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BUFFER SOLUTIONS AP Chemistry BUFFER SOLUTIONS AP Chemistry Solutions which resist large changes in pH when small amounts of acid or base are added are called buffer solutions. There are two general types of buffers, though they function in similar ways: 1. Acid buffer: a solution containing a weak acid and the conjugate base of the weak acid (for example, HF and NaF) 2. Base buffer: a solution containing a weak base and the conjugate acid of the weak base (for example, NH3 and NH4Cl) Notice each type of buffer contains a species that can function as a weak acid and a species that can function as a weak base. A buffer solution must have each of these components: a weak acid to neutralize additional base and a weak base to neutralize additional acid. Buffer capacity refers to the ability of a buffer system to resist changes in pH. Buffer solutions are most effective when the concentrations of the acid-base pair are about equal. When this is the case, the pH of the buffer solution will be near the pKa for the weak acid. This is important to the buffer’s ability to neutralize either additional acid or base. Buffer capacity is also affected by magnitude of the acid-base concentrations; where the greater the acid-base concentrations, the greater the buffer capacity. For example, a buffer solution of 0.1 M HF and 0.1 M NaF would have a greater buffer capacity than a solution of 0.01 M HF and 0.01 M NaF. This comes down to the number of particles available for neutralization. Let us now examine the reactions that will happen in a buffer solution when acid or base are added to it. EXAMPLE: Write the net ionic equations describing the reactions that occur in a CH3COOH and CH3COONa buffer system upon addition of a) HCl; b) NaOH; c) NaCl. - Here acetic acid is a weak acid, which will react with any additional base, and the acetate ion (CH3COO , from sodium acetate) is a weak base, which will react with any additional acid. a) The hydrochloric acid (HCl) will react with the acetate ion of the buffer. Note: HCl is a strong acid, thus in a net ionic equation it will be represented as simply H+ because of its complete dissociation into ions. + - H (aq) + CH3COO (aq) D CH3COOH(aq) b) The sodium hydroxide (NaOH) is a base and will react with the acetic acid of the buffer. Note: NaOH is a strong base, thus in a net ionic equation it will be represented as simply OH- because of its complete dissociation into ions. - - OH (aq) + CH3COOH(aq) D H2O(l) + CH3COO (aq) c) Sodium chloride is neither an acid nor a base, thus it will not react with the buffer solution. Next, let’s examine how to calculate the pH of a buffer solution. -5 EXAMPLE: A 1.00-L buffer solution contains 0.100 M NH4Cl and 0.0200 M NH3. Kb for NH3 is 1.8 x 10 . What is the pH of the solution? First, let’s consider the equilibrium system and the corresponding equilibrium constant expression. For this example it would be: " # + - "#$! %['$ ] )* NH3(aq) + H2O(l) D NH4 (aq) + OH (aq) �! = = 1.8 × 10 [#$$] Second, we can setup an ICE table to calculate equilibrium concentrations. + - NH3 D NH4 OH Initial 0.0200 M 0.100 M 0 Change -x +x +x Equilibrium 0.0200 – x 0.100 + x x The value of Kb indicates that the process is reactant favored. Furthermore, the presence of the common ion (ammonium ion, + NH4 , from NH4Cl) causes the equilibrium to shift even farther to the left. Therefore, the value of x is so small as to cause a + negligible difference in [NH3] and [NH4 ]. Subbing these equilibrium values into the equilibrium constant expression allows us to solve for [OH-] and thus the pOH. (0.100)(�) = 1.8 × 10)* 0.0200 � = [��)] = 3.6 × 10)+ ��� = − log[��)] = 5.44 The pH can then be calculated by using the relationship between pH and pOH. �� + ��� = 14.00 �� = 14.00 − ��� = 8.56 Next, let’s look at the effect additional base has on this same buffer