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Acid-Base Reactions Conjugates do not react!! Part 1.

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1 Acid-Base Reactions Conjugates do not react!! Part 1

2 2 Stomach Acidity & Acid-Base Reactions

3 3 ACIDS-BASE REACTIONS For any acid-base reaction where only one hydrogen ion is transferred, the equilibrium constant for the reaction can be calculated and is often called K net.For any acid-base reaction where only one hydrogen ion is transferred, the equilibrium constant for the reaction can be calculated and is often called K net. For the general reaction:For the general reaction: K net is always K a (reactant acid) / K a (product acid)K net is always K a (reactant acid) / K a (product acid) When K net >> 1, products are favored.When K net >> 1, products are favored. When K net << 1, reactants are favored.When K net << 1, reactants are favored. See 16.5 for manipulating K

4 4 There are four classifications or types of reactions: strong acid with strong base, strong acid with weak base, weak acid with strong base, and weak acid with weak base.There are four classifications or types of reactions: strong acid with strong base, strong acid with weak base, weak acid with strong base, and weak acid with weak base. NOTE:For all four reaction types the limiting reactant problem is carried out first. Once this is accomplished, one must determine which reactants and products remain and write an appropriate equilibrium equation for the remaining mixture.NOTE:For all four reaction types the limiting reactant problem is carried out first. Once this is accomplished, one must determine which reactants and products remain and write an appropriate equilibrium equation for the remaining mixture. ACIDS-BASE REACTIONS

5 5 The net reaction is: The product, water, is neutral. STRONG ACID WITH STRONG BASE

6 6 STRONG ACID WITH WEAK BASE The net reaction is: The product is HB + and the solution is acidic.

7 7 WEAK ACID WITH STRONG BASE The net reaction is: The product is A - and the solution is basic.

8 8 WEAK ACID WITH WEAK BASE The net reaction is: Notice that K net may even be less than one. This will occur when K a HB + > K a HA.

9 9 Acid-Base Reactions QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na + + Bz - + H 2 O QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na + + Bz - + H 2 O C 6 H 5 CO 2 H = HBz Benzoate ion = Bz - K a = 6.3 × K b = 1.6 ×

10 10 Acid-Base Reactions The product of the titration of benzoic acid, the benzoate ion, Bz -, is the conjugate base of a weak acid. The final solution is basic. The product of the titration of benzoic acid, the benzoate ion, Bz -, is the conjugate base of a weak acid. The final solution is basic. + + K b = 1.6 x Bz - + H 2 O HBz + OH -

11 11 Acid-Base Reactions Strategy — find the concentration of the conjugate base Bz - in the solution AFTER the titration, then calculate pH. This is a two-step problem: 1 st. stoichiometry of acid-base reaction 2 nd. equilibrium calculation Strategy — find the concentration of the conjugate base Bz - in the solution AFTER the titration, then calculate pH. This is a two-step problem: 1 st. stoichiometry of acid-base reaction 2 nd. equilibrium calculation QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution?

12 12 STOICHIOMETRY PORTION 1. Calculate moles of NaOH required. (0.100 L HBz)(0.025 M) = mol HBz This requires mol NaOH 2. Calculate volume of NaOH required mol (1 L / mol) = L 25 mL of NaOH required Acid-Base Reactions QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution?

13 13 STOICHIOMETRY PORTION 25 mL of NaOH required 3. Moles of Bz - produced = moles HBz = mol Bz - 4. Calculate concentration of Bz -. There are mol of Bz - in a TOTAL SOLUTION VOLUME of 125 mL There are mol of Bz - in a TOTAL SOLUTION VOLUME of 125 mL [Bz - ] = mol / L = M [Bz - ] = mol / L = M Acid-Base Reactions QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution?

