Buffers/Titration Aqueous Equilibria - I

Transcription

1 Slide 1 / 113 Slide 2 / 113 uffers/titration queous Equilibria - I Review hydrolysis of salts Slide 3 / The hydrolysis of a salt of a weak base and a strong acid should give a solution that is. Slide 4 / solution of one of the following is acidic. The compound is. weakly basic neutral strongly basic weakly acidic Nal H 3 OONa NH 4 l NH 3 Slide 5 / Which salt would undergo hydrolyzes to form an acidic solution? Slide 6 / Which substance when dissolved in water will produce a solution with a ph greater than 7? Kl Nal NH 4 l Lil H 3 OOH Nal Na 2 H 3 O 2 Hl

2 Slide 7 / basic solution would result from the hydrolysis of one of the ions in this compound. The compound is. NaNO 3 NH 4 l H 3 OONa al 2 Slide 8 / water solution of which compound will turn blue litmus red? K 2 O 3 NH 4 l NaOH Nal Slide 9 / 113 Slide 10 / 113 onsider an aqueous solution of acetic acid: H 3 OOH(aq) + H 2 O(l) H 3 O + (aq) + H 3 OO - (aq) If acetate ion is added to the solution, Le hâtelier 's Principle predicts that the equilibrium will shift to the left. The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte. In the following Sample Problem, compare the ph of a 0.20 M solution of HF with the ph of the same solution after some NaF is added to it. Slide 11 / 113 SMPLE PROLEM #1 a) alculate the ph of a 0.20 M HFsolution. The K a for HF is HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) Slide 12 / 113 K a = [H 3O + ][F - ] [HF] 6.8 x 10-4 = = 6.8 x 10-4 x 2 (0.20) HF(aq) H 2 O(l) H 3 O + (aq) F (aq) Initial 0.20 M 0 0 hange -x +x +x t Equilibrium 0.20-x x x x 2 = (0.20) ( ) x = So, [H + ] = [F - ] = and ph = 1.93

3 Slide 13 / 113 SMPLE PROLEM #1 (con't) b) alculate the ph of a for the same 0.20 M HF solution that also contains 0.10 M NaF. K a = [H 3O + ][F - ] [HF] K a for HF is = 6.8 x 10-4 Slide 14 / 113 b) alculate the ph of a for the same 0.20 M HF solution that also contains 0.10 M NaF. We use the same equation and the same K a expression, and set up a similar IE chart. ut because NaF is a soluble salt, it completely dissociates, so F - is not initially zero. HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) HF(aq) H 2 O(l) H 3 O + (aq) F (aq) Initial 0.20 M M hange at Equilibrium Slide 15 / 113 Slide 16 / 113 Therefore, HF(aq) H 2 O(l) H 3 O + (aq) F (aq) Initial 0.20 M M hange -x +x +x t Equilibrium 0.20-x x x and x = [H 3 O + ] = 1.4 x 10 3 ph = log (1.4 x 10 3 ) ph = 2.87 Slide 17 / 113 K a = 6.8 x 10-4 = (0.10)(x) (0.20) (0.20)(6.8 x 10-4 ) (0.10) = x = x Remember what the "x" is that you are solving for! Slide 18 / 113 HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) onsider the two solutions we just examined in Sample Problem #1. ompare the following: Solution Final [H 3 O + ] ph HF 0.12 HF and NaF 1.4 x 10-3 How do these results support Le hatelier's Principle?

