Unit 19 Acid Base Equilibria: Titrations CHM 1046: General Chemistry and Qualitative Analysis Unit 19 Acid Base Equilibria: Titrations Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL Textbook Reference: Chapter 19 (sec. 5-8) Module 9

Titration Soln-Unknown Concentration (M): Acid A volumetric technique in which one can determine the concentration of a solute in a solution of unknown concentration, by making it react with another solution of known concentration (standard). Standard-of known Conc.(M) Known Volume (V) Measure Vol to reach end pt. MolesB = M x V MolesA = M x V If: MolesA = MolesB {*TitrationMovie} Then: (M x V)A = (M x V)B

Determining the Concentration of Solutions by Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base) until neutralization occurs. (Standard) Example: HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) xa HA (aq) + xb MOH (aq)  MA (aq) + H2O (l) H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + H2O(l) Neutralization: equivalence point # mol1(acid) = # mol2(base) 1 mol1(acid) = 1 mol2(base) xa xb

Titration Calculations: Stoichiometry using Molarities xA HN + xB MOH  MN + HOH Neutralization: 1 moles(acid) = 1 moles(base) xA xB Since moles = MV = moles x Liter Liter ηA ηB MA x VA MB x VB = = = = = Where = coefficients from balanced equations * Equation Useful for determining Molarities and Volumes at the Equivalence Point of a Titration *

Solution Stoichiometry Problems: Molarity Problem: A volume of 16.3 mL of a 0.30M NaOH solution was used to titrate 25.00 mL H2SO4. What is the concentration of H2SO4 in the solution of unknown concentration? H2SO4 + 2 NaOH  2HOH + Na2SO4 MA x VA = MB x VB 1 2 = 0.098 M H2SO4 Titration of a Strong Acid with a Strong Base

Titration Graph: pH vs. Volume Titration Data: Excess base mL of NaOH pH 1.2 10 1.4 20 1.6 30 1.7 40 1.9 50 7.0 60 12.0 70 12.2 80 12.3 SB Phenolphthalein Indicator Acid = Base Methyl Red Indicator pH meter Excess acid SA {Titration2}

Titrations: The Strength of Acids & Bases Strong Base with Strong Acid SA Weak Base with Strong Acid SA Phenolphthalein 8-10 SB WB Weak Base with Weak Acid Strong Base with Weak Acid WA WA SB WB

(2) Titration of a WA with a SB With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

Selecting Appropriate Indicators Select appropriate indicator for following: Phenolph Meth Red Meth Red Phenolph ????? an indicator is chosen so that it will change color at a pH just beyond the equivalence point (mid point of the steep vertical portion of the graph). The first point at which the indicator permanently changes color marks the end of the titration and is called the indicator end-point. Dropping a perpendicular from the indicator end-point to the x-axis is a very close estimation of the equivalence point.

(1) Titration of a SA with a SB (or SB with a SA)

Acid-Base Neutralization Equations (1) Strong Acids and Bases are represented in completely dissociated state: as H+ and OH- (2) Weak Acids and Bases are represented in undissociated state: as HA and B (or MOH)

(H2CO3 + K+) (H2CO3 + Ca2+ + C2H3O2- ) (H2C O3 + Zn2+) (H2CO3 + Zn2+ + SO42-)

Equations and Tables used in solving A-B Titration Problems (1) Acid Base Neutralization: when you reach the end-point using SA or SB (2) Acid Base Neutralization: when not at the end-point or using WA or WB (3) WA or WB Equilibrium problems HA ↔ H+ + A- HA + OH-  H2O + A- Use: Mole ICEnd Table Use: [ICE] Table Mole HA + MOH → MA + H20 HA ↔ H+ A- (.15M)(.025L) (.10M)(.030L) II 0.061 C -x +x E 0.061 - x x II 0.0038 η 0.0030 η 0 η C - 0.0030 η + 0.0030 η End 0.0008 η

