Slide 39 / 113 11 If a buffer is made of HNO 2 and KNO 2 , which species will counteract the addition of a strong acid? HNO 2 A K + B - NO 2 C Both HNO 2 and K + D Both K + and NO 2 - E
Slide 40 / 113 Buffer capacity and pH Two important characteristics of buffers: 1- Its resistance to change in pH 2- Its buffer capacity Buffer 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 Buffer capacity and pH Buffer A Buffer B 1.0 M CH 3 COOH and 0.1 M CH 3 COOH and 1.0 M CH 3 COONa 0.1 M CH 3 COONa Total volume = 1.0 L Total volume = 1.0 L K a = [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] - ] [C 2 H 3 O 2 Since = 1 [HC 2 H 3 O 2 ] [ H 3 O + ] = Ka The above two combination of solution will have the same [H + ]. The pH will depend on k a 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 CH 3 COOH and NaCH 3 COO - . It can neutralize more of the acid/base added. The greater the amounts, greater resistance to pH change.
Slide 42 / 113 12 Of the following solutions, which has the greatest buffering capacity? A 0.1M NaF + 0.1MHF B 0.5M NaF + 0.55M HF C 0.8M NaF + 0.8M HF D 0.2M NaF + 0.2M HF E 1M NaF + 1M HF
Slide 43 / 113 Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H 2 O ↔ H 3 O + + A - K a = [H 3 O + ][A - ] [HA]
Slide 44 / 113 Buffer Calculations Rearranging slightly, this becomes [A - ] K a = [H 3 O + ] [HA] Taking the negative log of both sides, we get [A - ] -log K a = -log [H 3 O + ] +(-log ) [HA] pK a = pH - log [base] [acid]
Slide 45 / 113 Buffer Calculations pK a = pH - log [base] Since [acid] Rearranging this: pH = pK a + log [base] [acid] This is the Henderson-Hasselbalch equation
Slide 46 / 113 Buffer Calculations Buffer 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) Because 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 47 / 113 Buffer Calculations SAMPLE PROBLEM #6 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4 ´ 10 −4 . Method 1 - Traditional approach Write the K a expression. Solve for H + and pH using ICE chart HC 3 H 5 O 3 (aq) + H 2 O(l) ↔ H 3 O + (aq) + C 3 H 5 O 3 - (aq) 0.12 0 0.10 -x +x +x 0.12-x x 0.10+x 1.4 ´ 10 −4 = 0.10 x 0.12 x = 0.000168 P H =3.77
Slide 48 / 113 Buffer Calculations SAMPLE PROBLEM #6 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4 ´ 10 −4 . 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 = 3.85 + (-0.08) pH = 3.77
Slide 49 / 113 13 The pH of a buffer solution that contains 0.818 M acetic acid (K a = 1.76x10 -5 ) and 0.172 M sodium acetate is __________. 4.077 A 5.434 B 8.571 C 8.370 D 9.922 E
Slide 50 / 113 14 The pH of a buffer solution containing 0.145 M HF and 0.183 M KF is _____. (K a of HF is 3.5x10 -4 ) 3.56 A 2.19 B 2.66 C 3.35 D 4.32 E
Slide 51 / 113 Addition of Strong Acid or Strong Base to a Buffer 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.
Slide 52 / 113 Addition of Strong Acid or Strong Base to a Buffer Remember that all of the strong acid or base is consumed in the reaction. X - + H 3 O + ⇒ HX + H 2 O Add strong acid Use K a , [HX] and X - Recalculate pH HX and X- to calculate [H+] Add strong base HX + OH - ⇒ X - + H 2 O equilibrium calculation stoichiometric calculation
Slide 53 / 113 15 If a buffer is made of NH 3 and NH 4 Cl, which component of the buffer will neutralize a small amount of HCl that is added? NH 3 A + NH 4 B Cl - C Both A and B D Both B and C E
Slide 54 / 113 16 If a buffer is made of HNO 2 and KNO 2 , which component of the buffer will neutralize a small amount of Ba(OH) 2 that is added? HNO 2 A K + B - NO 2 C Ba 2+ D Both K + and NO 2 - E
Slide 55 / 113 17 If a buffer is made of NH 3 and NH 4 Cl, which component will neutralize any KOH that is added? NH 3 A + NH 4 B Cl - C Both HNO 2 and K + D Both K + and NO 2 - E
Slide 56 / 113 Addition of Strong Acid or Strong Base to a Buffer - H O M · Determine how the 2 0 . 0 pH =4.80 neutralization reaction affects d d a the amounts of the weak acid and its conjugate base in Buffer solution. pH= 4.74 0.3M HC 2 H 3 O 2 · Use the Henderson– 0.3M NaC 2 H 3 O 2 Hasselbalch equation to a determine the new pH of the d d 0 solution. . 0 2 M H + pH= 4.68
Slide 57 / 113 Calculating pH Changes in Buffers SAMPLE PROBLEM #7 A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Assume volume change is negligible.
