Chapter 17 Review

# 6 compare the values of ph initial ph 252 final ph

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6. Compare the values of pH: Initial pH = 2.52 Final pH = 3.05 Therefore, adding 1.00x10 -3 mol of NaOH raised the pH by 0.53 Solution Example, part 2: Without buffering

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37 Key Points on Buffered Solutions 1. They are weak acids or bases containing a common ion . 2. After addition of strong acid or base, deal with neutralization first , then equilibrium . CHM1311 Acid-Base Equilibria
38 The Henderson-Hasselbalch Equation CHM1311 Acid-Base Equilibria K a = [H + ][A " ] [HA] # [H + ] = K a [HA] [A " ] " log[H + ] = " log K a [HA] [A " ] \$ % & ' ( ) " log[H + ] = " logK a " log [HA] [A " ] pH = pK a + log [A " ] [HA]

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39 The Henderson-Hasselbalch Equation …useful for calculating pH when the [A - ]/[HA] ratios are known! pH = pK a +log [base]/[acid] CHM1311 Acid-Base Equilibria
40 The Henderson-Hasselbalch Equation CHM1311 Acid-Base Equilibria note that a buffer solution is most effective when [HA] [A - ], or: therefore, a buffer solution is most effective when pH pK a log [A " ] [HA] # log(1.0) # 0.0 pH = pK a + log [A " ] [HA] # pK a

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41 The Henderson-Hasselbalch Equation CHM1311 Acid-Base Equilibria 2 key criteria for buffer solutions: 0.1 < [HA] < 10 [A - ] [A - ] > 100 x K a and [HA] > 100 x K a 1 2
CHM1311 Acid-Base Equilibria 42 Solution Example: Using the H-H equation 25.0 g of propionic acid (CH 3 CH 2 COOH) and 36.2 g of sodium propionate (CH 3 CH 2 COONa) are mixed in a 1.00 L of solution. What is the pH? K a = 1.30x10 -5 1. What are the concentrations of HA and A - ? 25.0 g / 74.1 (g/mol) = 0.337 mol / 1.00 L = 0.337 M propionic acid 36.2 g / 96.1 (g/mol) = 0.377 mol/ 1.00 L = 0.377 M sodium propionate

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CHM1311 Acid-Base Equilibria 43 25.0 g of propionic acid (CH 3 CH 2 COOH) and 36.2 g of sodium propionate (CH 3 CH 2 COONa) are mixed in a 1.00 L of solution. What is the pH? K a = 1.30x10 -5 2. Use the H-H equation to find the pH. pH = -log(1.30x10 -5 ) + log (0.377 / 0.337) = 4.89 + 0.0487 = 4.94 pH = pK a +log [A - ]/[HA] Solution Example: Using the H-H equation
44 Six Methods of Preparing Buffer Solutions CHM1311 Acid-Base Equilibria

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CHM1311 Acid-Base Equilibria 45 How would you prepare a so-called carbonate buffer solution with a pH of 10.10? You are supplied with H 2 CO 3 , NaHCO 3 , and Na 2 CO 3 . Example: Preparing a buffer Solution K a of H 2 CO 3 = 4.2 x 10 -7 pK a = 6.38 K a of HCO 3 - = 4.8 x 10 -11 pK a = 10.32 The second pK a is closer to the desired pH, therefore, we should use the CO 3 2- / HCO 3 - conjugate pair for our buffer system.
CHM1311 Acid-Base Equilibria 46 How would you prepare a so-called carbonate buffer solution with a pH of 10.10? You are supplied with H 2 CO 3 , NaHCO 3 , and Na 2 CO 3 . pH = pK a + log [CO 3 2 " ] [HCO 3 " ] log [CO 3 2 " ] [HCO 3 " ] = pH " pK a = 10.10 " 10.32 = " 0.22 [CO 3 2 " ] [HCO 3 " ] = 10 " 0.22 = 0.60 We must add Na 2 CO 3 and NaHCO 3 to pure water in a molar ratio of 0.60:1.00. Example: Preparing a buffer Solution

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47 Buffer Capacity and Range Buffer capacity is the amount of acid or base that a buffer can neutralize before its pH changes appreciably. Maximum buffer capacity exists when [HA] and [A - ] are large and approximately equal to each other. Buffer range is the pH range over which a buffer effectively neutralizes added acids and bases.
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