# Weak Acid Titration Calculations Paul Gilletti Ph.D. Mesa

Weak Acid Titration Calculations Paul Gilletti Ph.D. Mesa Community College Mesa, AZ 1 Strong and Weak acids Equilibrium and pH Strong Acids: HClO4 H2SO4 HNO3 HI HBr HCl HClO3 Weak Acids: The Rest 2 Titration Curve of a Weak Acid with a Strong Base equivalence point note volume ratio between these points -5 a

buffer region 25.0 mL of 0.10M Acetic acid K =1.8 x 10 14 12 10 Starting pH 8 6 Equivalence point 4 2 0 0 5 10 15 20 25

30 35 Volume of Strong Base Added 40 45 50 0.10 M NaOH 3 Acetic acid is a weak acid. What would be the starting pH of the solution if it were a strong acid? (25.0 mL of 0.10M Acetic acid) Do a quick calculation as if acetic acid were a strong acid and compare this value to that on the graph. How do the two values compare? Weak Acid: Graph starting pH = 2.87 Calculated as Strong acid: pH = -log(0.10) = 1.0 How and Why are they different? Lets investigate! 4 pH 14

12 10 8 6 4 2 0 Weak Strong 0 20 40 60 mL of Base Added 5 How do the two values compare? Graph starting pH = 2.87 Strong acid calculation: pH = -log(0.10) = 1.0 What is the significance of this difference? Since the two values differ by ~2 pH units and pH is a log scale, the concentration of H3O+ in the strong acid calculation is 1 x 102 or 100 times greater than that observed on the graph. Lets see WHY they are different?

6 Strong Acids: 100% ionized (completely dissociated) in water. HCl + H2O H3O+ + ClNote the one way arrow. Weak Acids: Only a small % (dissociated) in water. HC2H3O2 + H2O H3O+ + C2H3O2Note the 2-way arrow. Why are they different? 7 Strong Acids: HCl HCl HCl HCl HCl (H2O) ADD WATER to MOLECULAR ACID 8 Strong Acids: Cl(H2O) H3O+ H3O+ H3O+

H3O+ H3O+ ClCl Cl- - Cl- Note: No HCl molecules remain in solution, all have been ionized in water. 9 Weak Acid Ionization: HC2H3O2 HC2H3O2 HC2H3O2 HC 2H 3O 2 HC2H3O2 (H2O) Add water to MOLECULES of WEAK Acid

10 Weak Acid Ionization: HC2H3O2 HC2H3O2 HC2H3O2 HC 2H 3O 2 HC2H3O2 (H2O) H30+ C2H3O2- Note: At any given time only a small portion of the acid molecules are ionized and since reactions are running in BOTH directions the mixture composition stays the same. This gives rise to an Equilbrium expression, K a 11 pH calculations: To do pH calculations one must

consider both the nature and condition (amount of ionization) of the species in the solution and then calculate the concentration of the hydronium ion (H+ or H3O+). 12 Strong acid: [H3O+] = concentration of acid so: pH = -log [H3O+] = -log[acid] Weak Acid: one must calculate the [H3O+] from an equilibrium ionization expression. HA + H2O H3O + A + Ka = _ [H3O+][A-] [HA] - These are equilibrium concentrations. 13 Titration Curve of a Weak Acid with a Strong Base equivalence point note volume ratio between these points

-5 a buffer region 25.0 mL of 0.10M Acetic acid K =1.8 x 10 14 12 10 2 8 4 3 6 4 1. Starting pH, only acid and water. 2 0 0 5 10

15 20 25 30 35 Volume of Strong Base Added 40 45 50 0.10 M NaOH 14 Titration Curve of a Weak Acid with a Strong Base buffer region 14 Approximations: equivalence point note

volume ratio between these points 12 [H ][ A ] Ka [HA ] 10 [H+][A-] 8 2 4 pOH=-Log[XS OH-] 3 6 4 1

2 [OH ][HA] Kb [A ] [H ][ A ] Ka [HA ] [OH-]=[HA] [H+]=[A-] 0 0 5 10 15 20 25 30

35 40 45 50 Volume of Strong Base Added 15 Problem: Calculate the pH of 25.0 mL of 0.10M acetic acid (HOAc). The Ka of HOAc = 1.8 x 10-5 Which region of a titration curve would this be? 16 Problem: Calculate the pH of 25.0 mL of 0.10M acetic acid (HOAc). The Ka of HOAc = 1.8 x 10-5 Ka = [H+][OAc-] [HOAc] Ka = [x][x] 0.10-x Ka = x 2 0.10 HOAc H+ + OACI 0.10 0

0 +x +x C -x x x E 0.10-x Small,drop, WHY? x = 0.00134 = [H+] pH = -log 0.00134 =2.87 17 Titration Curve of a Weak Acid with a Strong Base equivalence point note volume ratio between these points -5 a buffer region 25.0 mL of 0.10M Acetic acid K =1.8 x 10 14 12

2.Some base has been added 10 This is also the buffer region 8 4 3 6 4 pKa = pH at this point (halfway point) 1. 2 0 0 5 10 15 20