4 Slide 19 / 113 Slide 20 / 113 The K a for acetic acid, H 3 OOH, is 1.8 x The K a for acetic acid, H 3 OOH or H 2 H 3 O 2, is 1.8 x SMPLE PROLEM #2 SMPLE PROLEM #2 - nswers a) alculate the ph of a 0.30 M acetic acid solution. a) alculate the ph of a 0.30 M acetic acid, H 3 OOH. b) alculate the ph of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate. K a = [H 3O + ][ 2 H 3 O 2 - ] [H 2 H 3 O 2 ] 1.8 x 10-5 = x 2 (0.30) = 1.8 x 10-5 x = [H 3 O + ] = and ph = Slide 21 / 113 The K a for acetic acid, H 3 OOH or H 2 H 3 O 2, is 1.8 x SMPLE PROLEM #2 - nswers (con't) b) alculate the ph of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate. We use the same K a expression, except now the acetate ion concentration is not the same as [H 3 O + ]. K a = [H 3O + ][ 2 H 3 O 2 - ] [H 2 H 3 O 2 ] = 1.8 x 10-5 Slide 22 / 113 H 2 H 3 O 2 (aq) + H 2 O (l) 2 H 3 O 2 - (aq) + H 3 O + (aq) onsider the two solutions we just examined in Sample Problem #2. ompare the following: Solution [OH - ] ph H 2 H 3 O x 10-5 = So now, [H 3 O + ] (0.10) (0.30) [H 3 O + ] = H 2 H 3 O 2 and Na 2 H 3 O 2 How do these results support Le hatelier's Principle? and ph = Slide 23 / 113 Slide 24 / The ionization of HF will be decreased by the addition of: SMPLE PROLEM #3 alculate the ph of the following solutions: a) 0.85 M nitrous acid, HNO 2 b) 0.85 M HNO 2 and 0.10 M potassium nitrite, KNO 2 The K a for nitrous acid is 4.5 x E Nal NaF Hl oth and oth and nswers: a) b)

5 Slide 25 / The dissociation of l(oh) 3 will be decreased by the addition of: KOH ll 3 Mg(OH) 2 E oth and, and Slide 26 / Which of the following substances will not decrease the ionization of H 3 PO 4? K 3 PO 4 Hl Na 3 PO 4 E None of them will decrease the ionization ll of them will decrease the ionization Slide 27 / 113 Slide 28 / 113 The ommon-ion effect can also be observed with weak bases. onsider the ionization of ammonia, NH 3, which is a weak base. NH 3 + H 2 O <---> NH OH - Suppose a salt of the conjugate base is added to a solution of ammonia. What effect would this have on ph? NH 3 + H 2 O <---> NH OH - s with the previous Sample Problems, let us calculate the ph of the following: a) a 0.45 M solution of NH 3 b) a 0.45 M solution of NH 3 that also contains 0.15 M NH 4 l, ammonium chloride -- a soluble salt that readily yields NH 4 + ions. Slide 29 / 113 SMPLE PROLEM #4 alculate the ph of the following solutions: a) a 0.45 M solution of NH 3 b) a 0.45 M solution of NH 3 that also contains 0.15 M NH 4 l The K b for NH 3 is 1.8 x Slide 30 / 113 SMPLE PROLEM #4 - nswers a) alculate the ph of a 0.45 M solution of NH 3. The K b for NH 3 is 1.8 x K b = 1.8 x 10-5 = NH 3 + H 2 O <---> NH OH - [NH 4 + ][OH - ] [NH 3 ] 1.8 x 10-5 = x 2 (0.45) x = [OH - ] = ph =

6 Slide 31 / 113 SMPLE PROLEM #4 - nswers (con't) b) alculate the ph of a 0.45 M solution of NH 3 that also contains 0.15 M NH 4 l. The K b for NH 3 is 1.8 x NH 3 + H 2 O <---> NH OH - K b = 1.8 x 10-5 = So, 1.8 x 10-5 = [NH 4 + ][OH - ] [NH 3 ] (0.15) (x) (0.45) x = [OH - ] = Slide 32 / 113 SMPLE PROLEM #5 alculate the ph of the following solutions: a) M pyridine, 5 H 5 N b) M 5 H 5 N and M pyridinium chloride, 5 H 5 NHl Note that 5 H 5 NHl,( salt) dissociates into 5 H 5 NH + and l -. The K b for pyridine is 1.7 x nswers: a) b) and ph = Slide 33 / 113 uffers Slide 34 / 113 uffers uffers, or buffered solutions, are special mixtures that are resistant to large ph changes, even when small amounts of strong acid or strong base are added. uffers are able to resist large ph changes because they contain both an acidic component and a basic component. uffers are prepared by mixing either: 1) a weak acid and a salt containing its conjugate base OR 2) a weak base and a salt containing its conjugate acid Slide 35 / 113 uffers onsider a buffer solution composed of HF and NaF. Slide 36 / 113 uffers onsider a buffer solution composed of HF and NaF. uffer after addition of OH - uffer with equal conc. of weak acid and its conj. base The acidic component is HF. This component aids in the neutralization of any strong base that is added to the buffer. HF F - HF F - OH - H + HF F - The basic component is the fluoride ion, F -. This component aids in the neutralization of any strong acid that is added to the buffer. F - HF F - + H2O <-- HF + OH - H + + F - --> HF If a small amount of KOH is added to this buffer, for example, the HF reacts with the OH to make.