Slide 58 / 113 Calculating pH Changes in Buffers 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. Phase 2 - equilibrium · Write the relevant dissociation equation. · Create the ICE chart with the new [M] values(phase1) of acid and base components of the buffer. · Use Henderson-Hasselbalch equation to solve for pH.
Slide 59 / 113 Calculating pH Changes in Buffers SAMPLE PROBLEM #7 - Answer A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 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? OH - (aq) A − (aq) HA(aq) added H 2 O(aq) acid base substance Before 0.3 0.02 0.3 Neutralization -0.02 -0.02 0.3+0.02 After 0.28 0.0 0.32
Slide 60 / 113 Calculating pH Changes in Buffers SAMPLE PROBLEM #7 - Answer (con't) A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Phase 2 - equilibrium · Write the relevant dissociation equation. · Make a ICE chart using the amounts from Phase 1. HA(aq) H 2 O(l) H 3 O + (aq) A − (aq) Initial 0.28 M 0 0.32 M Change -x +x +x At Equilibrium 0.28-x x 0.32+x
Slide 61 / 113 Calculating pH Changes in Buffers Use the quantities from the ICE chart to calculate pH. You will need to look up the K a value. [H 3 O + ][A − ] K a = [HA] (x) 0.32) 1.8 x 10 -5 = (0.28) So, x = [H 3 O + ] = _______________ and pH = ___________
Slide 62 / 113 Calculating pH Changes in Buffers Alternatively, you can use the Henderson-Hasselbalch equation to calculate the new pH: pH = pK a + log [base] [acid] (0.320) pH = - log (1.8 x 10 -5 ) + log (0.280) pH = 4.74 + 0.06 pH = 4.80
Slide 63 / 113 Calculating pH Changes in Buffers SAMPLE PROBLEM #8 A buffer is made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO to enough water to make 1.00 L of solution. Assume volume changes are negligible. K a for cyanic acid = 3.5 x 10 -4 a) Calculate the pH of the buffer. b) Calculate the pH of the buffer after the addition of 0.015 mol KOH. c) Calculate the pH of the buffer after the addition of 0.018 mol HNO3. a) 3.351 Answers b) 3.46 c) 3.22
Slide 64 / 113 pH Range for Buffers · 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 p K a close to the desired pH. · After studying titration graphs, you will see why buffers work best when p K a is close to the desired pH.
Slide 65 / 113 Calculating pH Changes in Buffers SAMPLE PROBLEM #9 A buffer is made by adding 15.0 g ammonia, NH 3 , and 55.0 g ammonium chloride, NH 4 Cl. to enough water to make 1.00 L of solution. K b for NH 3 = 1.8 x 10 -5. Assume volume changes are negligible. a) Calculate the pH of the buffer. b) Calculate the pH of the buffer after the addition of 0.013 mol HClO 4 . c) Calculate the pH of the buffer after the addition of 0.015 mol KOH. a) 9.193 Answers b) 9.181 c) 9.206
Slide 66 / 113 Calculating pH Changes in Buffers SAMPLE PROBLEM #10 A buffer is made by adding 25.0 g ammonia, NH 3 , and 45.0 g ammonium chloride, NH 4 Cl. to enough water to make 1.75 L of solution. K b for NH 3 = 1.8 x 10 -5. Assume volume changes are negligible. a) Calculate the pH of the buffer. b) Calculate the pH of the buffer after the addition of 0.011 mol NaOH. c) Calculate the pH of the buffer after the addition of 0.016 mol HNO 3 . a) 9.50 b) 9.51 Answers c) 9.49
Slide 67 / 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. This is a quantitative analysis method.
