Page III-b-1 / Chapter Fourteen Part II Lecture Notes ACID-BASE REACTIONS Chapter (Part II A Weak Acid + Strong Base Titration Titrations In this technique a known concentration of base (or acid is slowly added to a solution of acid (or base A meter or indicators are used to determine when the solution has reached the equivalence point where moles of acid = moles of base Chemistry 3 Professor Michael Russell Titrations Acid-Base Titrations Adding NaOH from the buret to acetic acid: Initially the increases very slowly, then rises dramatically at equivalence point rises, then levels off as equivalence point passed Notice how [HOAc] diminishes with NaOH, but [OAc - ] increases with NaOH Important to consider conjugates in titrations where weak acids or bases dominate But first, buffers and the common ion effect QUESTION: What is the effect on the of adding NH Cl to.5 M NH 3 (aq? NH 3 (aq + H O qe NH +(aq + OH - (aq Here we are adding an ion COMMON to the equilibrium with NH3 (i.e. NH + Le Chatelier predicts that the equilibrium will shift to the. The will go. After all, NH + is an acid A special form of Le Chateliers Principle... A special form of Le Chateliers Principle... QUESTION: What is the effect on the of adding NH Cl to.5 M NH 3 (aq? NH 3 (aq + H O qe NH +(aq + OH - (aq Let us first calculate the of a.5 M NH 3 solution. [NH 3 ] [NH +] [OH - ] initial.5 change -x +x +x equilib.5 - x x x Page III-b-1 / Chapter Fourteen Part II Lecture Notes
Page III-b- / Chapter Fourteen Part II Lecture Notes A special form of Le Chateliers Principle... QUESTION: What is the effect on the of adding NH Cl to.5 M NH 3 (aq? NH 3 (aq + H O qe NH +(aq + OH - (aq K b = 1. x 1-5 = [NH + ][OH - ] [NH 3 ] Assuming x is <<.5, we have [OH - ] = x = [K b (.5] 1/ =.1 M This gives poh =.7 and so =. -.7 = 11.33 or: = + log [K b C b ] 1/ = 11.33 = x.5 - x A special form of Le Chateliers Principle... Problem: What is the of a solution with.1 M NH Cl and.5 M NH 3 (aq? NH 3 (aq + H O qe NH +(aq + OH - (aq We expect that the will decline on adding NH Cl. Lets test that [NH 3 ] [NH +] [OH - ] initial.5.1 change -x +x +x equilib.5 - x.1 + x x A special form of Le Chateliers Principle... Problem: What is the of a solution with.1 M NH Cl and.5 M NH 3 (aq? NH 3 (aq + H O qe NH +(aq + OH - (aq K b = 1. x 1-5 = [NH + ][OH - ] [NH 3 ] Because equilibrium shifts left, x is MUCH less than.1 M, the value without NH Cl. = x(.1 + x.5 - x Problem: What is the of a solution with.1 M NH Cl and.5 M NH 3 (aq? NH 3 (aq + H O qe NH +(aq + OH - (aq K b = 1. x 1-5 = [NH + ][OH - ] [NH 3 ] A special form of Le Chateliers Principle... [OH - ] = x = (.5 /.1K b =.5 x 1-5 M This gives poh =.35 and = 9.5 drops from 11.33 to 9.5 on adding a common ion (NH +. = x(.1 + x.5 - x The function of a buffer is to resist changes in the of a solution. Buffers invoke a special case of the common ion effect. Buffer Composition: Weak Acid + Conj. Base HOAc + OAc - H PO - + HPO - Weak Base + Conj. Acid NH 3 + NH + HCl is added to pure water. HCl is added to a solution of a weak acid H PO - and its conjugate base HPO -. Buffers and Henderson-Hasselbalch Guide Page III-b- / Chapter Fourteen Part II Lecture Notes