7 Slide 37 / 113 Slide 38 / 113 uffer after addition of OH - uffers uffer with equal conc. of weak acid and its conj. base uffer after addition of H + 10 If a buffer is made of H (weak acid) and Na, which species will counteract the addition of a strong base? HF F - HF F - OH - H + HF F - H Na + - F - + H2O <-- HF + OH - H + + F - --> HF oth H and Na + E oth H and - Similarly, if a small amount of strong acid (H + ) is added, the F reacts with it to form. Slide 39 / If a buffer is made of HNO 2 and KNO 2, which species will counteract the addition of a strong acid? HNO 2 K + NO 2 - oth HNO 2 and K + E oth K + and NO 2 - Slide 40 / 113 uffer capacity and ph Two important characteristics of buffers: 1- Its resistance to change in ph 2- Its buffer capacity uffer capacity: The amount of acid or base the buffer can neutralize before the ph begins to change. It depends on the amount of acid and base from which the buffer is made. Slide 41 / 113 Slide 42 / 113 [ H3O + ] = Ka uffer capacity and ph uffer 1.0 M H3OOH and 1.0 M H3OONa Total volume = 1.0 L Ka = [H3O+ ][2H3O2 - ] [H2H3O2] uffer 0.1 M H3OOH and 0.1 M H3OONa Total volume = 1.0 L Since [2H3O2 - ] [H2H3O2] = 1 12 Of the following solutions, which has the greatest buffering capacity? E 0.1M NaF + 0.1MHF 0.5M NaF M HF 0.8M NaF + 0.8M HF 0.2M NaF + 0.2M HF 1M NaF + 1M HF The above two combination of solution will have the same [H + ]. The ph will depend on ka and the relative concentration of acid and base only. The buffering capacity will be higher for the first buffer. It contains greater number of moles of H3OOH and NaH3OO -. It can neutralize more of the acid/base added. The greater the amounts, greater resistance to ph change.

8 Slide 43 / 113 uffer alculations onsider the equilibrium constant expression for the dissociation of a generic acid, H: H + H 2 O H 3 O K a = [H 3O + ][ - ] [H] Slide 44 / 113 uffer alculations Rearranging slightly, this becomes K a = [H 3 O + ] [ - ] [H] Taking the negative log of both sides, we get -log K a = -log [H 3 O + [ - ] ] +(-log [H] ) pk a = ph - log [base] [acid] Since Rearranging this: Slide 45 / 113 uffer alculations pk a = ph - log [base] [acid] ph = pk a + log [base] [acid] This is the Henderson-Hasselbalch equation SMPLE PROLEM #6 Slide 47 / 113 uffer alculations What is the ph of a buffer that is 0.12 M in lactic acid, H 3 H(OH)OOH, and 0.10 M in sodium lactate? K a for lactic acid is Method 1 - Traditional approach Write the K a expression. Solve for H + and ph using IE chart H 3 H 5 O 3 (aq) + H 2 O(l) H 3 O + (aq) + 3 H 5 O - 3 (aq) x +x +x 0.12-x x 0.10+x = 0.10 x 0.12 x = P H =3.77 Slide 46 / 113 uffer alculations uffer calculations typically require you to calculate one or more of the following: i) the ph of the buffer alone (Sample Prob #6) ii) the ph of the buffer after a small amount of strong base has been added (and neutralized) (Sample Prob #7) iii) the ph of the buffer after a small amount of strong acid has been added (and neutralized) (Sample Prob #8) ecause of the buffer system, the ph will not change drastically; it is usually less than a factor of 1.0 on the ph scale. Slide 48 / 113 uffer alculations SMPLE PROLEM #6 What is the ph of a buffer that is 0.12 M in lactic acid, H 3 H(OH)OOH, and 0.10 M in sodium lactate? K a for lactic acid is Method 2 -Using the Henderson-Hasselbalch equation Henderson Hasselbalch Equation ph = pk a + log [base] [acid] ph = -log (1.4 x 10-4 ) + log (0.10) (0.12) ph = (-0.08) ph = 3.77