(1) Titration of a SA with a SB Example: 25 mL of 0.15M HCl with 0.10M NaOH. (1) What volume of NaOH is required to reach the equivalence point? (2) What is the pH of the initial strong acid? (strong acid problem) In strong acid [HA]=[H+], so pH=-log [H+] = -log (0.15) = 0.82 (3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (excess SA problem) *what is used for neutralization Rx?* Mole HX + MOH → MX + H20 (.15M)(.025L) (.10M)(.030L) II 0.0038 η 0.0030 η C - 0.0030 η + 0.0030 η End 0.0008 η pH = -log (1.4 x 10-2) = 1.9 Salt of SA & SB: will not Hydrolyze

(1) Titration of a SA with a SB Example: 25 mL of 0.15M HCl with 0.1M NaOH. (4) What is the pH at the equivalence point? Only salt + water present and salt will not hydrolyze water since it is derived from SA & SB. So pH = 7 (5) What is the pH after the equivalence point? Lets say after 40. mL of NaOH. (excess SB problem) *what is used for neutralization Rx?* mole HX + MOH → MX + H20 (.15M)(.025L) (.10M)(.040L) II 0.0038 η 0.0040 η 0 η C - 0.0038 η + 0.0038 η End 0 η 0.0002 η Salt of SA & SB: will not Hydrolyze pOH = -log(3.1x10-3) = 2.5 pH + pOH = pKw pH = pKw - pOH pH = 14 – 2.5 = 11.5

(2) Titration of a WA with a SB SB (OH-) HA ↔ H+ + A- HA + OH-  H2O + A- *what is used for neutralization Rx?* Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. The pH >7 at the equivalence point. *what is used for equilibrium Rx?*

(2) Titration of a WA with a SB Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH. (1) What volume of NaOH is required to reach the equivalence point? (2) What is the initial pH of the acetic acid? (Before titration, WA Equilibrium problem) HA ↔ H+ A- II 0.15 C -x +x E 0.15 - x x

(2) Titration of a WA with a SB Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH. (3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (WA Buffer problem) mole HA + MOH → MA + H20 (.15M)(.025L) (.10M)(.030L) II 0.0038 η 0.0030 η 0 η C - 0.0030 η + 0.0030 η End 0.0008 η Salt of WA & SB: WILL Hydrolyze H2O

(2) Titration of a WA with a SB Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.1M NaOH. (4) What is the pH at the equivalence point? This happens @ 37.5 mL (Hydrolysis of Salt derived from a WA & SB) mole HA + MOH → MA + H20 (.15M)(.025L) (.10M)(.038L) II 0.0038 η 0 η C - 0.0038 η + 0.0038 η End 0 η Salt of WA & : will Hydrolyze water Salt is NaC2H3O2 Na+ = derived from SB (NaOH), will not hydrolyze. C2H3O2- = derived from WA (acetic acid) it WILL hydrolyze water.

(2) Titration of a WA with a SB Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.1M NaOH. (4) What is the pH at the equivalence point? This happens @ 37.5 mL (Hydrolysis of Salt derived from a WA & SB) mole HA + MOH → MA + H20 (.15M)(.025L) (.10M)(.038L) II 0.0038 η 0 η C - 0.0038 η + 0.0038 η End 0 η [C2H3O2-] =. Salt is NaC2H3O2 C2H3O2- + H2O ↔ HC2H3O2 + OH- II 0.061 C -x +x E 0.061 - x x pH + pOH = pKw pH = pKw - pOH pH = 14 – 5.2 = 8.8

2005B Q1

Mole ICEnd Table:

(3) Titration of a WB with a SA SA (H3O+) MOH ↔ M+ + OH- H3O+ + MOH  2H2O + M+ The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

(3) Titration of a WB with a SA: Calculation of pH (1) Acid Base Neutralization (2) WB Equilibrium problem B+ H2O ↔ BH+ + OH- B + H+  H2O + A- Weak base and strong acid Use: Mole ICEnd Table Use: [ICE] Table Mole B + H+ → A- + H20 B + H2O ↔ BH+ OH- (.15M)(.025L) (.10M)(.030L) II 0.061 C -x +x E 0.061 - x x II 0.0038 η 0.0030 η 0 η C - 0.0030 η + 0.0030 η End 0.0008 η