14 14 Acid-Base Reactions QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na + + Bz - + H 2 O QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na + + Bz - + H 2 O C 6 H 5 CO 2 H = HBz Benzoate ion = Bz - K a = 6.3 × K b = 1.6 ×

15 15 EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x [Bz - ] [HBz] [OH - ] [Bz - ] [HBz] [OH - ] initial initial change change equilib equilib Acid-Base Reactions QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution? x +x+x -x +x+x x x x x x x

16 16 EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x [Bz - ] [HBz] [OH - ] [Bz - ] [HBz] [OH - ] equilib x x x equilib x x x EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x [Bz - ] [HBz] [OH - ] [Bz - ] [HBz] [OH - ] equilib x x x equilib x x x K b  1.6 x = x x Acid-Base Reactions QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution? Solving in the usual way, we find x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, and pH = 8.25

17 17 Acid-Base Reactions HBz + H 2 O H 3 O + + Bz - K a = 6.3 x [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x pH = 4.20 pH = 4.20 QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH What is the pH at the half-way point?

18 QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) Here we are adding an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the ____________. The Common Ion Effect 18 LEFT Closer to 0 The pH will go _______. After all, NH 4 + is an acid!

19 19 Let us first calculate the pH of a 0.25 M NH 3 solution. [NH 3 ][NH 4 + ][OH - ] initial change -x +x +x equilib x x x QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) The Common Ion Effect

20 20 Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = M This gives pOH = 2.67 and so pH = = for 0.25 M NH 3 K b  1.8 x = [NH 4 + ][OH - ] [NH 3 ] = x x The Common Ion Effect QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq)

21 21 We expect that the pH will decline on adding NH 4 Cl. Let’s test that at 0.10M ! We expect that the pH will decline on adding NH 4 Cl. Let’s test that at 0.10M ! [NH 3 ][NH 4 + ][OH - ] initialchangeequilib The Common Ion Effect QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) x +x +x -x +x +x x x x

22 22 Because equilibrium shifts left, x is MUCH less than M, the value without NH 4 Cl. Because equilibrium shifts left, x is MUCH less than M, the value without NH 4 Cl. K b  1.8 x = [NH 4 + ][OH - ] [NH 3 ] = x(0.10+ x) x The Common Ion Effect QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) [OH - ] = x = (0.25 / 0.10)K b = 4.5 x M This gives pOH = 4.35 and pH = 9.65 pH drops from to 9.65 on adding a common ion.

23 23 Buffer Solutions HCl is added to pure water. HCl is added to pure water. HCl is added to a solution of a weak acid H 2 PO 4 - and its conjugate base HPO HCl is added to a solution of a weak acid H 2 PO 4 - and its conjugate base HPO 4 2-.

24 24 The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the common ion effect. Buffer Composition Weak Acid+Conj. Base HC 2 H 3 O 2 +C 2 H 3 O 2 - H 2 PO 4 - +HPO 4 2- Weak Base+Conj. Acid NH 3 +NH 4 + Buffer Solutions

25 25 Consider HOAc/OAc - to see how buffers work. ACID USES UP ADDED OH -. We know that OAc - + H 2 O HOAc + OH - has K b = 5.6 x Therefore, the reverse reaction of the WEAK ACID with added OH - has K reverse = 1/ K b = 1.8 x 10 9 K reverse is VERY LARGE, so HOAc completely uses up the OH - !!!! Buffer Solutions

26 26 Consider HOAc/OAc - to see how buffers work. CONJUGATE BASE USES UP ADDED H + HOAc + H 2 O OAc - + H 3 O + has K a = 1.8 x Therefore, the reverse reaction of the WEAK BASE with added H + has K reverse = 1/ K a = 5.6 x 10 4 K reverse is VERY LARGE, so OAc - completely uses up the H + ! Buffer Solutions

27 27 Problem: What is the pH of a buffer that has [HOAc] = M and [OAc - ] = M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x [HOAc] [OAc - ] [H 3 O + ] [HOAc] [OAc - ] [H 3 O + ]initialchangeequilib Buffer Solutions x +x +x -x +x +x x x x