9 Slide 49 / The ph of a buffer solution that contains M acetic acid (K a = 1.76x10-5 ) and M sodium acetate is E Slide 50 / The ph of a buffer solution containing M HF and M KF is. (K a of HF is 3.5x10-4 ) E 4.32 Slide 51 / 113 ddition of Strong cid or Strong ase to a uffer Recall that buffers resist drastic changes in ph, even when small amounts of either a strong acid or a strong base are added to it. When this happens, we assume that all of the strong acid or strong base is consumed in the reaction. In other words, all of the strong acid or base gets completely neutralized by ONE of the components of the buffer system. dd strong acid dd strong base X - + H 3O + HX + H 2O HX + OH - X - + H 2O Slide 52 / 113 ddition of Strong cid or Strong ase to a uffer Remember that all of the strong acid or base is consumed in the reaction. Recalculate HX and X- Use K a, [HX] and X - to calculate [H+] ph stoichiometric calculation equilibrium calculation Slide 53 / If a buffer is made of NH 3 and NH 4 l, which component of the buffer will neutralize a small amount of Hl that is added? Slide 54 / If a buffer is made of HNO 2 and KNO 2, which component of the buffer will neutralize a small amount of a(oh) 2 that is added? NH 3 + NH 4 l - oth and E oth and HNO 2 K + - NO 2 a 2+ E oth K + - and NO 2

10 Slide 55 / If a buffer is made of NH 3 and NH 4 l, which component will neutralize any KOH that is added? Slide 56 / 113 ddition of Strong cid or Strong ase to a uffer NH 3 + NH 4 l - oth HNO 2 and K + E oth K + - and NO 2 uffer add 0.02M OH - 0.3M H2H3O2 0.3M Na2H3O2 add 0.02M H + ph= 4.74 ph =4.80 etermine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. Use the Henderson Hasselbalch equation to determine the new ph of the solution. Slide 57 / 113 alculating ph hanges in uffers SMPLE PROLEM #7 buffer is made by adding mol H 2 H 3 O 2 and mol Na 2 H 3 O 2 to enough water to make 1.00 L of solution. The ph of the buffer is alculate the ph of this solution after mol of NaOH is added. ssume volume change is negligible. ph= 4.68 Slide 58 / 113 alculating ph hanges in uffers Problem-solving strategy Phase 1 - stoichiometric neutralization Identify the added substance. Identify the component of the buffer that will neutralize the added substance. Write down the equation for nutralization. Solve for the new concentration of the buffer components. Slide 59 / 113 alculating ph hanges in uffers SMPLE PROLEM #7 - nswer buffer is made by adding mol H 2 H 3 O 2 and mol Na 2 H 3 O 2 to enough water to make 1.00 L of solution. The ph of the buffer is alculate the ph of this solution after mol of NaOH is added. Phase 1 - Neutralization strong acid or a strong base added? Which component of the buffer will neutralize the added substance? H(aq) acid OH - (aq) added substance H 2 O(aq) efore (aq) base Neutralization fter Phase 2 - equilibrium Write the relevant dissociation equation. reate the IE chart with the new [M] values(phase1) of acid and base components of the buffer. Use Henderson-Hasselbalch equation to solve for ph. Slide 60 / 113 alculating ph hanges in uffers SMPLE PROLEM #7 - nswer (con't) buffer is made by adding mol H 2 H 3 O 2 and mol Na 2 H 3 O 2 to enough water to make 1.00 L of solution. The ph of the buffer is alculate the ph of this solution after mol of NaOH is added. Phase 2 - equilibrium Write the relevant dissociation equation. Make a IE chart using the amounts from Phase 1. H(aq) H 2 O(l) H 3 O + (aq) (aq) Initial 0.28 M M hange -x +x +x t Equilibrium 0.28-x x 0.32+x

