Section14_Buffers

Section14_Buffers - Buffers By the end of this section, you...

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Unformatted text preview: Buffers By the end of this section, you will be able to: (1) Define the word buffer (2) Understand how buffers are prepared, and how they work (3) Calculate the pH of a buffer (4) How to mix a buffer to achieve a target pH (5) Summarize general relationships in acid-base calculations Page 1 of 13 Buffers (MH5, Chapter 14.1) An aqueous solution containing either MH does not phrase it in these terms a weak acid and a salt of its conjugate base or a weak base and a salt of its conjugate acid is called a buffer. A buffer resists changes in pH brought about by the addition of acids or bases. For example, the pH of blood is buffered at 7.4. Strenuous exercise produces lactic acid that the body must counteract. Examples (a) Carbonic acid and sodium hydrogen carbonate an acidic buffer (in the blood): weak acid H2CO3 + H2O H2O HCO3 + H3O+ HCO3Na HCO3aq Na+ + aq under equillibrium control goes to completion salt of conjugate base The dissolution of HCO3Na suppresses the ionization of H2CO3 (by Le Chateliers Principle.) This leaves "acid in reserve" to react with added base; HCO3 takes care of added acid. Page 2 of 13 (b) Ammonia and ammonium chloride a basic buffer weak base NH3 + H2O NH4+ + OH NH4Cl NH4+ + Cl salt of conjugate acid The dissolution of NH4Cl suppresses the protonation of NH3 by H2O (by Le Chateliers Principle.) This leaves "base in reserve" to react with added acid; NH4+ takes care of added base. Na+ and Cl are often used as counterions in acidic and basic buffers, respectively, because they are spectator ions, i.e., they do not affect pH. And also, salts of these ions are generally soluble ACIDIC BUFFERS An acidic buffer is made by mixing a weak acid solution (HA), and a salt of its conjugate base (A). This mixture can react with strong base, OH: HA + OH H2O + A (1) and with strong acid, H3O+: A + H3O+ HA + H2O (2) Page 3 of 13 Recall: Hydroxide (OH) is a strong base; therefore, reaction (1) goes to completion. The conjugate base of a weak acid (A) is a strong base; therefore, reaction (2) goes to completion. Equations (1) and (2) show that any added H3O+ or OH are "removed" from the solution, and the pH change is small, i.e., the pH is "buffered." If all the HA or A has reacted, the buffer capacity is reached and the buffer no longer works. BASIC BUFFERS A basic buffer is made by mixing a weak base solution (B) and a salt of its conjugate acid (BH+). This mixture can react with strong acid, H3O+: B + H3O+ BH+ + H2O (3) and with strong base, OH: BH+ + OH B + H2O (4) Page 4 of 13 Recall: The hydronium ion (H3O+) is a strong acid; therefore, reaction (3) goes to completion. The conjugate acid of a weak base (BH+) is a strong acid; therefore, reaction (4) goes to completion. Like (1) and (2), equations (3) and (2) show that any added H3O+ or OH are "removed" from the solution, and the pH change is small, i.e., the pH is "buffered." A buffered solution is more resistant to pH changes than one that is unbuffered. Buffers and pH We need to be able to calculate the pH of a buffer solution. Consider the ionization equilibrium of a weak acid: AH + H2O [H3O]+ + A- Page 5 of 13 The equilibrium constant, Ka, is given by: [H 3O+ ][A- ] Ka = [AH] Solving for [H3O+] we obtain: [AH] [H 3O+ ] = K a - [A ] Take log10 of both sides and multiply though by -1 => H-H equation (5) This is a general equation; it holds for all buffer systems. To simplify our calculations, we assume: [AH] = [AH]0 [A] = [A]0 concentration of salt of conjugate base i.e., that at equilibrium there is very little ionization of the weak acid, and very little protonation of the conjugate base; the concentrations of these are essentially what we put into the flask to start