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Acid-Base-Buffers Review

# Acid-Base-Buffers Review - CHAPTER 8 CHARACTERISTICS OF...

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Unformatted text preview: CHAPTER 8. CHARACTERISTICS OF BUFFER SOLUTIONS CHEM. 1, VAN KOPPEN 1. Buffer solutions consist of either: a weak acid and the salt of its conjugate base (e.g. HF and NaF) or: a weak base and the salt of its conjugate acid (e.g. NHS and NH4Cl) Buffer solutions are resistant to pH change despite small additions of acid or base. Buffer systems are very important in living systems (e.g. constant blood pH is vital). 2. When H+ is added to a buffered solution, it reacts completely with the weak base present: H* + A‘ —> HA or H+ + B —> BH+ e.g. H+ + F' —> HF or H+ + NH3 —> NH4+ 3. When OH— is added to a buffered solution, it reacts completely with the weak acid present: OH‘ + HA —> A“ + HZO 0r OH‘+ BH+ —> B + H20 e.g. OH‘ + HF —> F‘ + H20 or V OH‘ + NH4+ —> NH3 + H20 4. Steps (2) and (3) are stoichiometry problems: In step (2), H* is completely consumed, leaving excess A‘ (or B). In step (3), OH' is completely consumed, leaving excess HA (or BH+ ). You must determine which species remain and how much of each remains in solution. Once [A‘] and [HA] are calculated, the pH of a solution can be calculated from the Henderson-Hasselbalch equation: pH = pKa + log([A‘]/ [HA] ). 5. The pH of a buffered solution will be determined by the ratio [A‘] / [HA], (or [B] / [BH+]). As long as this ratio remains constant, the pH remains constant. This will be the case if [HA] and [A'], (or [B] and [BH"]) are la_rge relative to [H‘] and [OH‘]. Optimum buffering occurs when [A‘] 2 [HA]. In this case, the ratio [A'] / [HA] is most resistant to pH change when H+ or OH' is added. [A] [HA] The pKa of the weak acid selected for the buffer should be as close as possible to the desired pH. pH = pKa + log If [A'] = [HA] then [A‘]/ [HA] : 1 and pH = pI<a TITRATION AND TITRATION CURVES Titration and titration curves are used to analyze the amount or the concentration of acid or base in solution: A solution of known concentration (titrant) is added from a burette into an unknown solution until the equivalence point is reached. At the Equivalence point: The substance being analyzed is just consumed, as signaled by a color change of an indicator. To obtain a titration curve: Calculate pH of solution at various points in the titration: (moles of H+) IH+l ‘ . . pH = —log [H*] (total volume of solutlon at that pomt) STRONG ACID-STRONG BASE TITRATIONS: Example: 50 mL 0.2 M HCl titrated with 0.1 M NaOH. Major species in solution: H+, Cl", Na*, OH', H20 Net Ionic Reaction: H+ + OH‘ —> HZO Neutralization reaction. This is a stoichiorrLtry problem: One of the species, OH‘ or H+ will be completely consumed, leaving an excess of the other. You need to determine which species remains and how much remains in solution, calculate [H"] and pH. For the above example, determine the volume of NaOH required to reach the equivalence point. At the equivalence point: moles of H* 2 moles of OH‘ [H*] Vch = [OH' ] VNaOH or VNaOH = 50 ml (0.2/0.1) = 100 mL It takes 100 mL of 0.1 M NaOH to neutralize 50 mL of 0.2 M HCl. In a neutral solution [H+] = [OH‘] = 1 x 10'7 M, pH = 7.0. WEAK ACID-STRONG BASE TITRATIONS: This is a two step problem: a stoichiometry and an equilibrium problem. 1. Stoichiometry problem: The OH’ reacts completely with the weak acid (OH‘ + HA —> A' + HZO) Determine concentrations [HA] and [A'] after OH' is completely consumed in the reaction. 