11 Slide 61 / 113 alculating ph hanges in uffers Use the quantities from the IE chart to calculate ph. You will need to look up the K a value. Slide 62 / 113 alculating ph hanges in uffers lternatively, you can use the Henderson-Hasselbalch equation calculate the new ph: K a = 1.8 x 10-5 = [H 3 O + ][ ] [H] (x) 0.32) (0.28) ph = pk a + log [base] [acid] ph = - log (1.8 x 10-5 ) + log (0.320) (0.280) ph = So, x = [H 3 O + ] = ph = 4.80 and ph = Slide 63 / 113 alculating ph hanges in uffers SMPLE PROLEM #8 buffer is made by adding mol cyanic acid, HNO, and mol potassium cyanate, KNO to enough water to make 1.00 L of solution. ssume volume changes are negligible. K a for cyanic acid = 3.5 x 10-4 a) alculate the ph of the buffer. b) alculate the ph of the buffer after the addition of mol KOH. c) alculate the ph of the buffer after the addition of mol HNO3 Slide 64 / 113 ph Range for uffers The ph range is the range of ph values over which a buffer system works effectively. It is best to choose an acid with a pk a close to the desired ph. fter studying titration graphs, you will see why buffers work best when pk a is close to the desired ph. nswers a) b) 3.46 c) 3.22 Slide 65 / 113 alculating ph hanges in uffers SMPLE PROLEM #9 buffer is made by adding 15.0 g ammonia, NH 3, and 55.0 g ammonium chloride, NH 4l. to enough water to make 1.00 L of solution. K b for NH 3 = 1.8 x ssume volume changes are negligible. Slide 66 / 113 alculating ph hanges in uffers SMPLE PROLEM #10 buffer is made by adding 25.0 g ammonia, NH 3, and 45.0 g ammonium chloride, NH 4 l. to enough water to make 1.75 L of solution. K b for NH 3 = 1.8 x ssume volume changes are negligible. a) alculate the ph of the buffer. b) alculate the ph of the buffer after the addition of mol HlO 4. c) alculate the ph of the buffer after the addition of mol KOH. a) alculate the ph of the buffer. b) alculate the ph of the buffer after the addition of mol NaOH. c) alculate the ph of the buffer after the addition of mol HNO 3. nswers a) b) c) nswers a) 9.50 b) 9.51 c) 9.49

12 Slide 67 / 113 Titration Slide 68 / 113 Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base). The concentration of the unknown will then be determined. Important things to remember in getting ready for titration: Rinse the buret with distilled water Rinse with the titrant Fill the buret with the titrant rain a small portion of the titrant so that air bubbles near the tip of the burete is expelled and filled to the tip. This is a quantitative analysis method. Slide 69 / 113 Titration Important things to remember in getting ready for titration: Titration 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. Slide 70 / 113 Rinse the pipette with distilled water Rinse with the solution to be pipeted Pipette the solution, adjust the level and dispense the fixed volume to a beaker or flask. t the equivalence point, # of moles of acid = # of moles of base MaVa = MbVb y using an indicator you could visually see the sudden change in color of the indicator at this point. This point is also known as the "end point" where you will stop adding the titrant. The end point is the appearance of the first permanent color change of the indicator. Slide 71 / 113 Note that this point will have a slight excess of the titrant in the solution Slide 72 / 113 t the equivalence point, ( stoichiometric) # of moles of acid = # of moles of base MaVa = MbVb Titration Equivalence point and End point y using an indicator you could visually see the sudden change in color of the indicator at this point. This point is also known as the "end point" where you will stop adding the titrant. The end point is the appearance of the first permanent color change of the indicator. Note that this point will have a slight excess of the titrant in the solution This minute amount of the titrant is making the indicator color visible.