28 28 [H 3 O + ] = 2.1 x and pH = 4.68 K a  1.8 x = [H 3 O + ](0.600) Problem: What is the pH of a buffer that has [HOAc] = M and [OAc - ] = M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x Buffer Solutions [HOAc] [OAc - ] [H 3 O + ] [HOAc] [OAc - ] [H 3 O + ] equilib x x x Assuming that x << and 0.600, we have

29 29 Notice that the expression for calculating the H + concentration of the buffer is Notice that the expression for calculating the H + concentration of the buffer is [H 3 O + ] = Orig. conc. of HOAc Orig. conc. of OAc - K a [OH  ]  [Base] [Conj. acid] K b [H 3 O  ]  [Acid] [Conj. base] K a Buffer Solutions This leads to a general equation for finding the H + or OH - concentration of a buffer. This leads to a general equation for finding the H + or OH - concentration of a buffer. Notice that the H + or OH - concentrations depend on K and the ratio of acid and base concentrations. Notice that the H + or OH - concentrations depend on K and the ratio of acid and base concentrations.

30 30 Henderson-Hasselbalch Equation This is called the Henderson-Hasselbalch equation. This is called the Henderson-Hasselbalch equation. pH = pK a + log [Conj. base] [Acid] pH = pK a - log [Acid] [Conj. base] Take the negative log of both sides of this equation Take the negative log of both sides of this equation OR [H 3 O + ] = [Acid] [Conj. base] K a

31 31 Henderson-Hasselbalch Equation This shows that the pH is determined largely by the pK a of the acid and then adjusted by the ratio of acid and conjugate base. This shows that the pH is determined largely by the pK a of the acid and then adjusted by the ratio of acid and conjugate base. pH  pK a + log [Conj. base] [Acid]

32 32 Adding an Acid to a Buffer Problem: What is the new pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68) Solution to Part (a) Calculate [HCl] after adding 1.00 mL of HCl to 1.00 L of water M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2 M 2 = 1.00 x M M 2 = 1.00 x M pH = 3.00 pH = 3.00

33 33 Adding an Acid to a Buffer 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 pH when 1.00 mL of 1.00 M HCl is added to: a)1.00 L of pure water (after HCl, pH = 3.00) b)1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)

34 34 Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H 3 O + ][OAc - ][HOAc] [H 3 O + ][OAc - ][HOAc] Before rxn Change After rxn What is the pH when 1.00 mL of 1.00 M HCl is added to: a)1.00 L of pure water (after HCl, pH = 3.00) b)1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)

35 35 Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O H 3 O + + OAc - HOAc + H 2 O H 3 O + + OAc - [HOAc] [OAc - ] [H 3 O + ] [HOAc] [OAc - ] [H 3 O + ] Before rxn Change -x +x +x After rxn0.710-x0.599+x x What is the pH when 1.00 mL of 1.00 M HCl is added to; a)1.00 L of pure water (after HCl, pH = 3.00) b)1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)

36 36 Adding an Acid to a Buffer [HOAc] [OAc - ] [H 3 O + ] After rxn0.710-x0.599+x x Because [H 3 O + ] = 2.1 x M BEFORE adding HCl, we again neglect x relative to and Because [H 3 O + ] = 2.1 x M BEFORE adding HCl, we again neglect x relative to and What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (after HCl, pH = 3.00) b)1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O H 3 O + + OAc - HOAc + H 2 O H 3 O + + OAc -

37 37 Adding an Acid to a Buffer [H 3 O + ] = 2.1 x M > pH = 4.68 [H 3 O + ] = 2.1 x M > pH = 4.68 The pH has not changed significantly upon adding HCl to the buffer! The pH has not changed significantly upon adding HCl to the buffer! What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (after HCl, pH = 3.00) b)1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68) [H 3 O  ]  [HOAc] [OAc - ] K a  (1.8 x ) Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O OAc - + H 3 O + HOAc + H 2 O OAc - + H 3 O +

38 38 Preparing a Buffer You want to buffer a solution at pH = This means [H 3 O + ] = 10 -pH = 5.0 x M It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ­ pK a ). It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ­ pK a ). You get the exact [H 3 O + ] by adjusting the ratio of acid to conjugate base.

