Buffers (2).notebook. March 14, 2016

Transcription

1 In this activity we will develop principles involved in making a buffer solution and then determine how a buffer minimizes changes in the ph upon addition of strong acid or base. Buffers consist of a weak acid and its conjugate (acidic buffer) or a weak base and its conjugate (basic buffer). Let s prepare a buffer! The first step involves obtaining a weak acid. Let s pick ml of a 0.200M solution of benzoic acid, HC 7 H 5 O 2. The K a for benzoic acid is 6.4 X Write the equation for the ionization (dissociation) of benzoic acid in water and indicate the Bronsted Lowry acid base conjugate pairs: Now calculate the ph of this solution and the percent ionization of the acid: ph = % ionization = 1

2 The second step in making our acidic buffer is to add a salt of the acid. In this case let s pick sodium benzoate, NaC 7 H 5 O 2. What ions does sodium benzoate dissociate into in solution? Which of these ions do you predict would be a spectator ion? Which of these ions is common to the acidic solution? Rewrite the reaction for the ionization benzoic acid in water below: Note this is an equilibrium reaction. What do you predict will happen if you added some sodium benzoate to this equilibrium solution? Will the equilibrium shift and if so in what direction? How will this shift affect the concentrations of unionized benzoic acid and hydronium ions in the solution and the ph? predict [HC 7 H 5 O 2 ] will predict [H 3 O + ] will predict ph will 2

3 Now let s add exactly mole worth of solid sodium benzoate to the ml of the benzoic acid solution. Assume the addition of the solid does not affect the volume of the solution. Calculate the molar concentration of benzoate ion (C 7 H 5 O 2 ) in the solution: Rewrite the equation for the ionization of benzoic acid in water. Note the original concentration of benzoic acid was 0.200M and the concentration of benzoate ion in this solution is now what you calculated above. Calculate the ph and percent ionization of the acid after the salt has been added: ph = % ionization = Do the new ph and percent ionization values match your previous prediction? 3

4 We have just created ml of an acidic buffer solution in which the concentration of benzoic acid and benzoate ion are both essentially 0.200M. This is a buffer system because both the benzoic acid and its conjugate base are now in relatively high concentrations. This means that there is a relatively large amount of weak acid (HC 5 H 7 O 2 ) to neutralize any strong base (OH ) that might be added to it, and there is a relatively large amount of weak base (C 5 H 7 O 2 ) that can neutralize any strong acid (H 3 O + ) that might be added to it. Let s review, in general formulas, how an acidic buffer is made. Again, a buffer consists of a weak acid and a salt of the acid (or weak base and its salt). weak acid: HA + H 2 O = H 3 O + + A add salt: XA = X + + A (X + is spectator ion) create buffer: HA + H 2 O = H 3 O + + A Note the large amounts of HA and A. The ph is governed by the ratio of [HA]/[A ]. The real utility of buffers comes when we add a strong acid or base. A buffered solution should resist ph change when a substantial amount of H + or OH is added. ("Substantial amount" will vary from solution to solution, depending on the "buffer capacity".) 4

5 In the case of making the benzoic acid buffer: weak acid: HC 5 H 7 O 2 + H 2 O = H 3 O + + C 5 H 7 O 2 add salt: NaC 5 H 5 O 2 = Na + + C 7 H 5 O 2 (Na + is spectator ion) create buffer: HC 7 H 5 O 2 + H 2 O = H 3 O + + C 7 H 5 O 2 5

6 Now that we have created a buffer, let s see how it actually works. Suppose we add 5.00 ml of 1.00M hydrochloric acid to our buffer. That s a lot of strong acid! What component of our buffer will the strong acid react with? If you said the benzoate ion (weak base) you are correct! The reaction between the strong acid (HCl) and weak base (C 7 H 5 O 2 ) will be complete and continue until one runs out (hopefully the strong acid). This part of the problem involves writing the equation for the neutralization reaction and doing the stoichiometry (math part) to calculate the molar concentrations of benzoic acid and benzoate ion after the strong acid has been neutralized. Write the equation for the reaction between the strong acid added and the weak base in our buffer below: note, recall any strong acid is represented by H 3 O + (the Cl is a spectator ion). Under each species in the reaction, indicate the initial moles of each species present (mole = M X V), the change in the moles after the reaction, and the new equilibrium moles after the reaction is complete. Then convert to molar concentrations by dividing moles by total volume (in liters). Just to keep track of the numbers, we are starting with ml of our buffer in which both the benzoic acid and benzoate ion concentrations are 0.200M. We are adding 5.00 ml of 0.100M HCl. 6

7 Neutralization reaction I C E [HC 7 H 5 O 2 ] = [C 7 H 5 O 2 ] = 7

8 Now go back to the original buffer and put in the new concentrations and solve for x (the [H 3 O + ]) and ph using the K a expression. buffer: HC 7 H 5 O 2 + H 2 O = H + + C 7 H 5 O 2 ph after addition of strong acid = ph of original buffer = Did the buffer work well? 8

9 To review in general terms, if we add a strong acid to a buffer, the reaction is complete in which the strong acid reacts with the weak base: A (aq) + H 3 O + (aq) HA(aq) + H 2 O(l) Note that the [A ] will decrease and the [HA] will increase but the relative [H 3 O + ] will change only a little. Keep in mind that the ph isn t a measure of how much acid (HA) is present. ph is a measure of the concentration of H 3 O +. 9