13 Slide 73 / 113 Titration - Equivalence Point t the equivalence point, # of moles of acid = # of moles of base This equation is only applicable for reactions in which the ratio of moles acid to moles base is 1:1. Slide 75 / 113 Titration - Equivalence Point onsider some examples of reactions in which the ratio of moles acid to moles base is 2:1. 2 Hl + a(oh) 2 --> al H 2 O 2 Hr + Sr(OH) 2 --> Srr H 2 O 2 HNO 3 + a(oh) 2 --> a(no 3 ) H 2 O moles acid moles base = M a V a M b V b = ross-multiplying, we obtain M a V a = 2 M b V b 2 1 Slide 77 / 113 Titration Summary of Equations for Equivalence Point Equation cid ase 2 M a V a = 2 M b V b Hl, HNO 3, Hr, H 3 OOH, HlO 4 H 2 SO 4, H 2 SO 3, H 2 2 O 4 Hl, HNO 3, Hr, H 3 OOH, HlO 4 NaOH, KOH, LiOH NaOH, KOH, LiOH a(oh) 2, Sr(OH) 2, a(oh) 2 Slide 74 / 113 Titration - Equivalence Point Here are some examples of reactions in which the ratio of moles acid to moles base is 1:1. Hl + NaOH --> Nal + H 2 O Hr + KOH --> Kr + H 2 O HNO 3 + LiOH --> LiNO 3 + H 2 O H3OOH + NaOH --> H3OONa + H 2 O So for reactions such as these, use the equation above to calculate the molarity or volume of acid or base at the equivalence point. Slide 76 / 113 Titration - Equivalence Point onsider some examples of reactions in which the ratio of moles acid to moles base is 1:2. H 2 SO KOH --> K 2 SO H 2 O H 2 SO LiOH --> Li 2 SO H 2 O H 2 2 O NaOH --> Na 2 2 O H 2 O moles acid moles base = M a V a M b V b = ross-multiplying, we obtain Slide 78 / What is the concentration of hydrochloric acid if 20.0 ml of it is neutralized by 40mL of 0.10M sodium hydroxide? M M 0.10 M 0.20 M 2 M a V a = 2 M b V b

14 Slide 79 / What is the concentration of KOH if 60 ml of it is neutralized by 20 ml 0.10M Hl? M M M 0.10 M E 0.30 M 2 M a V a = 2 M b V b Slide 80 / What is the concentration of sulfuric acid if 50mL of it is neutralized by 10mL of 0.1M sodium hydroxide? 0.005M 0.01M 0.25M 0.5M 2 M a V a = 2 M b V b Slide 81 / How much 0.5 M HNO 3 is necessary to titrate 25 ml of 0.05M a(oh) 2 solution to the endpoint? E 2.5 ml 5.0 ml 10 ml 20 ml 25 ml Slide 83 / M a V a = 2 M b V b 23 How much 1.5 M NaOH is necessary to exactly neutralize 20.0 ml of 2.5 M H 3 PO 4? E 11 ml 12 ml 33 ml 36 ml 100 ml 2 M a V a = 2 M b V b Slide 82 / How much 3.0 M NaOH is needed to exactly neutralize 20.0 ml of 2.5 ml H 2 SO 3? E 8.3 ml 17 ml 24 ml 33 ml 48 ml Slide 84 / M a V a = 2 M b V b 24 What is the molarity of a NaOH solution if 15 ml is exactly neutralized by 7.5 ml of a 0.02 M H 2 H 3 O 2 solution? M M M M 2 M a V a = 2 M b V b