solution. EXAMPLE: A 1.00-L buffer solution contains 0.100 M NH4Cl and 0.0200 M NH3. What is the pH of the solution if 0.0010 mol of NaOH is added? The additional base will react with the weak acid of the buffer system, which is the NH4+ in this case. - + OH (aq) + NH4 (aq) à H2O(l) + NH3(aq) We assume that the additional base will react completely with the weak acid as shown in the table below. - + OH + NH4 à H2O + NH3 Initial 0.0010 0.100 0.0200 Reaction (100%) -0.0010 -0.0010 +0.0010 New conc. 0 0.099 0.0210 At this point, we must examine what species remain in solution to determine the pH. If residual H+ or OH- remain, they will determine the pH. If a weak acid and/or weak base remains (without H+/OH-), then we must use equilibrium calculations to determine the pH. In this example, we have the latter. Therefore, we will need to setup an ICE table for the following equilibrium. + - -5 NH3(aq) + H2O(l) D NH4 (aq) + OH (aq) Kb = 1.8 x 10 + - NH3 + H2O D NH4 + OH Initial 0.0210 0.099 0 Change -x +x +x Equilibrium 0.0210 – x 0.099 + x x [0.099][�] = 1.8 × 10)*; � = [��)] = 3.8 × 10)+ Note: the addition of the NaOH changes the pH 0.0210 of the buffer solution by only 0.02. This is the ��� = − log(3.8 × 10)+) = 5.42; �� = 14.00 − 5.42 = 8.58 benefit of a buffered solution – maintaining stable pH even with additional acid or base. EXAMPLE: Which buffer solution has the larger buffer capacity toward additional acid, 0.010 M CH3COOH and 0.010 M CH3COONa or 0.010 M CH3COOH and 0.0020 M CH3COONa? Because the question asks about the buffer capacity for additional acid, we would want to choose the buffer system that has a greater amount of base particle available to neutralize the added acid. Here the correct choice would be the buffer with 0.010 M CH3COOH and 0.010 M CH3COONa. Now that we have seen how to perform calculations for the pH of a buffered solution, let’s examine a shortcut known as the Henderson-Hasselbalch equation. The equation is used to approximate the pH of a buffered solution by assuming there is little dissociation of the weak acid, and thus the equilibrium concentrations will be identical to the initial concentrations. We were already making this assumption, so we might as well make the calculations easy, right? The Henderson-Hasselbalch equation is as follows. [�)] �� = ��� + log > @ [��] + -4 EXAMPLE: Find [H ] in a solution of 1.0 M HNO2 and 0.225 M NaNO2. The Ka for HNO2 is 7.2 x 10 . 0.225 �� = − log(7.2 × 10),) + log > @ = 2.49 1.0 [�-] = 10).$ = 10)/.,1 = 3.2 × 10)2� The Henderson-Hasselbalch equation can also be useful for designing a buffered solution for a specific pH. EXAMPLE: What concentrations of acetic acid and sodium acetate are required to prepare a buffer solution with a pH of 4.60? -5 The Ka for acetic acid is 1.8 x 10 . Here we will want to plug in the desired pH and solve for the weak acid/base ratio. [�)] 4.60 = − log(1.8 × 10)*) + log C D [��] [�)] 4.60 = 4.74 + log C D [��] [�)] −0.14 = log C D [��] [�)] 10)3.4, = C D [��] [�)] 0.72 = C D [��] Based on this calculation, any number of combinations could work, as long as the concentration of sodium acetate is 0.72 time that of the acetic acid. Suppose we are given a buffer solution that is formed by a weak base and its conjugate weak acid. There is not a Ka value for a weak base. Therefore, a slight modification must be made. Remember, )4, �5 = �6 × �! = 1.0 × 10 Therefore, )4, �5 1.0 × 10 �6 = = �! �! We now have a value of Ka that can be plugged into the Henderson-Hasselbalch equation. EXAMPLE: Determine the pH of a solution prepared by dissolving 0.35 mole of ammonium chloride, NH4Cl, in 1.0 L of 0.25 M -5 aqueous ammonia, NH3. Kb for ammonia is 1.77 x 10 . 1.0 × 10)4, � = = 5.65 × 10)43 6 1.77 × 10)* [��2] �� = ��� + log C - D [��, ] 0.25 �� = − log(5.65 × 10)43) + log > @ 0.35 �� = 9.248 − 0.14 = 9.11 .
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