1. 16.3-16.4 Buffer Effectiveness – Titrations and pH Curves

2. Buffer Capacity A buffer can neutralize small amounts of added acids or bases, but can be destroyed if too much is added. Capacity of a buffer: How much added acid or base that the buffer can neutralize. Range of the buffer: the pH range over which the a particular acid and its conjugate base can be effective.

3. Buffer Effectiveness Two factors that have to be taken into consideration when determining the effectiveness of the buffer are: – 1.) The relative amounts of acid and conjugate base The buffer is most effective when the amounts of acid and conjugate base are close to each other. (the further the amounts are, the less effective the buffer). An effective buffer must have an [base]/[acid] ratio of 0.10-10. – 2.) The absolute concentrations of acid and conjugate base A buffer is most effective when the concentrations of acid and conjugate base are high. Buffer range: since the [base]/[acid] ratio should not differ by more than a factor of 10, the effective range for a buffering system is one pH unit on either side of pK a (so…pK a -1, or pK a + 1). This can be found using the Henderson-Hasselbalch equation. In order to choose an appropriate buffer system if given the buffered pH, we would need to use the given pKa values of the acids in question, and the pH that we would buffer at, and we would plug those values into the H-H equation and solve for the [base]/[acid]. The acid with the pKa value closest to the desired buffer pH is most appropriate.

4. Let’s Try a Practice Problem! Which acid would you choose to combine with its sodium salt to make a solution buffered at a pH 7.35? chlorous acid(HClO 2 ) pK a = 1.35 nitrous acid (HNO 2 ) pK a = 3.34 formic acid (HCHO 2 ) pK a = 3.74 hypochlorous acid (HClO) pK a = 7.54 Hypochlorous acid, since its pK a value is closest to the desired buffer pH. pH = pK a + log ([base]/[acid]) log ([base]/[acid]) = pH – pK a = 7.35 – 7.54 = -0.19 [base]/[acid] = 10 -19 = 0.646 If you have 500.0 mL of a 0.10 M solution of the acid, what mass of the corresponding sodium salt of the conjugate base do you need to make the buffer? 2.4 g NaClO

5. Let’s Try Another!!! A 1.0 L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a)Adding 0.050 mol of HCl (b)Adding 0.050 mol of NaOH (c)Adding 0.050 mol of NaF (d)None of the above (a) Adding 0.050 mol of HCl because it will react with all of the NaF, leaving no conjugate base in the mixture.

6. Titrations and pH Curves Acid-Base Titration: a basic (or acidic) solution of unknown concentration reacts with an acidic (or basic) solution of known concentration. The known solution is slowly added to the unknown one while the pH is monitored. – pH meter or indicator (an indicator is a substance whose color depends on the pH). – The point at which the indicator changes color is known as the end point (this is the observable event that happens during a titration). At the equivalence point, the point in the titration when the number of moles or base is stoichiometrically equal to the number of moles of acid, the titration is complete. Here, neither reactant is in excess.

7. College board  Titration ↓ pH curve ↓ ( titration curve )

8. Constructing a pH Curve (Titrating a Strong Acid with a Strong Base) First, a pH curve is a plot of pH vs. volume of titrant (what you are titrating the analyte (the unknown) with.) In this case, when titrating HCl with NaOH, before any base is added, the pH = -log[H + ]  or [H 3 O + ], and since this is a strong acid, the concentration of the acid = [H 3 O + ]. Remember, M = mol solute / L solution, so if you wanted to, or need to, find the number of moles added, just rearrange the formula above. Another important formula to remember is the titration formula: M A V A = M B V B

9. Constructing a pH Curve (Titrating a Strong Acid with a Strong Base) The scenario that they mention in the text is the titration between 25.0 mL of 0.100 M HCl with 0.100 M NaOH. Now, using the titration formula, we can find the volume of base that needs to be added to reach the equivalence point: M A V A = M B V B V b = (0.100)(25.0) / (0.100) = 25.0 mL or 0.025 L The initial pH, the point that you would start the pH curve would be 1, at 0 mL of base added. pH = -log[H 3 O + ] = -log[0.100] = 1.00