Consider HOAc/OAc - to see how buffers work The weak acid in buffers consumes strong base We know: OAc - + H O qe HOAc + OH - K b = 5. x 1-1 Therefore, the reverse reaction of the weak acid with added OH - (strong base would be: HOAc + OH - qe OAc - + H O and: K reverse = 1/ K b = 1. x 1 9 K reverse is VERY large, so HOAc completely consumes the OH - Page III-b-3 / Chapter Fourteen Part II Lecture Notes Consider HOAc/OAc - to see how buffers work Conjugate base in buffers consumes strong acid HOAc + H O qe OAc - + H 3 O + K a = 1. x 1-5 Therefore, the reverse reaction of the weak base with added H 3 O + (strong acid would be: OAc - + H 3 O + qe HOAc + H O K reverse = 1/ K a = 5. x 1 K reverse is VERY LARGE, so OAc - completely consumes the H 3 O + Problem: What is the of a buffer that has [HOAc] =.7 M and [OAc - ] =. M? HOAc + H O qe OAc - + H 3 O + K a = 1. x 1-5 [HOAc] [OAc - ] [H 3 O + ] initial.7. change -x +x +x equilib.7 - x. + x x Problem: What is the of a buffer that has [HOAc] =.7 M and [OAc - ] =. M? HOAc + H O qe OAc - + H 3 O + K a = 1. x 1-5 [HOAc] [OAc - ] [H 3 O + ] equilib.7 - x. + x x Assuming that x <<.7 and., we have K a = 1. x 1-5 = [H 3 O+ ](..7 [H 3 O + ] = 1. x 1-5 * (.7 /. =.1 x 1-5 and =. Notice that the expression for calculating the H + conc. of the buffer is: Orig. conc. of HOAc [H 3 O + ] = K Orig. conc. of OAc - a [H 3 O + ] = [Acid] or [Conj. base] K a [OH - ] = Page III-b-3 / Chapter Fourteen Part II Lecture Notes [Base] [Conj. acid] K b Notice that [H 3 O + ] or [OH - ] depend on K and the ratio of acid and base concentrations
Page III-b- / Chapter Fourteen Part II Lecture Notes Henderson-Hasselbalch Equation [H 3 O + ] = [Acid] [Conj. base] K a Take the negative log of both sides of this equation: [Acid] = pk a - log [Conj. base] The Henderson - [Conj. base] Hasselbalch Eq = pk a + log [Acid] The is determined largely by the pk a of the acid and then adjusted by the ratio of acid and conjugate base. Important equation Henderson-Hasselbalch Equation [Conj. base] = pk a + log [Acid] Note that the CONCENTRATIONS of the acid and conjugate base are not important. It is the RATIO of the NUMBER OF MOLES that affects Result: diluting a buffer solution does not change its Result: You can use moles or molarity when using Henderson-Hasselbalch See: Buffers and Henderson-Hasselbalch Guide Problem: What is the when 1. ml of 1. M HCl is added to a 1. L of pure water (before HCl, = 7. b 1. L of buffer that has [HOAc] =.7 M and [OAc-] =. M ( =. Solution to Part (a of strong acid = - log [H 3 O + ] = - log [HCl] M 1 V 1 = M V 1. M * 1. ml = M 11 ml M = 9.99 x 1 - M = [H 3 O + ] = 3. What is the when 1. ml of 1. M HCl is added to a 1. L of pure water (after HCl, = 3. b 1. L of buffer that has [HOAc] =.7 M and [OAc-] =. M ( =. Solution to Part (b Step 1 - do the stoichiometry H 3 O + (from HCl + OAc - (from buffer ---> HOAc (from buffer The reaction occurs completely because K is very large. What is the when 1. ml of 1. M HCl is added to a 1. L of pure water (after HCl, = 3. b 1. L of buffer that has [HOAc] =.7 M and [OAc-] =. M ( =. What is the when 1. ml of 1. M HCl is added to a 1. L of pure water (after HCl, = 3. b 1. L of buffer that has [HOAc] =.7 M and [OAc-] =. M ( =. Solution to Part (b: Step 1-Stoichiometry [H 3 O + ] [OAc - ] [HOAc] Initial.999..7 Change -.999 -.999 +.999 After rxn.599.71 Now we need to calculate using our new buffer solution Solution to Part (b: Step - Equilibrium HOAc + H O qe H 3 O + + OAc - [HOAc] [OAc - ] [H 3 O + ] Initial.71.599 Change -x +x +x Equilibrium.71-x.599+x x Page III-b- / Chapter Fourteen Part II Lecture Notes