39 39 Preparing a Buffer Solution Buffer prepared from HCO 3 - weak acid CO 3 2- conjugate base HCO H 2 O H 3 O + + CO 3 2-

40 40 Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x M You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x M POSSIBLE ACIDSK a POSSIBLE ACIDSK a HSO 4 - / SO x HSO 4 - / SO x HOAc / OAc x HOAc / OAc x HCN / CN x HCN / CN x Best choice is acetic acid / acetate.

41 41 You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x M Preparing a Buffer [H 3 O  ]  5.0 x = [HOAc] [OAc - ] (1.8 x ) Solve for [HOAc]/[OAc - ] ratio = 2.78/ 1 Therefore, if you use mol of NaOAc and mol of HOAc, you will have pH = Therefore, if you use mol of NaOAc and mol of HOAc, you will have pH = 4.30.

42 42 A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. This simplifying approximation will be correct for all buffers with 3

43 43 REVIEW PROBLEMS Calculate the pH of a 0.10 M HNO 2 solution before and after making the solution 0.25 M in NaNO 2.Calculate the pH of a 0.10 M HNO 2 solution before and after making the solution 0.25 M in NaNO 2. Calculate the pH of L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).Calculate the pH of L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

44 44 Calculate the pH of a solution that is 0.18 M in Na 2 HPO 4 and 0.12 M in NaH 2 PO x 10 -8Calculate the pH of a solution that is 0.18 M in Na 2 HPO 4 and 0.12 M in NaH 2 PO x Suggest an appropriate buffer system for pH 5.0.Suggest an appropriate buffer system for pH 5.0. REVIEW PROBLEMS SOLUTIONS

45 TitrationsTitrations pHpH Titrant volume, mL 45

46 46 Acid-Base Titrations Adding NaOH from the buret to acetic acid in the flask, a weak acid. Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly. In the beginning the pH increases very slowly.

47 47 Acid-Base Titrations Additional NaOH is added. Additional NaOH is added. pH rises as equivalence point is approached. pH rises as equivalence point is approached.

48 48 Acid-Base Titrations Additional NaOH is added. Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point. pH increases and then levels off as NaOH is added beyond the equivalence point.

49 49 QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution? Equivalence point pH of solution of benzoic acid, a weak acid

50 50 EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x EQUILIBRIUM PORTION Bz - + H 2 O HBz + OH - K b = 1.6 x QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution? K b  1.6 x = x x Acid-Base Titrations [Bz - ] [HBz] [OH - ] [Bz - ] [HBz] [OH - ] equilib x x x equilib x x x [Bz - ] [HBz] [OH - ] [Bz - ] [HBz] [OH - ] equilib x x x equilib x x x Solving in the usual way, we find: x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, and pH = 8.25

51 51 QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH to the equivalence point. What is the pH of the final solution? Half-way point

52 52 Acid-Base Reactions HBz + H 2 O H 3 O + + Bz - K a = 6.3 x [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x pH = 4.20 QUESTION: You titrate 100. mL of a M solution of benzoic acid with M NaOH What is the pH at the half-way point?