10. What about after 5.00 mL of NaOH is added…what would the new pH be? In order to answer this question, we must first set up a table that shows what the addition of hydroxide ions, do to the amount of hydronium ions. mol NaOH added = 5.00x10 -3 L NaOH x (0.100 mol/ L) = 5.00x10 -4 mol NaOH Now, in order to calculate the new pH, we have to calculate the new H 3 O + concentration. To do this, we have to divide the total number of remaining hydronium ions by the total volume (the initial volume + the added volume). 0.00200 mol H 3 O + [H 3 O + ] =--------------------------- = 0.0667 (0.025 L +0.005 L) pH = -log(0.0667) = 1.18 (This would be the same procedure all the way to the equivalence point. After the equivalence point, the will be approximately no hydronium ions left because they would be neutralized by the hydroxide ions and the pH would be 7. After this point, you would have to calculate the [OH - ] and then ultimately convert it to pH. The overall pH curve of a strong acid with a strong base has an S-shape. If the curve was for a strong base being titrated with a strong acid, the curve would be reversed.) OH - (aq) + H 3 O + (aq)  2H 2 O(l) Before addition~0 0.00250 mol Addition0.000500 mol- After Addition0.000500 mol0.00200 mol

11. Let’s Try a Practice Problem! A 50.0 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid. Calculate pH after the addition of 60.0 mL of 0.200 M HNO 3. Initial amount of NaOH = 0.050 L x (0.200 mol/L) = 0.010 mol mol added HNO 3 = 0.060 L x (0.200 mol/L) = 0.012 mol 0.002 mol H 3 O + [H 3 O + ] = ------------------------- = 0.0182 (0.050 L + 0.060 L) pH = -log(0.0182) = 0.963 OH - (aq) + H 3 O + (aq)  2H 2 O(l) Before addition0.010 mol ~0 Addition-0.012 mol After Addition~00.002 mol  # of hydronium ions remaining after neutralization.

12. The Titration of a Weak Acid with a Strong Base First off, the volume at the equivalence point in an acid base titration does not depend on whether the acid being titrated is strong or weak; it depends only on the amount (in moles) of acid present in solution before the titration begins, and the concentration of the added base. There are a couple of differences here. – 1 st.) The initial pH has to be calculated using an ICE table, and converting the [H 3 O + ] to pH. – 2 nd.) After the titration begins, the OH - being added, converts the weak acid into its conjugate base, so the H-H equation can be used. – 3 rd.) At the distance half-way to the equivalence point, one-half of the original amount will be converted to conjugate base, so [A - ]/[HA + ] = 1, and at this point the pH = pK a – 4 th.) At the equivalence point, the solution is no longer a buffer, and now we need to calculate the pH from the concentration of the conjugate base. In order to do so, we also need to calculate K b from K a, find [OH - ], and ultimately convert the hydroxide concentration to pH. When titrating the a weak acid with a strong base, the equivalence point will always be basic. – Continue to carry out the problem the same way, and what you will find is the curve will have that same signature S-shape.

13. Let’s Try a Practice Problem! A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the equivalence point for the titration of HNO 2 and KOH. (The K a of nitrous acid is: 4.6x10 -4 ) Initial amount of HNO 2 0.040 L x (0.100 mol/L) = 0.0040 mol HNO 2 Volume of KOH at equivalence = (0.100 M)(0.040L) / 0.200 M = 0.020 L KOH Addition of KOH = 0.020 L x (0.200 mol/L) = 0.004 mol KOH So now we need to calculate the [NO 2 - ] by dividing the number of moles by the total volume. Then we need to set up an ICE table, and set up an equilibrium expression (here you will also need to convert K a to K b ) to find [OH - ]. Next convert [OH - ] to pOH and pOH to pH. Phewwwwwwwwwwwwww!!!! OH - (aq) + HNO 2 (aq)  H 2 O(l) + NO 2 - (aq) Before Addition~00.0040 mol0 Addition0.0040 mol-- After Addition~0 0.0040 mol