Page III-b-5 / Chapter Fourteen Part II Lecture Notes What is the when 1. ml of 1. M HCl is added to a 1. L of pure water (after HCl, = 3. b 1. L of buffer that has [HOAc] =.7 M and [OAc-] =. M ( =. Solution to Part (b: Step - Equilibrium HOAc + H O qe H 3 O + + OAc - [HOAc] [OAc - ] [H 3 O + ] Equilibrium.71-x.599+x x Because [H 3 O + ] =.1 x 1-5 M BEFORE adding HCl, we again neglect x relative to.71 and.599. What is the when 1. ml of 1. M HCl is added to a 1. L of pure water (after HCl, = 3. b 1. L of buffer that has [HOAc] =.7 M and [OAc-] =. M ( =. Solution to Part (b: Step - Equilibrium HOAc + H O qe H 3 O + + OAc - [H 3 O + ] = [HOAc] [OAc - ] K a =.71.599 (1. x 1-5 [H 3 O + ] =.1 x 1-5 M ------> =. The has not changed on adding HCl to the buffer Use the Henderson-Hasselbalch equation to calculate the when 1. ml of 1. M HCl is added to 1. L of buffer that has [HOAc] =.7 M and [OAc-] =. M We can use an alternate form of the H-H equation: ( mol Conj base mol strong acid = pk a + log mol weak acid + mol strong acid pk a = - log (1. * 1-5 =.7 ( ( (..1 =.7 + log =.7.7 +.1 Answer in good agreement with ice method (. See: Buffers and Henderson-Hasselbalch Guide Adding Acids and Bases to Buffers For adding strong acids to buffers: ( mol Conj base mol strong acid = pk a + log mol weak acid + mol strong acid ( For adding strong bases to buffers: ( mol Conj base + mol strong base = pk a + log mol weak acid mol strong base ( Very useful for calculating changes in buffers See: Buffers and Henderson-Hasselbalch Guide Preparing a Buffer You want to create a buffer solution with a =.3. This means [H 3 O + ] = 1 - = 5. x 1-5 M It is best to choose an acid such that: * [H 3 O + ] K a, or * pk a You get the exact [H 3 O + ] (or by adjusting the ratio of weak acid to conjugate base. For a =.3 buffer, we will look for a value of K a 5. x 1-5 or a pk a.3 Preparing a Buffer You want to buffer a solution at =.3 or [H 3 O + ] = 5. x 1-5 M POSSIBLE ACIDS K a pk a HSO - / SO - 1. x 1 -. HOAc / OAc - 1. x 1-5.7 HCN / CN -. x 1-1 9. Best choice is acetic acid / acetate - closest in [H 3 O + ] to K a or to pk a Page III-b-5 / Chapter Fourteen Part II Lecture Notes
Page III-b- / Chapter Fourteen Part II Lecture Notes Preparing a Buffer You want to buffer a solution at =.3 or [H 3 O + ] = 5. x 1-5 M [H 3 O + ] = 5. x 1-5 = [HOAc] [OAc - ] (1. x 1-5 Solve for [HOAc] / [OAc - ] ratio: = [H 3 O + ] / K a =.7 / 1 Therefore, if you use.1 mol of NaOAc and.7 mol of HOAc, you will have =.3. Preparing a Buffer Solution Buffer prepared from. g NaHCO 3 weak acid 1. g Na CO 3 conjugate base HCO - 3 + H O qe H 3 O + + CO - 3 What is the? Check yourself Answer: 1.5 Titration Calculations Allow us to calculate at any point in a titration We will study four types of titrations: SA + SB (equivalence = 7 SB + SA (equivalence = 7 WA + SB (equivalence > 7 WB + SA (equivalence < 7 Titrations separated into regions each with their own formula Buffers used in WA + SB and WB + SA We will do examples of all four as practice in lab. See Titration Guide Titration Calculation Hints Strong acids and strong bases annihilate their opposites Weak acids and bases produce their conjugates K a and K b may be determined through titrations or half-equivalence values Remember: K a *K b = K w = 1.