What pH do you need the buffer to hold? z You have a weak acid (monoprotic) solution of known concentration and a strong base z How robust do you want the buffer to be? (monohydroxide) solution of unknown z What chemicals could be used to make a buffer? concentration. You wish to find out the concentration of the strong base solution. z What chemicals do you have available? z The concentration of weak acid is 0.100 M. –4 The Ka value is 4.5 u 10 z Which indicator is appropriate? z How should the titration proceed? (Which solution goes in the Erlenmeyer flask, and which goes in the burette?)

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Key points about titration and buffers

z Always ask yourself first if the acid and base are weak and/or Compare binary acids strong z There are four regions in a titration: Ordered according to Ka value (smallest Ka value on left) 1. Before any titrant has been added: concentrations of chemicals K K determined by a or b equation for whatever the analyte is z HF < HCl < HBr < HI 2. Buffer region where near equal molar amounts of conjugate acid K K and base present, range is r1 pH from p a (or p b): use equilibrium calculation/ICE chart or H-H equation z H2O < H2S < H2Se z Titration midpoint is halfway to equivalence point and is where perfect buffer exists z H N (usually written NH ) < H O < HF K K 3 3 2 z At titration midpoint, pH = p a (or pOH = p b) 3. Equivalence point where moles acid = moles base: if not strong acid + strong base, then use equilibrium for hydrolysis reaction 4. Beyond equivalence point where excess titrant: figure out how much titrant in excess

z Use an indicator that changes color near the equivalence pt 19

Molecular structure informs properties of acids Compare oxoacids (HOXOn) 1. Binary acids (HX where X is a nonmetal)

2. Oxoacids (formula HOXOn where X is a nonmetal) z Compare formulas HOX 3. Carboxylic acids (RCOOH where R is a carbon group) HOI < HOBr < HOCl The question is – how easily can the H+ be let go? Answer always depends on how much electron density gets drawn away from the H in the acid molecule z Compare HOXOn Look for the one feature that differs and therefore must be responsible HOCl < HOClO < HOClO2 < HOClO3 for the trend Arguments can depend on: Comparing sizes of atoms when H+ is to be removed from a single atom (leaving behind X¯) Comparing electronegativities of nonmetals when structures are the same Looking at “formal charges” of all atoms when structures differ to gauge where electron density is being drawn to 20

Metal cations form complexes with water in a Compare carboxylic acids “hydration sphere”

Effect of “acidifying” the H can be accomplished by placing electron- edition by Kotzby edition C 2003. Treichel. / hungry atoms within the R group th . Fax 800-730-2215. . Fax From Chemistry & Chemical Reactivity 5 Reactivity & Chemical Chemistry From Most metal cations have 6 waters in the hydration sphere Learning: Thomson of division a of Brooks/Cole, permission with Reprinted www.thomsonrights.com

3+ 3+ Al + 6 H2O o [Al(H2O)6] Lewis acid-base rxn

Since O’s electron lone pairs are busy being attracted to the metal cation, it + is relatively easy for an H2O to let go of an H

3+ + 2+ 21 [Al(H2O)6] + H2O ' H3O + [Al(H2O)5(OH)] Bronsted-Lowry rxn

Group Problem Lewis acids and bases –4 Formic acid (HCO H) has Ka = 1.9 u 10 . Begin with 0.200 z Lewis acid = accepts a pair of electrons 2 moles of formic acid in 1.00 L of solution. Gradually add z Lewis base = donates a pair of electrons NaOH to the solution while titrating the acid solution. Definition is more general than Bronsted-Lowry (Assume that the addition of titrant affects the total volume of solution negligibly.) a) What is the pH of the initial 0.200 M solution of formic acid? b) What is the pH after 0.100 moles of NaOH have been added to the solution? edition by Kotzby edition C 2003. Treichel. / th c) What is the pH after 0.200 moles of NaOH have been added to the solution? . Fax 800-730-2215. . Fax d) What is the pH after 0.300 moles of NaOH have been added to the solution? From Chemistry & Chemical Reactivity 5 Reactivity & Chemical Chemistry From Reprinted with permission of Brooks/Cole, a division of Thomson Learning: Learning: Thomson of division a of Brooks/Cole, permission with Reprinted www.thomsonrights.com e) Sketch the titration curve. Label the equivalence point.