15 Slide 85 / 113 Slide 86 / 113 Titration of a Strong cid with a Strong ase Typically, there are 4 "zones" in which you may be asked to calculate ph: efore any titrant is added from the buret fter a small amount of titrant has been added t the equivalence point fter the equivalence point Slide 87 / 113 Titration of a Strong cid with a Strong ase From the start of the titration to near the equivalence point, the ph goes up slowly. First, we will examine these zones on a titration graph. Then, we will review the ph calculation for each region. The strategy for calculating ph is different for each zone. Slide 88 / 113 Titration of a Strong cid with a Strong ase Just before (and after) the equivalence point, the ph increases rapidly. The low initial ph indicates that the substance being titrated is a strong acid. (1) efore any titrant is added from the buret (2) fter a small amount of titrant has been added* Slide 89 / 113 Titration of a Strong cid with a Strong ase Slide 90 / 113 Titration of a Strong cid with a Strong ase t the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. s more base is added, the increase in ph again levels off. (3) t the equivalence point (4) fter the equivalence point

16 Slide 91 / 113 Titration of a Strong cid with a Strong ase In summary, the four regions of any titration graph are: efore any titrant is added from the buret fter a small amount of titrant has been added t the equivalence point fter the equivalence point The ph calculation differs for each "zone," so we will consider each one separately. Slide 93 / 113 2) When some titrant is added from the buret, before the equivalence point is reached Example: 20 ml 0.5M Hl is titrated with 0.25M NaOH. What is the ph after 10 ml of base is added? It is recommended that you calculate the volume of the titrant needed to neutralize the acid at the beginning to see which segment of the titration we are in. write down the neutralization reaction Hl + NaOH --> Nal + H2O 0.01 mol mol mol mol Strong cid - Strong ase mol 0 mol left [H+] after the 10 ml base had been neutralized by the acid is = mols /0.030L = 0.25M ph = 0.6 Slide 95 / 113 Slide 92 / 113 Strong cid - Strong ase 1) efore any titrant is added from the buret The ph depends up on the concentration of the acid or base. Use the molarity of the acid or base to determine the ph. ph = - log [H3O + ] OR ph = 14 - (-log [OH - ] Remember to check the acid or base is polyprotic or polyhydroxy. For example: 20 ml 0.5M Hl is titrated with 0.25M NaOH The original acid is 0.5M; The concentration of H+ = 0.5M MaVa = MbVb 20 ml 0.5M H 2SO 4 is titrated with 0.25M NaOH The original acid is 0.5M; The concentration of H+ = 0.5x2 M 2MaVa = MbVb 20 ml 0.5M Hl is titrated with 0.25M a(oh) 2 The original acid is 0.5M; The concentration of H+ = 0.5 M MaVa = 2MbVb2 Slide 94 / 113 Strong cid - Strong ase 3) t the equivalence point: Example: 20 ml 0.5M Hl is titrated with 0.25M NaOH. What is the ph at the equivalence point? You will know you have reached the equivalence point when MOLES of acid = MOLES of base Since there is no excess [H + ] or [OH - ], we must look to the salt tha is formed to see if it will affect ph. Since it is from a strong base and strong acid, will not undergo hydrolysis to change the ph. In cases such as these, when a strong acid and strong base are titrated, the pure water will have a ph = 7.0 Slide 96 / 113 Strong cid - Strong ase Strong cid - Strong ase 4) eyond the equivalence point, when excess titrant has been added from the buret Example: 20 ml 0.5M Hl is titrated with 0.25M NaOH. What is the ph after 45 ml of base had been added? 0.5 x 20ml = 0.25 x V2ml V2 = 40 ml NaOH need to neutralize the 20 ml acid. write down the neutralization reaction Hl + NaOH --> Nal + H2O 0.01 mol mol -0.01mol -0.01mol 0 mol mol excess [OH-] after the 45 ml base added = = mols /0.065L = M poh = ph = cid in the flask and titrant is the base. The ph at eqequivalence point will be =7