*1 - (5 C Often helpful to find pk a or pk b Remember: pk a + pk b = Need to know initial concentration and volume of titrant Strong Acid + Strong Base Strong Acid + Strong Base 1 1 post-equivalence (n sb > n sa equivalence (n sa = n sb, = 7 A strong acid + strong base titration 1 1 post-equivalence (n sb > n sa equivalence (n sa = n sb, = 7 initial (only n pre-equivalence (n sa > n sb sa 1 5 9 5 51 55 1 Volume (ml Initial: Pre-equivalence: Equivalence: Post-equivalence: initial (only n pre-equivalence (n sa > n sb sa 1 5 9 5 51 55 1 Volume (ml Page III-b- / Chapter Fourteen Part II Lecture Notes = - log [nsa / Vsa] = - log n sa - n sb $ # V sa + V sb & = 7 (neutral salt + water = + log n sb - n sa $ # V sb + V sa &
Page III-b-7 / Chapter Fourteen Part II Lecture Notes QUESTION: You titrate 5. ml of a.1 M solution of HCl with.1 M NaOH. What is the of the initial solution? HCl + H O ---> Cl - + H 3 O + SOLUTION: HCl is a strong acid NaOH is a strong base Hence, this is a strong acid + strong base titration Other common strong acids: HNO 3, HBr, HI, HClO Other common strong bases: KOH, LiOH QUESTION: You titrate 5. ml of a.1 M solution of HCl with.1 M NaOH. What is the of the initial solution? HCl + H O ---> Cl - + H 3 O + SOLUTION: For strong acids, = - log[h 3 O + ] ( = - log.1 = 1. QUESTION: You titrate 5. ml of a.1 M solution of HCl with.1 M NaOH. What is the after 5 ml of NaOH have been added? HCl + H O ---> Cl - + H 3 O + SOLUTION: First, determine which region of a strong acid and strong base titration to use by comparing mol HCl to mol NaOH n sa =.5 L *.1 M =.5 mol HCl n sb =.5 L *.1 M =.5 mol NaOH Since n sa > n sb, this will fall in the pre-equivalence region QUESTION: You titrate 5. ml of a.1 M solution of HCl with.1 M NaOH. What is the after 5 ml of NaOH have been added? HCl + H O ---> Cl - + H 3 O + From before: n sa =.5 mol, n sb =.5 mol = - log n sa - n sb $ # V sa + V sb &.5 -.5 = - log$ #.5 +.5 & =. QUESTION: You titrate 5. ml of a.1 M solution of HCl with.1 M NaOH. What is the at the equivalence point? At equivalence, HCl + NaOH ---> Na + + Cl - + H O Only species present at equivalence are Na +, Cl - and water. Na + and Cl - are conjugates of strong bases and acids; hence, they have no effect on the at equivalence point will be equal to 7 (neutral due to presence of water The volume required to get to equivalence:.5 mol HCl * (mol NaOH / mol HCl * (L/.1 mol NaOH =.5 L or 5. ml QUESTION: You titrate 5. ml of a.1 M solution of HCl with.1 M NaOH. What is the after 55 ml of NaOH have been added? SOLUTION: (from before, n sa =.5 mol It took 5. ml of NaOH to reach equivalence; hence, we are now in the post-equivalence region n sb =.1 M *.55 L =.55 mol = + log n sb - n sa $ # V sb + V sa &.55 -.5 = + log$ #.55 +.5 & = + (-.3 = 11. Page III-b-7 / Chapter Fourteen Part II Lecture Notes