53 53 Sample Problem Titration Curve mL 0.30 M HC 2 H 3 O 2 is titrated with 0.30 M NaOH. Calculate the pH at 0 mL NaOH added HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x + x + x - x + x + x 0.30 x x K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 1.8 x x 2 (0.30) X = 2.3 x M = [H + ] pH = 2.63

54 54 Sample Problem HC 2 H 3 O OH - ---> C 2 H 3 O HOH HC 2 H 3 O OH - ---> C 2 H 3 O HOH [C 2 H 3 O 2 - ] == M [HC 2 H 3 O 2 ] == 0.18 M mL NaOH added

55 55 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x + x + x - x + x + x Sample Problem 0.18 x K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 1.8 x x(0.060) (0.18) X = 5.4 x M = [H + ] pH = 4.27

56 56 Sample Problem HC 2 H 3 O 2 + OH - ---> C 2 H 3 O HOH HC 2 H 3 O 2 + OH - ---> C 2 H 3 O HOH [C 2 H 3 O 2 - ] == 0.15 M mL NaOH added

57 57 C 2 H 3 O HOH OH - + HC 2 H 3 O 2 C 2 H 3 O HOH OH - + HC 2 H 3 O x + x + x - x + x + x Sample Problem 0.15 x x K b = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 5.6 x x 2 (0.15) X = 9.2 x M = [OH - ] pH = 8.96

58 58 Sample Problem HC 2 H 3 O 2 + OH - ---> C 2 H 3 O HOH HC 2 H 3 O 2 + OH - ---> C 2 H 3 O HOH [OH - ] == M mL NaOH added pOH = 1.22 pH = 12.78

59 59 Sample Problem [H + ] = M pH = mL NaOH added Titration Curve mL M HCl is titrated with M NaOH. Calculate the pH at:

60 60 Sample Problem HCl + NaOH ---> NaCl + HOH HCl + NaOH ---> NaCl + HOH [H + ] = = M mL NaOH added pH = 0.535

61 61 Sample Problem HCl + NaOH ---> NaCl + HOH HCl + NaOH ---> NaCl + HOH mL NaOH added pH = 7.00

62 62 Sample Problem HCl + NaOH ---> NaCl + HOH HCl + NaOH ---> NaCl + HOH [OH - ] = = M mL NaOH added pH =

63 63 Sample Problem Titration Curve mL M NH 3 is titrated with M HCl. Calculate the pH at: 0 mL HCl added NH 3 + HOH NH OH - NH 3 + HOH NH OH x + x + x - x + x + x x x K b = [NH 4 + ][OH - ] [NH 3 ] = = 1.8 x x 2 (0.150) X = 1.6 x M = [OH - ] pH = 11.20

64 64 Sample Problem NH 3 + H + ---> NH 4 + NH 3 + H + ---> NH [NH 4 + ] == M [NH 3 ] = = M mL HCl added

65 65 NH 4 + NH 3 + H + NH 4 + NH 3 + H x + x + x - x + x + x Sample Problem x K a = = = 5.6 x (0.0667)x (0.0333) X = 2.8 x M = [H + ] pH = 9.55 [NH 3 ] [H + ] [NH 4 + ]

66 66 Sample Problem = M 30.0 mL HCl added NH 3 + H + ---> NH 4 + NH 3 + H + ---> NH [NH 4 + ] =

67 67 Sample Problem NH 4 + NH 3 + H + NH 4 + NH 3 + H x + x + x - x + x + x x x K a = = x 2 (0.0600) X = 5.8 x M = [H + ] pH = 5.24 [NH 3 ] [H + ] [NH 4 + ] = 5.6 x

68 68 Sample Problem 40.0 mL HCl added = pH = NH 3 + H + ---> NH 4 + NH 3 + H + ---> NH [H + ] =

69 69 Practice Problems 1. Determine the pH of a solution made by mixing 25.0 mL of 0.20 M nitric acid with 25.0 mL of 0.10 M potassium hydroxide. 2. Determine the pH at the equivalence point if 20.0 mL of 0.30 M HCN is titrated with 0.20 M sodium hydroxide. 3. Determine the pH at the equivalence point if 15.0 mL of 0.20 M nitric acid is titrated with 0.20 M ammonia.

