Chap 16A - Chem 210 / WANG / Chapter 16A Acid Base...

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Unformatted text preview: Chem 210 / WANG / Chapter 16A Acid Base Equilibria: Definitions & pH Chapter 16.1 to 16.9 Acid – Base Equilibria 16.1 Acids and Bases: A Review Acids and bases A very important class of chemicals. 16.2 Bronsted-Lowry Acids and Bases Bronsted- • Control CO2 transport in the blood 16.3 Auto-Ionization of Water Auto- (buffers) 16.4 The pH Scale • Amino acids building blocks of protein 16.5 Strong Acids and Bases (contain both and acid and a base) 16.6 Weak Acids 16.7 Weak Bases • Acids (sour) are one of the 4 tastes. 16.8 Relationship between Ka and Kb • Bases feel slippery like soap 16.9 Acid-Base Properties of Salt Solutions Acid- Arrhenius theory Swedish scientist, … the first Swede to be awarded the Nobel Prize in chemistry (1903) .. (1903) first to predict that emissions of CO2 from from burning fossil fuels would cause global warming. S. Arrhenius (1859-1927) linked acid behavior 1859-1927) with the presence of H+ ions and base behavior -1 ions in aqueous solutions. with the OH Acid Anything that produces hydrogen (H+1 or H3O+1) ions in water. HCl + H2O H3O+ + ClBase Anything that produces hydroxide ions in water. NaOH Na+(aq) + OH-(aq) OH aq) Acids Acids = Proton Donor Bases = Proton Acceptor H2O + HCl How H+1 exist in water A proton (H+) in water can only exist in the hydrated hydrated state, because: H+, being a small + charge, is strongly attracted to a non-bonding electron pair on Oxygen: non- = H+ = The The hydrated proton is called Hydronium Hydronium ion with the formula: H3O+1. H3O+1 is experimentally observed in water. experimentally (acid, H+ (base, H+ donor) Lone pair acceptor) + binds H H3O+ + Cl- BronstedBronsted-Lowry theory Extending Arrhenius’ definition of acid-base beyond aqueous solutions, aqueous • J. Brönsted (Danish) and T. Lowry (English), proposed the Brönsted - Lowry concept of acids Brönsted and bases. They independently arrived at this concept in the same year (1923). • The concept is based on the fact that acid-base transfer reactions involve the transfer of H+1 (proton) from one substance to another: – The reactant which donates H+ (proton donor) donates Brönstedin the reaction is a Brönsted-Lowry acid, – The reactant which accepts H+ (proton accepts Brönstedacceptor) in the reaction is a Brönsted-Lowry base. • Proton transfer as the essential feature of a Brønsted-Lowry acid-base reaction. Label the acid and base in this reaction. Lone pair binds H+ + + Cl-H OH2 Cl- + H 3O + CH3NH2 B-L ACID: HCl donates proton B-L BASE: H2O accepts proton H 3 O+ = hydronium ion Fall 2009 UM-SJTU JI Lone pair binds H+ + + NH3 (base, H+ acceptor) H 2O methylamine H 2O (acid, H+ donor) NH4+ + OHCH3NH3+ OH- methylammonium ion Page 1 Chem 210 / WANG / Chapter 16A Acid Base Equilibria: Definitions & pH AcidAcid-base equilibria NH3 + H2O NH4+ + OHNH Brønsted acid (proton donor) & Brønsted base (proton acceptor) designations work for both forward and reverse forward and reverse reactions in equilibrium base acid acid base base acid conjugate acid conjugate base Both the forward and the reverse reactions involve proton transfer, proton • the proton donor in the forward reaction proton • becomes the proton acceptor in the proton reverse reaction. Conjugated acidConjugated acid-base pair = compounds or ions that differ only in the presence or absence of one H+1. one • Every acid has a conjugate base, and conjugate • Every base has a conjugate acid. conjugate Conjugate Acid - Base Pairs Give conjugate bases: (one less H+) (a) HCl - conjugate base: Cl-1 The base: N2H4 is conjugated with the acid: N2H5+ The Conjugate Pairs in