Page III-b- / Chapter Fourteen Part II Lecture Notes HCl + NaOH Titration Graph at 55 ml = 11. Acid-Base Reaction Strong Base + Strong Acid Notice similarities between SA + SB and SB + SA titrations A strong base + strong acid titration at 5 ml =. Equivalence point (5 ml = 7 1 1 initial (only n sb pre-equivalence (n sb > n sa initial = 1. equivalence (n sb = n sa, = 7 post-equivalence (n sa > n sb 1 9 5 51 1 Volume (ml 1 1 initial (only n sb Strong Base + Strong Acid pre-equivalence (n sb > n sa equivalence (n sb = n sa, = 7 Weak Acid + Strong Base Equivalence point dominated by conjugate base of weak acid 1 9 5 51 1 Volume (ml Initial: Pre-equivalence: Equivalence: Post-equivalence: post-equivalence (n sa > n sb = + log [nsb / Vsb] = + log n sb - n sa $ # V sb + V sa & = 7 (neutral salt + water = - log n sa - n sb $ # V sa + V sb & A weak acid + strong base titration 1 1 s Pre-equivalence has weak acid and its conjugate base, acts like a buffer pk a At half-equivalence, = pk a initial (only n wa half-equivalence 5 1 15 5 3 35 Volume (ml post-equivalence (n sb > n wa equivalence (n wa = n sb > 7 pre-equivalence (n wa > n sb Weak Acid + Strong Base 1 s 1 pk a initial (only n wa half-equivalence 5 1 15 5 3 35 Volume (ml post-equivalence (n sb > n wa equivalence (n wa = n sb > 7 pre-equivalence (n wa > n sb Initial: = - log K a C a Pre-equivalence: n = pk a + log sb $ # n wa - n sb & Equivalence: = + log K w n wa $ $ # K a &# V wa + V sb & = + log n sb - n Post-equivalence: wa $ # V sb + V wa & Page III-b- / Chapter Fourteen Part II Lecture Notes QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH. What is the of the initial solution? K a =.3 x 1-5 HBz + H O qe Bz - + H 3 O + C H 5 CO H = HBz Benzoic acid Benzoate ion = Bz -
Page III-b-9 / Chapter Fourteen Part II Lecture Notes QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH. What is the of the initial solution? K a =.3 x 1-5 HBz + H O qe Bz - + H 3 O + SOLUTION: For weak acids, = - log K a C a = - log $.3 *1-5.5 # & =.9 ( QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH. What is the after 1. ml of NaOH have been added? K a =.3 x 1-5 HBz + H O qe Bz - + H 3 O + SOLUTION: First, determine which region of a weak acid (benzoic acid and strong base (NaOH titration to use by comparing mol HBz to mol NaOH n wa =.1 L *.5 M =.5 mol HBz n sb =.1 L *.1 M =.1 mol NaOH Since n wa > n sb, this will fall in the pre-equivalence region Note that because n sb is converted to mol Bz - (the conjugate base, this region acts like a buffer QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH. What is the after 1. ml of NaOH have been added? K a =.3 x 1-5 From before: n wa =.5 mol, n sb =.1 mol pk a = -log K a = - log (.3*1-5 =. n = pk a + log sb $ # n wa - n sb &.1 =. + log$ #.5 -.1& =. -.17 =. A version of the Henderson-Hasselbalch equation QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH. What is the after 1.5 ml of NaOH have been added? K a =.3 x 1-5 From before: n wa =.5 mol, pk a =. n sb =.15 L *.1 M =.15 mol NaOH Since n wa > n sb, this is the pre-equivalence region n = pk a + log sb $ # n wa - n sb &.15 =. + log$ =. + log(1 #.5 -.15& =. + =. QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH. What is the after 1.5 ml of NaOH have been added? K a =.3 x 1-5 Before: n wa =.5 mol, pk a =., n sb =.15 mol When n sb = 1 / n wa, this is called the half-equivalence region = pk a at the half-equivalence point Useful method of finding K