70 70 Practice Problems 4. a) Calculate the pH of a solution made by mixing 50.0 mL of 0.15 M formic acid and 0.41 g of sodium formate. b) Calculate the pH if 10.0 mL of 0.10 M NaOH is added. 5. What is the pH of a solution made by mixing 25.0 mL of 0.20 M benzoic acid and 45.0 mL of 0.10 M sodium benzoate? 6. How many moles of sodium carbonate must be added to 0.20 mole sodium hydrogen carbonate in 250. mL to obtain a pH of 10.00?

71 71 Practice Problems 7. How many mL’s of 0.30 HCl must be added to 25.0 mL of M sodium phosphate to produce a solution with a pH of 13.00? 8. A mL sample of M HNO 3 is titrated with M KOH. Calculate the pH of the solution: a) before the titration begins. b) when mL of base have been added. c) when mL of base have been added. d) at the equivalence point. e) when mL of base have been added.

72 72 Practice Problems 9. A mL sample of M formic acid is titrated with M sodium hydroxide. Calculate the pH of the solution: a) before the titration begins. b) when mL of base have been added. c) when mL of base have been added. d) at the equivalence point. e) when mL of base have been added.

73 73 Practice Problems 10. A mL sample of M ammonia is titrated with M HCl. Calculate the pH of the solution: a) before the titration begins. b) when mL of acid have been added. c) when mL of acid have been added. d) at the equivalence point. e) when mL of acid have been added.

74 74 Practice Problems Answers , ,.77, 2.0, 7.00, , 3.74, 5.03, 8.57, , 9.25, 7.66, 5.07, 2.48 The End!!

75 75 Acid-Base Reactions 1. Calculate the pH if 25.0 mL 0.20 M of HCl is added to 40.0 mL of 0.20 M NaOH. HCl + NaOH ---> NaCl + HOH [OH - ] == M pOH = 1.34 pH =

76 76 Acid-Base Reactions 2. Calculate the pH at the equivalence point if mL 0.20 M of HCl is titrated with 0.20 M NH 3. HCl + NH 3 ---> NH Cl - HCl + NH 3 ---> NH Cl [NH 4 + ] == 0.10 M

77 77 Acid-Base Reactions 2. [NH 4 + ] = 0.10 M NH 4 + NH 3 + H + NH 4 + NH 3 + H + O O x + x + x - x + x + x 0.10 x x K a = [H + ][NH 3 ] [NH 4 + ] = = 5.6 x x 2 (0.10) X = 7.5 x M = [H + ] pH = 5.12

78 78 Acid-Base Reactions 3. Calculate the pH at the equivalence point if mL 0.20 M of HC 2 H 3 O 2 is titrated with 0.30 M NaOH. HC 2 H 3 O 2 + OH - ---> HOH + C 2 H 3 O 2 - HC 2 H 3 O 2 + OH - ---> HOH + C 2 H 3 O [C 2 H 3 O 2 - ]== 0.12M

79 79 Acid-Base Reactions 3. [C 2 H 3 O 2 - ]= 0.12 M C 2 H 3 O HOH HC 2 H 3 O 2 + OH - C 2 H 3 O HOH HC 2 H 3 O 2 + OH x + x + x - x + x + x 0.12 x x K b = [HC 2 H 3 O 2 ][OH - ] [C 2 H 3 O 2 - ] = = 5.6 x x 2 (0.12) X = 8.2 x = [OH - ] pH = 8.91

80 80 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x + x + x - x + x + x Sample Problem Calculate the pH of a M HC 2 H 3 O 2 solution x x K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 1.8 x x 2 (0.100) X = 1.3 x M = [H + ] pH = 2.89

81 81 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x + x + x - x + x + x Sample Problem Calculate the pH of a solution that is M HC 2 H 3 O 2 and M NaC 2 H 3 O x K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 1.8 x x(0.100) (0.100) X = 1.8 x M = [H + ] pH = 4.74

82 82 Sample Problem 1.00 L H 2 O has a pH = 7.00 Calculate the pH if mole HCL is added. [H + ] = = pH = pH units. Adding mole HCl changes the pH from 7.00 to 2.00, 5.00 pH units.