Slide 68 / 113 Titration 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 Drain a small portion of the titrant so that air bubbles near the tip of the burete is expelled and filled to the tip.
Slide 69 / 113 Titration Important things to remember in getting ready for titration: 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.
Slide 70 / 113 Titration 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. At the equivalence point, # of moles of acid = # of moles of base M a V a = M b V b By 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
Slide 71 / 113 Titration Equivalence point and End point At the equivalence point, ( stoichiometric) # of moles of acid = # of moles of base MaVa = MbVb By 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.
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Slide 73 / 113 Titration - Equivalence Point At the equivalence point, # of moles of acid = # of moles of base M a V a = M b V b This equation is only applicable for reactions in which the ratio of moles acid to moles base is 1:1.
Slide 74 / 113 Titration - Equivalence Point M a V a = M b V b Here are some examples of reactions in which the ratio of moles acid to moles base is 1:1. HCl + NaOH --> NaCl + H 2 O HBr + KOH --> KBr + H 2 O HNO 3 + LiOH --> LiNO 3 + H 2 O CH3COOH + NaOH --> CH3COONa + 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 75 / 113 Titration - Equivalence Point Consider some examples of reactions in which the ratio of moles acid to moles base is 2:1. 2 HCl + Ca(OH) 2 --> CaCl 2 + 2 H 2 O 2 HBr + Sr(OH) 2 --> SrBr 2 + 2 H 2 O 2 HNO 3 + Ba(OH) 2 --> Ba(NO 3 ) 2 + 2 H 2 O moles acid = M a V a 2 = moles base M b V b 1 Cross-multiplying, we obtain M a V a = 2 M b V b
Slide 76 / 113 Titration - Equivalence Point Consider some examples of reactions in which the ratio of moles acid to moles base is 1:2. H 2 SO 4 + 2 KOH --> K 2 SO 4 + 2 H 2 O H 2 SO 3 + 2 LiOH --> Li 2 SO 3 + 2 H 2 O H 2 C 2 O 4 + 2 NaOH --> Na 2 C 2 O 4 + 2 H 2 O moles acid M a V a 1 = = moles base M b V b 2 Cross-multiplying, we obtain 2 M a V a = M b V b
Slide 77 / 113 Titration Summary of Equations for Equivalence Point Equation Base Acid HCl, HNO 3 , HBr, M a V a = M b V b NaOH, KOH, LiOH CH 3 COOH, HClO 4 H 2 SO 4 , H 2 SO 3, NaOH, KOH, LiOH 2 M a V a = M b V b H 2 C 2 O 4 HCl, HNO 3 , HBr, Ca(OH) 2 , Sr(OH) 2 , M a V a = 2 M b V b CH 3 COOH, HClO 4 Ba(OH) 2
Slide 78 / 113 18 What is the concentration of hydrochloric acid if 20.0 mL of it is neutralized by 40mL of 0.10M sodium hydroxide? 0.025 M A 0.050 M B 0.10 M C 0.20 M D M a V a = M b V b 2 M a V a = M b V b M a V a = 2 M b V b
Slide 79 / 113 19 What is the concentration of KOH if 60 mL of it is neutralized by 20 mL 0.10M HCl? 0.005 M A 0.025 M B 0.033 M C 0.10 M D 0.30 M E M a V a = M b V b 2 M a V a = M b V b M a V a = 2 M b V b
Slide 80 / 113 20 What is the concentration of sulfuric acid if 50mL of it is neutralized by 10mL of 0.1M sodium hydroxide? 0.005M A 0.01M B 0.25M C 0.5M D M a V a = M b V b 2 M a V a = M b V b M a V a = 2 M b V b
Slide 81 / 113 21 How much 0.5 M HNO 3 is necessary to titrate 25 mL of 0.05M Ca(OH) 2 solution to the endpoint? 2.5 mL A 5.0 mL B 10 mL C 20 mL D 25 mL E M a V a = M b V b 2 M a V a = M b V b M a V a = 2 M b V b
Slide 82 / 113 22 How much 3.0 M NaOH is needed to exactly neutralize 20.0 mL of 2.5 mL H 2 SO 3 ? 8.3 mL A 17 mL B 24 mL C 33 mL D 48 mL E M a V a = M b V b 2 M a V a = M b V b M a V a = 2 M b V b
Slide 83 / 113 23 How much 1.5 M NaOH is necessary to exactly neutralize 20.0 mL of 2.5 M H 3 PO 4 ? 11 mL A 12 mL B 33 mL C 36 mL D 100 mL E M a V a = M b V b 2 M a V a = M b V b M a V a = 2 M b V b
Slide 84 / 113 What is the molarity of a NaOH solution if 15 mL is 24 exactly neutralized by 7.5 mL of a 0.02 M HC 2 H 3 O 2 solution? 0.005 M A 0.010 M B 0.020 M C 0.040 M D M a V a = M b V b 2 M a V a = M b V b M a V a = 2 M b V b