Some Acid-Base Reactions Acid- Conjugate Pair + Base Base + Acid Conjugate Pair (a) CO3 - conjugate acid: Reaction 1 Reaction 2 F- HF + H2O HCOOH + CN- + H3O+ H3O+1 H 2O SAMPLE PROBLEM 1 The following reactions are important environmental processes. Identify the conjugate acid-base pairs. HPO42-(aq) + HCO3-(aq) + HCOO- + HCN (d) OH-1 - conjugate acid: O-2 (b) H2O(l) + The acid: HCN is conjugated with the base: CN-1 -2 (d) OH-1 - conjugate base: OH-(aq) H+ donor H+ Acceptor Give conjugate acids: (one more H+1) (c) H2O - conjugate acid: OH-1 SO32-(aq) N2H5+ + CN- HCN + N2H4 CH3NH3+1 (c) H2O - conjugate base: (a) H2PO4-(aq) + CO32-(aq) H+ Acceptor Acid (b) CH3NH2 - conjugate acid: NH3 becomes H+ acceptor in the acceptor reverse reaction. H+ donor Conjugate Acid - Base Pairs HCO3-1 (b) NH4+1 - conjugate base: Conjugate Acid – Base Pairs H+ Donor Donor HSO3-(aq) Acid, Base and BOTH By definition, for a substance to be a Brönsted-Lowry acid, it must have H+1 (to be donated) in its chemical structure. To be a Brönsted-Lowry base, a molecule or ion must have non-bonding e pair or anion (to nonaccept H+1). When a substance fulfills both of the above criteria, it can act either as an acid or a base. Such substances are called: amphoteric. Fall 2009 UM-SJTU JI Reaction 3 NH4+ + CO32- NH3 + HCO3- AcidAcid-base properties of water Water is amphoteric because: It can act either as an acid or a base as base, seen in its auto-ionization. • Also shown below: • HC2H3O2(aq) + H2O(l) acid base H2O(l) + NH3(aq) acid base H3O+ + C2H3O2-(aq) acid base NH4+(aq) + OH-(aq) acid base Page 2 Chem 210 / WANG / Chapter 16A Acid Base Equilibria: Definitions & pH Strength Strength of Acids and Bases Acid, Base and BOTH Consider the following molecules or ions: HCl, H2SO4, NH3, OH-1, H-1, F-1 and HCO3-1. (a) Which can act as acids? HCl, H2SO4, NH3, OH-1 HCO3-1. (b) Which can act as bases? NH3, OH-1, H-1, F-1 and HCO3-1 (c) Which can be amphoteric? N H 3, OH-1 HCO3-1. The stronger the acid the more ready it is to donate H+, the weaker its conjugate base the less ready it is to accept H+. For example, By common knowledge,HCl is the strongest acid below. The weakest base is therefore: Cl-1. (a) HCl - conjugate base: Cl-1 (b) NH4+1 - conjugate base: NH3 (c) H2O - conjugate base: OH-1 (d) OH-1 - conjugate base: O-2 SAMPLE PROBLEM 2 Ref. Figure 16.4 Predict the net direction and whether Kc is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species): Strengths of conjugate acid-base pairs (a) H2PO4-(aq) + NH3(aq) stronger acid Strength of Acids and Bases Among the bases given below, OH-1 is the strongest. The weakest acid below is therefore: H2O. The strongest acid below is H3O+1. The weakest base is therefore: H2O. (a) CO3-2 conjugate acid: HCO3-1 (b) CH3NH2 conjugate acid: CH3NH3+1 (c) H2O conjugate acid: H3O+1 (d) OH-1 conjugate acid: H2O Predict the products of these Acid-Base Reactions, identify Acidthe Conjugate Pairs, and predict whether Kc > 1 or Kc< 1. Conjugate Pair Acid + stronger base weaker base weaker acid Net direction is to the right with Kc > 1. Reaction a (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) weaker stronger weaker stronger base base acid acid Net direction is to the left with Kc < 1. AutoAuto-ionization of Water AcidAcid-base properties of water AutoAuto-ionization Water molecules donate/accept protons with one another to form ions. Accepts a proton = base base H2O(l) + H2O(l) acid donates a proton = acid H3O+(aq) + OH-(aq) H2O(l) H2O(l) HSO4- + N2H62+ Kc > 1 Reaction c HPO42- + SO32- Kc< 1 PO43- + HSO3- AutoAuto-ionization of water H3O+ + OHH3O+(aq) H3O+(aq) + OH-(aq) Kw = [ H3O+ ] [ OH- ] conjugate base Fall 2009 UM-SJTU JI HPO42- + H2O H2SO4 + N2H5+ H2O(l) + H2O(l) ion product of water [ H3O+ ] = [ OH- ] = 10-7 M in pure water at 25oC. Kw = 1.0 x 10-14 at 25oC + H2 O + H2 O H2PO4- + OH- Equilibrium Constant for Auto-ionization of water Auto- + conjugate acid Base + Acid Conjugate Pair Kc > 1 Reaction b See next slide Base HPO42-(aq) + NH4+(aq) OH-(aq) Note: H2O(l) as a pure liquid is not included in Kw. Page 3 Chem 210 / WANG / Chapter 16A Acid Base Equilibria: Definitions & pH Relations among [H3O+], pH, [OH-], and pOH. pH and pOH value 1000 500 100 50 10 5 1 0.5 0.1 0.05 0.01 0.005 0.001 We need to measure and use acids and bases over a very large concentration range. very pH and pOH are systems to keep track of these very large ranges. pH = -log[H3O+] OR: [H3O+] = 10-pH pOH = -log[OH-] OR: [OH-] = 10-pOH At 25oC, in pure water and in all aqueous solutions: Kw = [H3O+] [OH-] = 10-14 10 = 10-pH x 10-pOH = 10-(pH + pOH) pH + pOH = 14 pH = -log[H+] pH calculations [H+] = 10- pH Determine the following. pH of 6.70 x10-3 M [H+] 6.70 3 sig. fig. pH of 5.2x10-12 M [H+] 5.2 2 sig. fig. [H+], pH = -log[H+] pOH = -log[OH-] = 14 – pH [H+] pH = 2.170 3 digits after decimal pt. pH = 11.28 2 digits after decimal pt. if the pH is 4.5 1 digit after decimal pt. [H+] = 3 x10-5 M 1 sig. fig. = 10- pH [OH-] = mod. basic 14 2 sig. fig. after decimal pt 1.7 pOH of 1.7x10-4 M NaOH pOH = 3.77 pH = 10.23 2 sig. fig. after decimal pt pOH of 5.2x10-12 M [H+] 5.2 pH = 11.28 Logarithms The sign tells us if the number is > 1 or < 1. The number before the decimal point tells us the exponent (10n) in scientific notation. The number after the decimal point gives the number of significant significant figures in the original value. pH pH scale pOH examples 10- pOH Determine the following. log(value) 3.0 2.7 2.0 1.7 1.0 0.7 0.0 -0.3 -1.0 -1.3 -2.0 -2.3 -3.0 12 slight basic slight acidic 10 9 8 7 6 10-14 M strongly Basic mod. acidic 543210 10-7 M Neutral 1M strongly Acidic pOH = 2.72 [OH-] , if the pH is 4.50 pOH = 14 - 4.50 = 9.50 [OH [OH-] = 3.2 x10-10 M 3.2 2 sig. fig. SAMPLE PROBLEM 3 pH of some common materials Substance pH relative acidity strongly acidic very acidic slightly acidic Neutral slightly basic moderately basic strongly basic Coffee is 102 (100) times more acidic than pure water. 10 times acidic Blood is 100.35 to 100.45 (2.24 to 2.82) times more basic 10 (2.24 times basic than pure water. Fall 2009 UM-SJTU JI Refer to similar figure given in Figure 16.5 1 M HCl 0.0 Lemon juice 2.3 Coffee 5.0 Pure Water 7.0 Blood 7.35-7.45 Milk of Magnesia 10.5 1M NaOH 14.0 In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate 0.30M 0.0063M [H3O+], pH, [OH-], and pOH of the three solutions at 250C. For 2.0M HNO3, 2.0M [H3O+] = 2.0M -log [H3O+] = -0.30 = pH pH = - 0.30 [OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30 Page 4 Chem 210 / WANG / Chapter 16A Acid Base Equilibria: Definitions & pH SAMPLE PROBLEM 3 SAMPLE PROBLEM 3 In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 250C. In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 250C. For 0.3M HNO3, [H3 O+] = 0.30M -log [H3O+] = 0.52 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; Methods for measuring the pH of an aqueous solution For 0.0063M HNO3, [H3O+] = 0.0063M -log [H3O+] = 2.20 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 pH (indicator) paper pH meter = 1.6x10-12M; Ref. Figure 16.6 pOH = 13.48 Figure 16.7 pOH = 11.80 Close-up diagram showing the tip of the pH probe Fall 2009 UM-SJTU JI Page 5 ...
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This note was uploaded on 07/30/2011 for the course CHEM 210 taught by Professor Zhang during the Spring '09 term at Shanghai Jiao Tong University.

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