83 83 Sample Problem pH = L M HC 2 H 3 O 2 and M NaC 2 H 3 O 2 has a pH = 4.74 Calculate the pH if mole HCL is added. C 2 H 3 O H + ---> HC 2 H 3 O 2 C 2 H 3 O H + ---> HC 2 H 3 O [C 2 H 3 O 2 - ] == M [HC 2 H 3 O 2 ] = = M

84 84 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x + x + x - x + x + x Sample Problem x K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 1.8 x x(0.090) (0.110) X = 2.2 x M = [H + ] pH = pH units. Adding mole HCl changes the pH from 4.74 to 4.66, only 0.08 pH units.

85 85 Preparing a Buffer Preparing Buffers 1. Solid/Solid: mix two solids. (Example 1) 2. Solid/Solution: mix one solid and one solution.(Example 2) 3. Solution/Solution: mix two solutions. (Example 3) (Example 3) 4. Neutralization: Mix weak acid with strong base (Examples 4 and 5) or weak base with strong acid. or weak base with strong acid. (Example 6) (Example 6)

86 86 Sample Problem 1. Calculate the pH of a solution made by mixing 1.5 moles of phthalic acid and 1.2 moles of sodium hydrogen phthalate in 500. mL of sol’n. K a = 3.0 x HA H + + A - HA H + + A x + x + x - x + x + x x x x K a = [A - ] [H + ] [HA] = (2.4 + x)(x) ( ) x = 3.7 x M pH = 3.43 = 3.0 x Conjugates do not react!!

87 87 Sample Problem 2. How many moles of sodium acetate must be added to 500. mL of 0.25 M acetic acid to produce a solution with a pH of 5.50? X = moles NaC 2 H 3 O 2 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x/ x/ x x x x x x x x10 -6 x/0.500 K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = (x/0.500)(3.2x10 -6 ) (0.25) x = 0.70 mole NaC 2 H 3 O 2 = 1.8 x Conjugates do not react!!

88 88 Sample Problem 3. How many mLs of 0.10 M sodium acetate must be added to 20.0 mL of 0.20 M acetic acid to produce a solution with a pH of 3.50? X = mLs NaC 2 H 3 O 2 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O (20.0/20.0+x) (x/20.0+x) - 3.2x x x x x x x (20.0/20.0+x) 3.2x (x/20.0+x) Conjugates do not react!!

89 89 Sample Problem K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] {0.10(x/20.0+x)}(3.2x10 -4 ) 0.20(20.0/20.0+x) x = 2.2 mL NaC 2 H 3 O 2 = 1.8 x 10 -5

90 90 Sample Problem 4. How many moles of potassium hydroxide must be added to 500. mL of.250 M HCN to produce a solution with a pH of 9.00? X = moles KOH OH - + HCN ---> CN - + HOH OH - + HCN ---> CN - + HOH x x x - x + x - x - x + x x x [CN - ] = x.500 [HCN] = x 0.500

91 91 HCN H + + CN - HCN H + + CN - ( x)/0.500 x/0.500 ( x)/0.500 x/ x x x x x x Sample Problem 1.0 x ( x)/ x x/0.500 K a = [CN - ][H + ] [HCN] = = 4.0x x/.500 (1.0 x ) ( x)/.500) X = mole KOH

92 92 5. How many mLs of 0.30 M sodium hydroxide must be added to 25.0 mL of M acetic acid to produce a solution with a pH of 4.10? X =mL NaOH HC 2 H 3 O 2 + OH - ---> HOH + C 2 H 3 O 2 - HC 2 H 3 O 2 + OH - ---> HOH + C 2 H 3 O x x x x x x x x x x [C 2 H 3 O 2 - ] = 0.30x x Sample Problem [HC 2 H 3 O 2 ] = x x