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Slide 86 / 113 Titration of a Strong Acid with a Strong Base Typically, there are 4 "zones" in which you may be asked to calculate pH: · Before any titrant is added from the buret · After a small amount of titrant has been added · At the equivalence point · After the equivalence point 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 87 / 113 Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly. The low initial pH indicates that the substance being titrated is a strong acid. (1) Before any titrant is added from the buret
Slide 88 / 113 Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly. (2) After a small amount of titrant has been added*
Slide 89 / 113 Titration of a Strong Acid with a Strong Base At 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. (3) At the equivalence point
Slide 90 / 113 Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off. (4) After the equivalence point
Slide 91 / 113 Titration of a Strong Acid with a Strong Base In summary, the four regions of any titration graph are: · Before any titrant is added from the buret · After a small amount of titrant has been added · At the equivalence point · After the equivalence point The pH calculation differs for each "zone," so we will consider each one separately.
Slide 92 / 113 Solving Titration Problems: Strong Acid - Strong Base 1) Before 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 [H 3 O + ] OR pH = 14 - (-log [OH - ] Remember to check the acid or base is polyprotic or polyhydroxy. For example: · 20 ml 0.5M HCl 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 2 SO 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 HCl is titrated with 0.25M Ca(OH) 2 The original acid is 0.5M; The concentration of H+ = 0.5 M MaVa = 2MbVb2
Slide 93 / 113 Solving Titration Problems: Strong Acid - Strong Base 2) When some titrant is added from the buret, before the equivalence point is reached Example: 20 ml 0.5M HCl 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 HCl + NaOH --> NaCl + H2O 0.01 mol 0.0025mol -0.0025mol -0.0025mol 0.0075mol 0 mol left [H+] after the 10 ml base had been neutralized by the acid is = 0.0075 mols /0.030L = 0.25M pH = 0.6
Slide 94 / 113 Solving Titration Problems: Strong Acid - Strong Base 3) At the equivalence point: Example: 20 ml 0.5M HCl 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 that 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 95 / 113 Solving Titration Problems: Strong Acid - Strong Base 4) Beyond the equivalence point, when excess titrant has been added from the buret Example: 20 ml 0.5M HCl 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 HCl + NaOH --> NaCl + H2O 0.01 mol 0.0125mol -0.01mol -0.01mol 0 mol 0.0025mol excess [OH-] after the 45 ml base added = = 0.0025 mols /0.065L = 0.0385M pOH = 1.415 pH = 12.585
Slide 96 / 113 Solving Titration Problems: Strong Acid - Strong Base Acid in the flask and titrant is the base. The pH at eqequivalence point will be =7
Slide 97 / 113 Solving Titration Problems: Strong Acid - Strong Base Acid in the flask and base is the titrant Base in the flask and acid is the titrant
Slide 98 / 113 25 The moles of acid equals the moles of base at the equivalence point in a titration of _____ with _____. strong acid, weak base A strong base, weak acid B strong acid, strong base C weak acid, weak base D All of the above E
Slide 99 / 113 26 What is the pH of a titration between a weak acid and a strong base at the equivalence point? Less than 7 A Equal to 7 B Greater than 7 C
Slide 100 / 113 27 What is the pH of a titration between a weak base and a strong acid at the equivalence point? Less than 7 A Equal to 7 B Greater than 7 C
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