25 30 35 Volume of Strong Base Added 40 45 50 0.10 M NaOH 18 Region 2 Theory and Calculations: Base has been added and some of the acid has been neutralized. What has changed and WHY? 1. The moles of acid is decreased. 2. The moles of acid anion (A-) is increased. HA + OH- H2O + A3. The total volume has increased. HA + OH- H2O + A- + XS HA note: the 1:1 ratio 19 Region 2 Theory and Calculations:

Base has been added and some of the acid has been neutralized. intial RXN HA + OH- H O + A2 Equilibrium XS HA + H2O H3O+ + A- 2 sources always Leads to: (calculate from equilibrium) Ka= [H+][A-] [HA] and [H+][A-] Why? 20 Region 2 Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5). Solution: HOAc + OH- H2O + OAc- 1:1 ratio

1. Find initial moles of Acid: 0.02500L 0.10mol HOAc = 0.0025 mol HOAc 1L 2. find moles of OH-: 0.01400L NaOH 0.080molNaOH 1molOH0.00112 mol OH_____________________________= 1L 1molNaOH 21 Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10 -5). Solution: 1. We found initial mol of Acid: HOAc + OH- H2O + OAc- 1:1 ratio 0.02500L 0.10mol _____ HOAc = 0.0025 mol HOAc 1L 2. We found mol of OH-: which =s initial moles of OAC- anion. 0.01400LNaOH 0.080molNaOH 1molOH_____________________________________= 0.00112 mol OH1molNaOH 1L 3. Subtraction gives what? No, not the numerical answer!

moles of XS acid (initially present for OUR Eq. calculation). 22 Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10 -5). Solution: 1. Find initial moles of Acid: HOAc + OH- H2O + OAc- 1:1 ratio 0.02500L 0.10mol _____ HOAc = 0.0025 mol HOAc 1L 2. find moles of OH-: which =s initial moles of OAC- anion. 0.01400LNaOH 0.080molNaOH 1molOH_____________________________________= 0.00112 mol OH1L 1molNaOH 3. Subtraction gives moles of XS acid (initial for I.C.E.). 0.0025 - 0.00112 = 0.00138 mole acid INITIAL 4. and..we have 0.00112 mole OAc- INITIAL for I.C.E. 23 Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10 -5).

Solution: FROM PREVIOUS SLIDE. 1. initial M of Acid: HOAc + OH- H2O + OAc- 1:1 ratio 0.00138 mole/0.039L = 0.0354 M acid after RXN =INITIAL 2. M of OH-: which =s initial M of OAC- anion. 0.00112 mole/0.039L = 0.0287 M OAc- after RXN =INITIAL** [H ][ A ] Ka [HA] [x][0.0287 x] 1.8 x 10 [0.0354 x] -5 x = 0.0000222 =[H+] + [HOAc] [H ] + [OAc ] ** I 0.0354 0 0.0287 -x +x +x

C E 0.0354-x x 0.0287+x Can we drop these? pH = - Log [0.0000222] = 4.65 24 Region 2 summary: 1. write balanced equation for initial weak/strong RXN. 2. find moles of acid 3. find moles of base 4. subtract to determine XS (and limiting reactant which equals initial moles of salt formed) 5. Change XS to M and limiting Reactant to M 6. Write balanced Chemical Equilibrium Equation: - HA H+ + A 7. Use values from #5 a initial values in I.C.E. Chart 8. Work as Equilibrium problem. 25 Titration Curve of a Weak Acid with a Strong Base equivalence point note volume ratio between these points

-5 a buffer region 25.0 mL of 0.10M Acetic acid K =1.8 x 10 14 12 2. 10 4 8 3. Equivalence point 6 4 1. 2 0 0 5

10 15 20 25 30 35 Volume of Strong Base Added 40 45 50 0.10 M NaOH 26 Region 3. Theory and Calculations. Mole of acid = moles of OH- added. Equivalence pt. HA + OH- H2O + A- no XS of either

The titration curve indicates the pH to be above 7. What causes this? Hydrolysis of the A- anion: A- + H2O HA + OHSince this equilibrium involves OH- being formed, it is a Kb problem. [OH ][HA] Kb [A ] 27 Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. 1. Find moles of Acid (which =s moles of base at Eq pt.) 25.00mL 0.10mol HOAc 1000mL = 0.0025 mol HOac 2. Find mol of OH31.25mL 0.080mol NaOH 1molOH __ - = 0.0025 mol OH1000mL 1mol NaOH 3. Since the # of moles are the same, this is at the Eq. Pt. and therefore the initial moles of OAc- ion = 0.0025 mol

28 Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. moles of Acid = moles of base = 0.0025 = Eq. Pt. New Volume = 0.05625L..and [OAc ] = 0.0025/0.05625=0.0444M Question: What equilibrium(ia) gives rise to the pH? [OH ][HOAc] Kb [OAc ] [x 2 ] Kb [0.0444 x] 1. hydrolysis of the acetate anion:HOAc + OH OAc + H2O [OAc-] [HOAc] [OH-] 0 0 I 0.0444 +x +x