10 Now let s add some strong base to our original benzoic acid buffer. Suppose we add ml of 1.00M sodium hydroxide (that s a lot of strong base). What component of our buffer will the strong base react with? If you said the benzoic acid (weak acid) you are correct! The reaction between the strong base (NaOH) and weak acid (HC 7 H 5 O 2 ) will be complete and continue until one runs out (hopefully the strong base). This part of the problem involves writing the equation for the neutralization reaction and doing the stoichiometry (math part) to calculate the molar concentrations of benzoic acid and benzoate ion after the strong base has been neutralized. 10

11 Write the equation for the reaction between the strong base added and the weak acid in our buffer below: note, recall any strong base is represented by OH (the Na + is a spectator ion). Under each species in the reaction, indicate the initial moles of each species present (mole = M X V), the change in the moles after the reaction, and the new equilibrium moles after the reaction is complete. Then convert to molar concentrations by dividing moles by total volume (in liters). Just to keep track of the numbers, we are starting with ml of our buffer in which both the benzoic acid and benzoate ion concentrations are 0.200M. We are adding ml of 1.00M NaOH. 11

12 Neutralization reaction I C E [HC 7 H 5 O 2 ] = [C 7 H 5 O 2 ] = 12

13 Now go back to the original buffer and put in the new concentrations and solve for x (the [H 3 O + ] and ph using the K a expression. buffer: HC 7 H 5 O 2 + H 2 O = H + + C 7 H 5 O 2 ph after addition of strong base = ph of original buffer = Did the buffer work well? 13

14 If we add a strong base to a solution that contains a weak acid, the reaction is complete in which the strong base reacts with the weak acid): HA(aq) + OH (aq) A (aq) + H 2 O(l) Again, the [HA] will decrease and the [A ] will increase, but the [OH ] will change only a little, and consequently the [H 3 O + ] and ph will change only slightly Notice that in these equations, unless excess strong acid or base is added, no strong acid or base builds up. Only weak species remain. This is why a buffer works so well. Your textbook talks about using the Henderson Hasselbach equation in solving the ph of buffer problems. You can review this on the last page of this worksheet. 14

15 Buffer Problems 1. A solution consists of ml of 0.800M nitrous acid and 0.500M sodium nitrite. The K a for nitrous acid = 4.0 X a) a) write the equation for the ionization of nitric acid in water and indicate the acid base conjugate pairs: b) calculate the ph of this acidic buffer solution: c) Calculate the ph of the solution after the addition of 5.00 ml of 1.00M nitric acid: Neutralization reaxn & math part (strong acid reacts completely with weak base in buffer): 15

16 Now back to buffer equilibrium with new concentrations: 16

17 Now lets try a basic buffer! 2. A solution consists of ml of 0.500M ammonia (NH 3 ) and M ammonium chloride. The K b for ammonia = 1.8 X a) write the equation for this reaction and indicate the acid base conjugate pairs: b) calculate the ph of this basic buffer solution: c) Calculate the ph of the solution after the addition of 10.0 ml of 1.00M nitric acid: Neutralization reaxn & math part (strong acid reacts completely with weak base in buffer): 17

18 Now back to buffer equilibrium with new concentrations: 18

19 3. A buffered solution is made by adding 7.50 g of methylamine chloride, CH 3 NH 3 Cl, to ml of a M solution of methylamine (CH 3 NH 2 ). Assume no change in volume on the addition of the salt. The K b for methylamine is 4.38 X a) Determine the ph of this buffered solution. Show relevant equation. 19

20 b) Exactly 10.0 ml of a 2.00 M hydrochloric acid solution is added to this buffer. Calculate the resulting ph. Show relevant equations. 20

21 4. Exactly 15.0 g of sodium lactate, NaC 3 H 5 O 3, is added to 200.0mL of a 0.400M lactic acid HC 3 H 5 O 3, solution. Assume no change in volume upon the addition of the salt. The K a for lactic acid is 1.38 X a) Determine the ph of this buffered solution. Show relevant equation. 21

22 b) Exactly 10.0 ml of a 1.50 M potassium hydroxide solution is added to this buffer. Calculate the resulting ph. Show relevant equations. 22

23 All other things being equal, the closer the [HA]/[A ] ratio is to 1, the better will be the buffer capacity of the solution. This will happen when [H + ] = K a, or ph = pk a. This means that when selecting a buffer that will be used to maintain the ph of a solution in a particular small range, The pk a of the buffer should be as close as possible to the desired ph of the solution Problem: We wish to buffer a solution at ph = Which one of the following bases (and conjugate acid salts) would be most useful? a) NH 3 (K b = 1.8 X 10 5 ) b) C 6 H 5 NH 2 (K b = 4.2 X ) c) N 2 H 4 (K b = 9.6 X 10 7 ) 23

24 ph and pk a / the Henderson Hasselbach Equation H + + A HA [H + ] = K a log[h + ] = log(k a ) log ph = pk a log or, where inverting the log term reverses the sign: ph = pka + log Note that if [HA] = [A ], then ph = pk a (and poh = pk b ) 24