14. 0.00400 mol NO 2 - [NO 2 - ] = --------------------------- = 0.0667 M (0.040 L + 0.020 L) NO 2 - (aq) + H 2 O(l) HNO(aq) + OH - (aq) Now, use the K a of the acid that the conjugate base came from to solve for K b. K w 1.0x10 -14 K b = ----- = --------------- = 2.2x10 -11 K a 4.6x10 -4 x 2 2.2x10 -11 = -------------- x = 1.20x10 -6 0.0667 – x Since x = [OH - ] = 1.2x10 -6 M, we can now convert [OH - ]  pOH  pH!!! pOH = -log[1.2x10 -6 ] = 5.92 pH = 14.0 - 5.92 = 8.08 [NO 2 - ] M[HNO] M[OH - ] M Initial0.06670~0 Change-X+X Equilibrium0.0667-XXX

15. The Titration of a Weak Base with a Strong Acid This is very similar to the titration of a weak acid with a strong base, except here we are started with a base…so the curve is backwards. You can use the H-H equation within the buffer range (the range after the first addition of strong acid and up until the drastic change, and again after the drastic change) of the titration curve. Here the pKa is of the conjugate acid of the base being titrated.

16. Let’s Try a Practice Problem! What is the half-equivalence point in the titration of a weak base with a strong acid? The pKb of the weak base is 8.75. (a)8.75 (b)7.0 (c)5.25 (d)4.37 (c) 5.25. Remember, the half-equivalence point is when pKa = pH so… pK b + pK a = 14 pK a = 14.0 -8.75= 5.25 College Board:

17. Titration Curves of Polyprotic Acids Here’s what you need to know. – If you are dealing with a diprotic acid whose K a1 and K a2 are sufficiently different, you pH curve will have two equivalence points. Triprotic acids with large differences in K a values will have three equivalence points.

18. Let’s Try a Practice Problem! Consider these three titrations: (i)The titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii)The titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH (iii)The titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH Which statement is most likely to be true? (a)All three titrations have the same initial pH (b)All three titrations have the same pH at their first equivalent point. (c)All three titrations require the same volume of NaOH to reach their first equivalence point. (c) All three titrations require the same volume of NaOH to reach their first equivalence point. Since the volumes and concentrations of all three acid are the same, the volume of NaOH required to reach the first (and only) equivalence point is the same for all three titrations.

19. Indicators: pH Dependent Colors A pH meter can be use to track pH changes as either an acid or base is being titrated. However, with an indicator, a weak organic acid that is a different color than it’s conjugate base, we rely on the point in which the indicator changes color (the end point), to determine the equivalence point. In order for the indicator to correctly identify the equivalence point, we must choose the correct indicator. What is the name of the common indicator that is used in an acid base titration? Phenolphthalein

20. In the following equilibrium, HIn (the letter before the n is a capital i) represents the acid form of the generic indicator, and In - represents the conjugate base form. HIn(aq) + H 2 O(l) H 3 O + (aq) + In - (aq) When the amount of [H 3 0 + ] changes during the titration, the equilibrium shifts. A low pH, is the hydronium ion concentration high or low? Which way will the equilibrium shift? At a low pH, [H 3 O + ] is high and the equilibrium lies far to the left. This results in the indicators color 1. Near the equivalence point, there is a large change in [H 3 O + ], so one extra drop of base at this point would change the color of the indicator to its second color. The color of the solution containing an indicator depends on the concentrations on HIn and In -. The H-H equation can be used to find the ratio of ln - / [HIn], if the pH, and pKa are know. [In - ] If ------- = 1, the indicator solution will be intermediate in color. [HIn] [In - ] If ------- > 10, the indicator solution will be the color of In -. [HIn] [In - ] If ------- < 0.1, the indicator solution will be the color of HIn. [HIn]

21. 16.3-16.4 pgs. 805-807 #’s 54 (a&b), 58, 62, 64, 82 and 84 Read 16.5-16.6 pgs. 783-791