C -x x x E 0.0444-x 29 Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. moles of Acid = moles of base = 0.0025 = Eq. Pt. New Volume = 0.05625L...... and [OAc-] =0.0444M Question: What equilibrium(ia) gives rise to the pH? 1. hydrolysis of the acetate anion: HOAc + OH OAc + H 2O [OH ][HOAc] OAc HOAc OH Kb [OAc ] 0 0

I 0.0444 -x +x +x C [x 2 ] Kb x x E 0.0444-x [0.0444 x] Can we drop this x? How do we find Kb? Also: KaKb=Kw=1.0 x 10-14 @25oC so Kb= 5.56 x 10-10 30 Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. moles of Acid = moles of base = 0.0025 = Eq. Pt. New Volume = 0.05625L...... and [OAc-] =0.0444M Question: What equilibrium(ia) gives rise to the pH? 1. hydrolysis of the acetate anion: HOAc + OH OAc - + H2O

[OH ][HOAc] OAc HOAc OH Kb [OAc ] 0 0 I 0.0444 +x +x C -x 2 [ x ] 5.56 x 10-10 x x E 0.0444-x [0.0444 x] - x = 4.97 x 10-6 = [OH ] pOH = 5.30 pH =14 - 5.30 = 8.70 31

Titration Curve of a Weak Acid with a Strong Base equivalence point note volume ratio between these points -5 a buffer region 25.0 mL of 0.10M Acetic acid K =1.8 x 10 14 12 2. 10 4 8 6 XS base 3.

4 1. 2 0 0 5 10 15 20 25 30 35 Volume of Strong Base Added 40 45 50

0.10 M NaOH 32 Region 4 theory and Calculations: All the acid has been neutralized and XS base has been added HA + OH- H2O + A- + XS OHWhat is the dominating factor that controls the pH? The XS strong base (OH-) Calculations: Find moles of XS OH- and then use the new volume to find [OH-] and then the pOH and pH. 33 Region 4 problem: Calculate the pH of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 40.00mL of 0.080M NaOH. Ka(HOAc)=1.8 x 10-5. 1. Find moles of acid: 25.00mL 0.10mol HOAc 1000mL = 0.0025 mol HOAc 2. Find moles of OH-: 40.00mL 0.080molOH 1000mL 3. Subtract to find XS: = 0.0032 mol OH-

0.0032 - 0.0025 = 0.0007 mole XS OH4. Find OH- concentration, pOH, and pH: pOH = -Log[0.0007/0.06500] = 1.97 and pH = 12.03 34 Titration Curve of a Weak Acid with a Strong Base buffer region 14 Approximations: equivalence point note volume ratio between these points 12 [H ][ A ] Ka [HA ] 10 [H+][A-] 8

2 4 STUDY FOR QUIZ pOH=-Log[XS OH-] 3 6 4 1 2 [OH ][HA] Kb [A ] [H ][ A ] Ka [HA ] [OH-]=[HA] [H+]=[A-] 0 0

5 10 15 20 25 30 35 40 45 50 Volume of Strong Base Added 35 Titration Curves for Three Different Weak Acids Ka = 1.4 x 10-7 pH Ka = 1.4 x 10-6

Ka = 1.4 x 10-5 mL of Strong Base Added 36 37 CO32- + H+ HCO3- HCO3- + H+ HCO3 38 39 Titration Curve of a Weak Acid with a Strong Base molesOH molHA [OH ] totalvolum Region 4: equivalence point note volume ratio between these pointse

buffe2: r region Region acid + baseXS acid + salt HA+NaOH HA(XS)+NaA 14 12 moleOH [H ] totalLiters Ka molHA molOH totalLiters 10 8 pOH=-Log[OH-] Region 3: mol acid=mol base hydrolysis of the salt anion A- + H2OHA + OH

6 [OH ][HA] x2 Kb [ A ] [ HA] initialmoleHA totalLiters 4 2 Region 1: only acid (HA) Ka [H ][ A ] x2 x2 [ HA] [HA]0 x [HA]0 0 0 5

10 15 20 25 30 35 Volume of Strong Base Added 40 45 50 0.10 M NaOH 40 Titration Curve of a Weak Acid with a Strong Base Region 4: XS base molesOH molHA [OH ] totalvolum equivalence point

note volume ratio between these pointse buffer region 14 pOH=-Log[OH-] Region 2: acid + baseXSacid + salt HA+NaOH HA(XS)+NaA 12 10 moleOH [ H ] totalLiters Ka molHA molOH totalLiters

8 Region 3: mol acid=mol base hydrolysis of the salt anion A- + H2OHA + OH- 6 Kb 4 Region 1: only acid (HA) 2 Ka [OH ][HA] x2 [ A ] [HA] initialmoleHA totalLiters [H ][ A ] x2 x2

[ HA] [HA]0 x [HA]0 0 0 5 10 15 20 25 30 35 40 45 50 Volume of Strong Base Added 41

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