with. Page 6 of 13 From (5), it follows that: [base]/[acid] pH = pK a + log10 [A ]0 [HA]0 - This is known as the Henderson-Hasselbalch Equation. For pH calculations, it is often helpful to keep in mind that [HA] n HA /V n HA = = [A- ] n A - /V n A - Therefore (5) can be rewritten as: nacid/nbase [H 3O+ ] = K a n HA nA - The pH depends only on the mole ratio. Dilution of a buffer with water does not change the pH. For a basic buffer, it can likewise be shown that: [BH + ]0 pH = pK b + log10 [B]0 Page 7 of 13 The Henderson-Hasselbach equation shows that when the concentrations of the weak acid or base and salt are equal, i.e., [HA] = [A]0 or [B] = [BH+]0, the log term becomes zero, and the pH of the buffer equals the pKa or pKb of the acid or base, respectively. Here we have the greatest buffering capacity Let's consider the formic acid/sodium formate buffer system H2CO2 NaHCO2 pH Titrate with strong acid/base buffered region v. little pH change w. added acid or base (=pKa)6.4 of formic acid 1.0 .5 0 .5 1.0 equiv. OH- added (mol) equiv. H3O added (mol) Page 8 of 13 Example: (MH5, example 14.1) Lactic acid CH3-CH(OH)-CO2H (LacH) has a Ka of 1.4 x 10-4. A buffer is prepared by dissolving 1.00 mol LacH and 1.00 mol of sodium lactate (LacNa) in 0.550 L of water. What is the resulting pH? [H3O+] = Ka * (nLach/nLacNa) [H3O+] - 1.4*10^-4 pH = 3.85 (= pKa(LacH)) What happens to pH if an additional 2.0 mol of LacNa are added? [H3O+] = Ka * nLacH/nLacNa = 1.4*10^-4 *(1/3) pH = 4.33 How do we rationalize this? LacH + H2O <--> Lac- + H3O+ Force to LHS, decrease [H3O+] => raise pH Calculate pH after 0.08 mol H3O+ are added H3O+ + Lac- -------> LacH + H2O I 0.08 3.00 1.00 C -0.08 -0.08 +0.08 E 0 2.92 1.08 [H3O+] = Ka * nLacH/nLac= 1.4*10-4 * (1.08/2.92) => pH = 4.29 pH = -0.04 very small! Page 9 of 13 Dec. 2006 exam question Page 10 of 13 Choosing a Buffer System The equation shows that the pH of a buffer depends on two factors: (1) The Ka of the weak acid (2) The mole ratio nHA / nA Addition of more acid HA makes the buffer more acidic, addition of more base B- makes the buffer more basic. Example: (Similar to MH5, Example 14.2) Based on the following data, how would you prepare a buffer at pH 9.00? Acetic acid Ka = 1.8 10-5 Carbonic acid Ka = 4.4 10-7 AMmonium ion Ka = 5.5*10-10 pKa = 4.74 pKa = 6.36 pKa = 9.25 Greatest buffering capacity for pKa pH Could choose NH3/NH4Cl pH = pKa + log ([NH3]/[NH4+]) 9.00 9.25 ratio = 0.56 Page 11 of 13 Most often, buffers are prepared by mixing a weak acid with a salt of its conjugate base. Another possibility is the partial neutralization of a weak acid or a weak base. Example: 0.1 mol HCl is added to 0.2 mol NH3 H3O+ + NH3 0.1 0.2 0 0.1 NH4+ + H2O 0 0.1 initial: final: (H+ is limiting) The resulting solution contains both a weak base (NH3) and its conjugate acid (NH4+). It is a buffer! strong base Example: Which of the following mixtures are buffers? strong acid salt of NaOH + CH3CO2H 1.0 L of 0.50 M HCl + 1.0 L of 0.50 M CH3CO2Na reaction will go to completion "4 NOT a buffer weak acid 0.5 L of 0.50 M HCl + 1.0 L of 0.50 M CH3CO2Na Is a buffer. Only Partially titrated Page 12 of 13 Summary: pH Calculations pH = -log[H3O+] and pOH = -log[OH-] is always true! A) Strong acid HCl: HCl + H2O --> H3O+ + Cl- [H3O+] = [HCl] B) Strong base KOH [OH-] = [KOH] KOH(s) ---> K+(aq) + OH-(aq) C) Weak acid* AH + H2O <--> A- + H3O+ [AH] = Co [H3O+] Ka = [A-][H3O+]/[HA] [H3O+] (Ka C0)1/2 D) Weak base* [OH-] (Kb C0)1/2 B + H2O <--> BH+ + OH- Kb = [BH+][OH-]/[B] [OH-] E) Buffer Ka * ([HA]/[A-]) M+H * assuming that % ionization is < 5% [HA] = C [HA] Page 13 of 13 ...
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This note was uploaded on 06/20/2008 for the course CHEM 024b taught by Professor Jones during the Winter '07 term at UWO.

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