a (and K b values Also, the volume of titrant (NaOH at half equivalence is exactly half the volume necessary to reach equivalence We will need (*1.5 = 5 ml of NaOH to reach equivalence QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH to the equivalence point. What is the at the equivalence point? K a =.3 x 1-5 At equivalence, HBz + NaOH ---> Na + + Bz - + H O mol NaOH = mol HBz, or NaOH annihilates HBz mol HBz = mol Bz - at equivalence, and dominated by conjugate base of weak acid at equivalence point will be basic when titrating a weak acid with a strong base Page III-b-9 / Chapter Fourteen Part II Lecture Notes + qe +
Page III-b-1 / Chapter Fourteen Part II Lecture Notes QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH to the equivalence point. What is the at the equivalence point? Ka =.3 x 1-5 QUESTION: You titrate 1. ml of a.5 M solution of benzoic acid with.1 M NaOH. What is the after. ml of NaOH have been added? nwa =.1 L *.5 M =.5 mol HBz.5 mol HBz =.5 mol NaOH at equivalence.5 mol NaOH * (L /.1 mol =.5 L = Vsb It took 5 ml of NaOH to reach equivalence; hence, we are now in the post-equivalence region K nwa = + log $ w $ # K a &# Vwa + Vsb &.5 1- $ = + log $ $ -5 #.3 * 1 &#.1 +.5 & = + (-5.75 =.5 SOLUTION: (from before, nwa =.5 mol nsb =.1 M *. L =. mol n - nwa = + log$ sb # Vsb + Vwa &. -.5 = + log$ #. +.1 & = + (-3.1 = 1.9 Benzoic acid + NaOH Titration Graph at. ml = 1.9 Acid-Base Reaction at halfequiv (1.5 ml =. With weaker acids, the initial is higher and changes near the equivalence point are more subtle. Equivalence point (5 ml =.5 initial =.9 at 1. ml =. Polyprotic Acids Equivalence point dominated by conjugate acid of weak base A weak base + strong acid titration Pre-equivalence has weak base and its conjugate acid, acts like a buffer 1 7.5 5.5 first half equivalence ( = pka for H3PO3 equivalence (nwb = nsa < 7 post-equivalence (nsa > nwb half-equivalence pka x second half equivalence ( = pka for HPO3-1 pre-equivalence (nwb > nsa initial (only nwb x Titrations of polyprotic acids with bases have multiple equivalence points (and half equivalence points for each dissociation. Weak Base + Strong Acid At half-equivalence, = pka 1 Volume (ml Page III-b-1 / Chapter Fourteen Part II Lecture Notes 3
Page III-b-11 / Chapter Fourteen Part II Lecture Notes 1 Weak Base + Strong Acid 7.5 initial (only nwb Indicators pre-equivalence (nwb > nsa 5 half-equivalence pka equivalence.5 post-equivalence (nsa > nwb 1 Initial: Volume (ml Pre-equivalence: Equivalence: Post-equivalence: 3 = + log K bcb n - n $ = pk a + log # wb sa & n sa K $ n wb $ = -log # w &# & K b Vwb + Vsa n - n wb $ = - log # sa & Vsa + Vwb Equivalence point important in titration, can use an indicator to signify equivalence point Indicator color change must reflect equivalence point to be useful Color Changes for Bromthymol Blue Indicators Which indicator to use? Phenolphthalein or Methyl Red? The at the equivalence point in this titration is < 7. Methyl red is the indicator of choice here. <. =.-7.5 > 7.5 End of Chapter Part II See: Chapter Fourteen Part II Study Guide Chapter Fourteen Part II Concept Guide Types of Equilibrium Constants Titration Guide Buffers and Henderson-Hasselbalch Guide Page III-b-11 / Chapter Fourteen Part II Lecture Notes