17 Slide 97 / 113 Strong cid - Strong ase Slide 98 / The moles of acid equals the moles of base at the equivalence point in a titration of with. strong acid, weak base E strong base, weak acid strong acid, strong base weak acid, weak base ll of the above cid in the flask and base is the titrant ase in the flask and acid is the titrant Slide 99 / 113 Slide 100 / What is the ph of a titration between a weak acid and a strong base at the equivalence point? Less than 7 Equal to 7 Greater than 7 27 What is the ph of a titration between a weak base and a strong acid at the equivalence point? Less than 7 Equal to 7 Greater than 7 Slide 101 / 113 Weak - Strong 1) efore any titrant is added from the buret onsider the dissociation equation for the substance in the flask: For a weak acid in the flask For a weak base in the flask K a = x 2 / [acid] K b = x 2 / [base] ph = - log x ph = 14 - (-log x) Slide 102 / 113 Weak - Strong 2) When some titrant is added from the buret, before the equivalence point is reached example# 11: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 10 ml NaOH is added: ka = 1.8 x 10-5 H 3OOH + NaOH --> Na + H 3OO - + H 2O 0.01mol mol mol mol mol mol mol 0.25M (acid) 0.083M ( c.base) Remember you have a buffer situation now because of the salt produced from the neutralizationof the weak acid. Use the Henderson-Hasselbalch equation to solve for the ph of the solution at this point. ph = pka + log ( [base] / [acid] )

18 Slide 103 / 113 Weak - Strong Sample # 12 calculate the ph of the solution after adding 10 ml of 0.05M KOH to 40 ml of 0.025M enzoic acid. Ka for benzoic acid = 6.3 x 10-5 Slide 104 / 113 Weak - Strong sample #13 alculate the ph of the solution after adding 10 ml of 0.1M Hl to 20 ml of 0.1 M NH 3 Kb of NH 3 = 1.8 x 10-5 Slide 105 / 113 Weak - Strong 3) t the equivalence point example # 14: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 40 ml NaOH is added: ka = 1.8 x 10-5 H3OOH + NaOH --> Na + H3OO- + H2O 0.01mol 0.01mol -0.01mol -0.01mol mol 0.0mol 0.0 mol 0.01mol Slide 106 / 113 Weak - Strong sample # 15 alculate the ph at equivalence point when 40 ml of 0.025M benzoic acid is titrated with 0.05M KOH. Ka of benzoic acid is M (base) The salt Na + H3OO - has a strong conjugate base H3OO- The anion, H3OO- will undergo hyrolysis as follows: H3OO- + H2O --> H3OOH + OH M 0 0 -X +x +x X x x Kb = X 2 / ; Solve for x alculate ph Will be slightly basic! Remember Kb = Kw / Ka Slide 107 / 113 Weak - Strong Slide 108 / 113 Titration of a Weak cid with a Strong ase Sample # 16 calculate the ph at the equivalence point when 40 ml of 0.1M NH 3 is titrated with 0.2 M Hl. Kb of NH 3 is The ph at the equivalence point is above 7 strong acid vs strong base weak acid vs strong base

19 Slide 109 / 113 Titration of a Weak cid with a Strong ase With weaker acids, the initial ph is higher and ph changes near the equivalence point are more subtle. Slide 110 / 113 Titration of a Weak ase with a Strong cid The ph at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice. strong base with strong acid, ph =7 weak base NH3 with strong acid, ph = 5.5 Slide 111 / 113 Weak - Strong 4) eyond the equivalence point, when excess titrant has been added from the buret example: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 50 ml NaOH is added: Slide 112 / 113 Titrations of Polyprotic cids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. This is exactly the same as for strong acid-strong base titrations. The strong base added after the end point will increase the OHin the solution making it strongly basic. alculate the [OH-] from the excess base in the new volume and determine the ph. Slide 113 / 113