2. Equilibrium problem: Using concentrations [HA] and [A‘] determined in step (1) and the given Ka for the equilibrium: HA F A‘ + H+, calculate [H+] and pH of solution. Key points in the weak acid-strong base titration: 1. Halfway point: moles of OH" 2 (moles of HA)/2 . In this case, [HA] 2 [A'] , buffer region, pH = pKa 2. Equivalence point: moles of OH' = moles of HA. No weak acid remains, it is completely consumed by OH" : OH‘+ HA —> A' + HZO At this point, the major species in solution are: A' and H20 where A‘ is the conjugate base of a weak acid and has significant affinity for protons. A" reacts with water to produce a basic solution, pH > 7 due to hydrolysis: A‘ + H20 # AH + OH‘ 3. Past the equivalence point: Major species in solution: A‘ , OH“ and H20 but moles of OH‘ produced in the reaction of A' + H20 is negligible compared with OH‘ from NaOH. Stoichiometry problem: determine excess moles of OH’ [OH_] _ (excess moles of OH‘) . . [H] = Kw/[OH'] pH = 40g [m (total volume of solution at that pomt) RELATED QUESTIONS 1. BUFFER PROBLEM: A buffer solution is made by mixing 40 mL of 0.4 M CH3COOH with 30 mL of 0.5 M NaCH3COO. Calculate the pH when 10 mL of 0.1 M NaOH is added to this solution. Major species in solution: CH3COOH, Na*, CH3COO ‘, OH', and H20 What is the most convenient way to begin to solve this problem? The most reactive species in the solution is the OH'. OH' is a strong base and will react with the acetic acid, CH3COOH. This reaction will go to completion. CHSCOOH + OH’ —> CHsCOO‘ + H20 sec; {cl/tic me it re) (0.4M) (0.04L) (0.1 M) (0.01L) (0.5 M) (0.03 L) Pro b M m before rxn: 0.016 mol 0.001 mol 0.015 mol pt 5 (Z. M 0 l e 3 — 0.001 — 0.001 + 0.001 . ‘ . __ ___ __ smase arrow (—>) after rxn: 0.015 mol 0.0 0.016 mol The OH’ is completely consumed in the reaction and all you have left in solution is CH3COOH and CH3COO ‘. The concentrations of these will come to equilibrium according to the following equation. CH3COOH é H+ + CH3COO‘ Ka = 1.76x10'5 . ‘ brig m Initial 0.19M ’ 0 0.20M e01” ‘, . _ m Atequil. 0.19—x x 0.20+x Pmb le’ t . _ + (Age: Coﬂw ‘Om-S K=[CHCOO ] [H] = (0.20+x)x : 1'76X10_5 _______. [CH3COOH] 0.19 — x ,A ‘ ' ’ 0a } x : [H+] 21.67 x10-5 pH=_10g[H+] 2478 double, arr (c) 2. WEAK ACID/STRONG BASE TITRATION: Calculate the pH at the equivalence point when 25.0 mL of 0.200 M HF is titrated with 0.200 M NaOH. First you have to determine if HF is a weak or strong acid by looking up the acid ionization constant, Ka for HF (Ka = 6.6 x 10—4). Because Ka << 1, HF is a weak acid, it dissociates less than 1% in water. At the equivalence point: moles of HF = moles of OH‘ (0.025 L) (0.2 M) = VNaOH (0.2 M) VNaOH = 25.0 mL HF + OH' —> F + H20 T (0.2M) (0.025L) (0.2 M) (0.025L) before rxn: 0.005 mol 0.005 mol 0.0 mol "a — 0.005 — 0.005 + 0.005 after rxn: 0.0 0.0 0.005 mol VMaoH The OH' and HF are both completely consumed in the reaction and all you have left in solution is F'. F” is the conjugate base of HF. Because HF is moderately weak acid, F‘ is a fairly weak base. F‘ reacts with water to produce a basic solution. The concentration of F‘ is [F'] = 0.005 mol/ 0.05 L = 0.1 M. F' + HZO # HF + OH‘ Kb = KW / Ka = 1.0 x 10‘14/66 x 1041 = 1.52 x 10‘11 at equilibrium 0.1 —x x x 0 1X2 2 152 x 10-11 x =[0H-]=1,23 x 10-6 pOH = 5.91 pH=14—pOH=8.1 . —x THREE COMMON MISTAKES WERE MADE. 1) In the equilibrium constant expression you must use concentrations (mol 1 L) of reactants and products, not moles. 2) HF was treated as a strong acid rather than a weak acid. For a strong acid / strong base titration, pH = 7 at the equivalence point. 3) The Henderson-Hasselbalch equation, pH = pKa + log([A']/ [HA] ), was used to calculate the pH at the equivalence point for the HF/NaOH titration problem. The buffer region for a weak acid / strong base titration is at the half-way point, where the moles of OH" = (moles of HF)/ 2 . In this case, [HF] = [F‘] and pH = pKa. However, at the equivalence point all of the HF has been consumed and you are far away from the buffer region. Thus, the pH cannot be calculated using initial concentrations in the Henderson—Hasselbalch equation. ...
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Acid-Base-Buffers Review - CHAPTER 8 CHARACTERISTICS OF...

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