6-Acids-Bases-Buffers II.pdf

PreMed1 Chemistry for Medicine Acids, Bases, and Buffers II Dr. Alya A. Arabi College of Medicine and Health Sciences UAEU Sections to be covered: • The pH Scale • Buffers and Buffered Solutions pH = -log [H+] Henderson-Hasselbalch equation https://wps.prenhall.com/wps/media/objects/602/61651...

PreMed1 Chemistry for Medicine Acids, Bases, and Buffers II Dr. Alya A. Arabi College of Medicine and Health Sciences UAEU Sections to be covered: • The pH Scale • Buffers and Buffered Solutions pH = -log [H+] Henderson-Hasselbalch equation https://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter15/Text_Images/FG15_01.JPG The pH Scale In most solutions [H+(aq)] is quite small. We define p X = − log X pH = − log[H3O+ ] = − log[H + ] pOH = − log[OH- ] In neutral water at 25 C, pH = pOH = 7.00. In acidic solutions, [H+] > 1.0  10-7, so pH < 7.00. In basic solutions, [H+] < 1.0  10-7, so pH > 7.00. The higher the pH, the lower the pOH, the more basic the solution. The pH Scale p X = − log X log (ab) = log a + log b -log (ab) = (-log a) + (-log b) -log (ab) = -log a -log b 𝐾𝑤 = H + OH − 1.0 × 10−14 = H + OH − − log( 1.0 × 10−14 ) = − log H + OH − 14 = − log[ H + ] − log[ OH − ] 𝟏𝟒 = 𝐩𝐇 + 𝐩𝐎𝐇 pH + pOH = 14 Most pH and pOH values fall between 0 and 14. pH + pOH = 14 pH -log [H+] pOH 10-pH [H+] -log [OH-] 10-pOH [H+] [OH-] = 1.0 x 10-14 [OH-] The pH Scale The pH Scale Measuring pH New “accurate” methods: pH meter. Old “less accurate” methods: using indicators An indicator is a dye that changes its color as pH changes. Indicators are less precise than pH meters because: 1. Most indicators do not have a sharp color change as a function of pH. 2. Most indicators tend to be red in more acidic solutions. The pH Scale Indicators Guide Buffered Solutions Composition and Action of Buffered Solutions A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X-) (e.g. CH3COOH / CH3COO- buffer of HX/X-) that can resist drastic changes in pH upon the addition of small amounts of strong acid or base. HX(aq) The Ka expression is (remember state liquid and solid are always excluded from the Ka expression) + - H (aq) + X (aq) [ H + ][X - ] Ka = [ HX] [ HX] + [H ] = K a [X ] [ H + ][X - ] Ka = [ HX] [ HX] + [H ] = K a [X ] Take –log of both sides Remember that log (ab) = log a + log b (apply this rule to the right hand side of the eq.) [HX] + − log[H ] = − log K a − log [X ] [X ] Henderson pH = pK a + log [HX] Hasselbalch equation Buffered Solutions Buffer Capacity and pH Henderson- Hasselbalch equation Buffered Solutions What happens when [acid] = [conjugate base]? 0 because log 1=0 pH = pKa Buffered Solutions Buffer Capacity and pH Buffer capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree, i.e. it depends on the composition of the buffer. The buffer is at its best capacity when [HA]=[A-] Note: The greater the amounts of conjugate acid-base pair, the greater the buffer capacity. Example (acetic acid/acetate): compare 1 M CH3COOH / 1 M CH3COOto 0.1 M CH3COOH / 0.1 M CH3COOThe pH of a buffer depends on Ka as well as the relative concentrations of the acid and its conjugate base. Practice Example 1: Calculate the pH of a buffer composed of benzoic acid (0.12 M) and its conjugate base sodium benzoate (0.20 M). (Ka of benzoic acid = 6.3 x 10-5 ). pH = -log (6.3 x 10-5) + log (0.20/0.12) = 4.42 Practice Example 2: Calculate the concentration of sodium benzoate (conjugate base) that must be present in a 0.20 M solution of benzoic acid (HC7H5O2) to produce a pH of 4.0 (Ka of benzoic acid = 6.3 x 10-5 ). 4.0 = -log (6.3 x 10-5) + log ([A-]/0.2) -0.2=log ([A-]/0.2) Take 10x for both sides 0.63 = [A-]/0.2 [A-] = 0.13 M Buffered Solutions Composition and Action of Buffered Solutions A buffer solution resists a drastic change in pH when a small amount of OH- or H+ is added, as follows: 1. When OH - is added to the buffer, the OH- reacts with HX to produce X - and H2O. But, the [HX]/[X -] ratio remains more or less constant, so the pH is not significantly changed. HX + OH- → X- + H2O 2. When H + is added to the buffer, X- is consumed to react with H3O+. HX and H2O are produced. Once again, the [HX]/[X-] ratio is more or less constant, so the pH does not change significantly. X- + H3O+ → HX + H2O Buffered Solutions Addition of Strong Acids or Bases to Buffers The addition of a strong acid or base results in a neutralization reaction: X- + H3O+ → HX + H2O HX + OH- → X- + H2O. To solve for pH of the solution: 1. By knowing how much H3O+ or OH- was added (stoichiometry), calculate how much HX or X- is formed. 2. With the concentrations of HX and X- (note the change in volume of solution) we can calculate the pH from the: Henderson-Hasselbalch equation. Practice Example 3: A buffer solution was made by adding 0.300 mol propionic acid and 0.300 mol sodium propionate to enough water to make 1.00 L of solution. The Ka of the buffer is 1.8 x 10-5. What will be the pH of the solution as a result of the addition of 0.020 mol of HCl? Practice Example 4: Calculate the pH of the solution that would result from the addition of 0.020 mol of HCl to 1.00 L of pure water. What is the pOH of the solution? Practice Example 1: Calculate the pH of a buffer composed of benzoic acid (0.12 M) and its conjugate base sodium benzoate (0.20 M). (Ka of benzoic acid = 6.3 x 10-5 ). Practice Example 2: Calculate the concentration of sodium benzoate (conjugate base) that must be present in a 0.20 M solution of benzoic acid (HC7H5O2) to produce a pH of 4.0 (Ka of benzoic acid = 6.3 x 10-5 ). Buffered Solutions Addition of Strong Acids or Bases to Buffers The addition of a strong acid or base results in a neutralization reaction: X- + H3O+ → HX + H2O HX + OH- → X- + H2O. To solve for pH of the solution: 1. By knowing how much H3O+ or OH- was added (stoichiometry), calculate how much HX or X- is formed. 2. With the concentrations of HX and X- (note the change in volume of solution) we can calculate the pH from the: Henderson-Hasselbalch equation.