2007A Q1 Titration: weak acid with strong base

Use: Mole ICEnd Table (e) What’s pH? For F-: For HF: = 0.15 M F- OH- → A- + H20 (.40 M)(.025L) (.40M)(.015L) II 0.010 η 0.0060 η 0 η C - 0.0060 η + 0.0060 η End 0.0040 η (e) What’s pH? For F-: = 0.15 M F- For HF: = 0.10 M HF

Use: [ICE] Table II 0.10 0.15 C -x +x E 0.10 - x x 0.15 + x 0.15 M F- 0.10 M HF Use: [ICE] Table HA H+ ↔ A- II 0.10 0.15 C -x +x E 0.10 - x x 0.15 + x

Titrations of Polyprotic Acids Ka3 In these cases there is an equivalence point for each dissociation. Ka2 Ka1

Titration: weak acid strong base 2005B Q1

2005B Q1

Titration: weak base strong acid 2000 QA

Titration: weak acid strong base 2001 Q3

Answers 2001 Q3

Titration: weak acid strong base 2002A Q1

2003A Q1 Titration: weak base strong acid

2006B Q1 Titration: weak acid strong base

Expressing Concentrations of Solutions: Molarity (& Normality*) * For MDC students only!

Molarity (M) xA HN + xB MOH  MN + HOH mol mol = moles of solute Liters of solution = xA HN + xB MOH  MN + HOH mol mol A = B (mol/L)A x LA (mol/LB) x LB = MA x VA = MB x VB Where = coefficients for the acid (A) and the base (B) from the balanced neutralization equations

(mol/L)A x LA (mol/LB) x LB = mol mol A = B (mol/L)A x LA (mol/LB) x LB = MA x VA MB x VB = molesA MB x VB =

xA HN + xB MOH  MN + HOH For titrations: Since MB x VB =

Molarity (M) vs. Normality (N) Lesson for MDC students only: Molarity (M) vs. Normality (N) equiv of solute L of solution N = mol of solute L of solution M = Where: M = N When n = 1 That is when using HCl, KHP NaOH But not when using H2SO4, Ca(OH)2 A/B = # H+ or #OH- Redox = #e- involved in balanced redox equation.

Molarity (M) vs. Normality (N) Acid g-MM (g/) + H20 to HCl 36 g + 1L = H2SO4 98 g + 1L = H3PO4 98 g + 1L = Molarity (/L) Normality (eq/L) g-EW 36/1 =36 98/2 =49 98/3 =32.7 Eq(g/gEW) 36/36 98/49 98/33 1M = 1N 2N 3N

N = M or M = N NA x VA = NB x VB 2H3PO4 + 3 Ca(OH)2  6 HOH + Na3PO4 2M H3PO4 3M Ca(OH)2 Using Molarity 1N H3PO4 1N Ca(OH)2 Using Normality N = M or M = N NA x VA = NB x VB Using Normality for titrations:

Titration of a WA with a SB Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations.

Titration: measuring the Equivalence Point (H+ = OH-) Methyl red in base (range R4-6Y) Phenolphthalein in base (range C 8-10 F) A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. The end-point of a titration is when indicator changes color.

Titration of a SA with a SB xA HN + xB MOH  MN(aq) + HOH 1. From the start of the titration the pH goes up slowly. Just before the equivalence point, the pH increases rapidly. Excess base 2. At the equivalence point, moles H+ = moles OH-, and the solution contains only water and the salt from the cation of the base and the anion of the acid. H+ = OH- 3. Just after the equivalence point, the pH increases rapidly. As more base is added, the increase in pH again levels off. Excess acid

Titration xa xb Solution of known concentration (M2 x V2 = #mol2) Solution of unknown concentration (M1 x V1 = #mol1) Neutralization: 1 mola = 1 molb (equivalence point) xa xb {*TitrationMovie}

Stoichiometric/Volumetric Calculations xA HN + xB MOH  MN + HOH ACID ACID ACID xA xB BASE BASE BASE MA x VA MB x VB = = =

2006 (B)

2007 (A)