93 93 HC 2 H 3 O 2 C 2 H 3 O H + HC 2 H 3 O 2 C 2 H 3 O H + ( x)/(25.0+x) 0.30x/(25.0+x) x x x x x x ( x)/(25.0+x) 0.30x/(25.0+x) 7.9 x K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 1.8 x {0.30x/(25.0+x)} 7.9 x ( x)/(25.0+x)) X = 7.6 mL NaOH Sample Problem

94 94 6. Calculate the pH of a solution made by mixing 50.0 mL of 0.15 M NH 3 and 20.0 mL of 0.10 M HCl. HCl + NH 3 ---> NH Cl - HCl + NH 3 ---> NH Cl NA NA NA [NH 4 + ] == M Sample Problem [NH 3 ] == M

95 95 NH 4 + NH 3 + H + NH 4 + NH 3 + H x + x + x - x + x + x x K a = [H + ][NH 3 ] [NH 4 + ] = = 5.6 x x(0.079) (0.029) X = 2.1 x M = [H + ] pH = 9.68 Sample Problem For more practice do all General problems

96 96 Sample Problems 1. Calculate the pH of a 0.10 M HNO 2 solution before and after making the solution 0.25 M in NaNO 2.

97 97 HNO 2 H + + NO 2 - HNO 2 H + + NO x + x + x - x + x + x Sample Problem Calculate the pH of a 0.10 M HNO 2 solution x x x K a = [NO 2 - ][H + ] [HNO 2 ] = = 4.5 x x 2 (.10 - x) X = 6.5 x M = [H + ] pH = 2.19

98 98 HNO 2 H + + NO 2 - HNO 2 H + + NO x + x + x - x + x + x Sample Problem Calculate the pH of a 0.10 M HNO 2 and 0.25 M NaNO 2 solution x x x K a = [NO 2 - ] [H + ] [HNO 2 ] = = 4.5 x x( x) ( x) X = 1.8 x M = [H + ] pH = 3.74

99 99 Sample Problems 2. Calculate the pH of L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

100 100 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x + x + x - x + x + x Sample Problem Calculate the pH of L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid x 0.10 K a = [C 2 H 3 O 2 - ] [H + ] [HC 2 H 3 O 2 ] = = 1.8 x x(0.10) (0.15) X = 2.7 x M = [H + ] pH = 4.57

101 101 Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid. HC 2 H 3 O 2 + OH - ---> HOH + C 2 H 3 O 2 - HC 2 H 3 O 2 + OH - ---> HOH + C 2 H 3 O [HC 2 H 3 O 2 ] == [C 2 H 3 O 2 - ] == 0.15 M HC 2 H 3 O(.500L)(.15M) =.075 mole NaC 2 H 3 O(.500L)(.10M) =.050 mole NaOH(1.0g)(40.0g/mole) =.025 mole

102 102 HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O x + x + x - x + x + x Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid x 0.10 K a = [C 2 H 3 O 2 - ][H + ] [HC 2 H 3 O 2 ] = = 1.8 x x(0.15) (0.10) X = 1.2 x M = [H + ] pH = 4.92

103 Calculate the pH of a solution that is 0.18 M in Na 2 HPO 4 and 0.12 M in NaH 2 PO 4. Sample Problems

104 104 H 2 PO 4 - H + + HPO 4 2- H 2 PO 4 - H + + HPO x + x + x - x + x + x Sample Problem 0.12 x 0.18 K a = [HPO 4 2- ][H + ] [H 2 PO 4 - ] = = 6.2 x x(0.18) (0.12) X = 4.1 x M = [H + ] pH = 7.38

105 Suggest an appropriate buffer system for pH 5.0. Sample Problems Name of Acid K a pK a Oxalic Acid3.8 x Hydrogen Sulfate Ion1.2 x Phosphoric Acid7.1 x Formic Acid1.8 x Hydrogen Oxalate Ion5.0 x Acetic Acid1.8 x Dihydrogen Phosphate Ion6.3 x Boric Acid6.0 x Ammonium Ion9.6 x Hydrogen Carbonate Ion4.7 x Hydrogen Phosphate Ion4.4 x


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