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Acids and Bases

Acids and Bases. Acids taste sour (citric acid, acetic acid) Bases taste bitter (sodium bicarbonate) There are 3 ways to define acids and bases, you will learn 2 of these: Arrhenius: - Acids form H 3 O + in water (HCl + H 2 O  H 3 O + + Cl - )

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Acids and Bases

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  1. Acids and Bases • Acids taste sour (citric acid, acetic acid) • Bases taste bitter (sodium bicarbonate) • There are 3 ways to define acids and bases, you will learn 2 of these: Arrhenius: - Acids form H3O+ in water (HCl + H2O  H3O+ + Cl-) - Bases form OH- in water (NaOH  Na+ + OH-) Brønsted-Lowry (B-L): - Acids donate H+ and Bases accept H+ HCl + NaOH  H2O + NaCl HCl is the acid, it donates H+ to OH- (the base)

  2. B-L Acids and Formation of H3O+ • In an acid-base reaction, there is always an acid/base pair (the acid donates H+ to the base) • H+ is not stable alone, so it will be transferred from one covalent bond to another Example: formation of H3O+ from an acid in water HBr + H2O  H3O+ + Br-

  3. Identifying B-L Acids and Bases • Compare the reactants and the products - The reactant that loses an H+ is the acid - The reactant that gains an H+ is the base Examples: HCl + H2O  H3O+ + Cl- Acid = HCl and Base = H2O (HCl gives H2O an H+) NH3 + H2O  NH4+ + OH- Acid = H2O and Base = NH3 (H2O gives NH3 an H+) CH3CO2H + NH3 CH3CO2- + NH4+ Acid = CH3CO2H and Base = NH3 (CH3CO2H gives NH3 an H+)

  4. Conjugate Acids and Bases • When a proton is transferred from the acid to the base (in a B-L acid/base reaction), a new acid and a new base are formed: HA + B  A- + HB+ acid + base  conjugate base + conjugate acid • The acid (HA) and the conjugate base (A-) that forms when HA gives up an H+ are a conjugate acid/base pair • The base (B) and the conjugate acid (HB+) that forms when B accepts an H+ are another conjugate acid/base pair

  5. Identifying Conjugate Acid/Base Pairs • Identify the acid and base for the reactants • Identify the acid and base for the products • Identify the conjugate acid/base pairs acid conjugate base + + baseconjugate acid HF H3O+ F- H2O

  6. Acid and Base Strength • Strong acids give up protons easily and completely ionize in water: HCl + H2O  H3O+ + Cl- • Weak acids give up protons less easily and only partially ionize in water: CH3CO2H + H2O  CH3CO2- + H3O+ • Strong bases have a strong attraction for H+ and completely ionize in water: KOH(s)  K+ (aq) + OH-(aq) NaNH2 + H2O  NH3 + NaOH • Weak bases have a weak attraction for H+ and only partially ionize in water: HS- + H2O  H2S + OH-

  7. Direction of an Acid/Base Equilibrium • In general, there’s an inverse relationship between acid/base strength within a conjugate pair: - strong acid  weak conjugate base - strong base  weak conjugate acid (and vice-versa) • The equilibrium always favors the direction that goes from stronger acid to weaker acid Example 1: HBr + H2O ? H3O+ + Br- stronger acid (HBr)  weaker acid (H3O+) (equilibrium favors products) Example 2: NH3 + H2O ? NH4+ + OH- weaker acid (H2O)  stronger acid (NH4+) (equilibrium favors reactants)

  8. Dissociation Constants • Since weak acids dissociate reversibly in water, we can write an equilibrium expression: HA + H2O  H3O+ + A- Keq = [H3O+][A-]/[HA][H2O] • But, since [H2O] remains essentially constant we can write: Ka = Keq x [H2O] = [H3O+][A-]/[HA] • The acid dissociation constant (Ka) is a measure of how much the acid dissociates (A higher Ka = a stronger acid) • Example: CH3CO2H + H2O  CH3CO2- + H3O+ Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5 • Can also write dissociation constants for weak bases: NH3 + H2O  NH4+ + OH- Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5

  9. Ionization of Water • Since H2O can act as either a weak acid or a weak base, one H2O can transfer a proton to another H2O: H2O + H2O  H3O+ + OH- Keq = [H3O+][OH-]/[H2O][H2O] • Since [H2O] is essentially constant, we can write: Kw = Keq x [H2O]2 = [H3O+][OH-] (where Kw = the ion-product constant for water) • For pure water: [H3O+] = [OH-] = 1.0 x 10-7 M So, Kw = [H3O+][OH-] = (1.0 x 10-7 M)2 = 1.0 x 10-14 (units are omitted for Kw as for Keq and Ka)

  10. Using Kw • If acid is added to water, [H3O+] goes up - for an acidic solution [H3O+] > [OH-] • If base is added to water, [OH-] goes up - for a basic solution [OH-] > [H3O+] • Kw is constant (1.0 x 10-14) for all aqueous solutions • Can use Kw to calculate either [H3O+] or [OH-] if given the other concentration • Example: if [H3O+] = 1.0 x 10-4 M, what is the [OH-]? Kw = [H3O+][OH-] [OH-] = Kw/ [H3O+] = 1.0 x 10-14/1.0 x 10-4= 1.0 x 10-10 M Is this an acidic or a basic solution? Since [H3O+] > [OH-], it’s an acidic solution

  11. The pH Scale • pH is a way to express [H3O+] in numbers that are easy to work with • [H3O+] has a large range (1.0 M to 1.0 x 10-14 M) so we use a log scale: pH = - log [H3O+] • The pH scale goes from 0 - 14 • Each pH unit = a ten-fold change in [H3O+] • pH 7 = neutral, pH < 7 = acidic, pH > 7 = basic • Can use an indicator dye (on paper or in solution) that changes color with changes in pH, or a pH meter, to measure pH

  12. Calculating pH and pOH • Can calculate pH from [H3O+]: If [H3O+] = 1.0 x 10-3 M, what is the pH? pH = - log [H3O+] = - log(1.0 x 10-3) = 3.00 • Note: sig. figs. in [H3O+] = decimal places in pH • Can also calculate [H3O+] from pH: If pH is a whole number, [H3O+] = 1 x 10-pH So, if pH = 2, then [H3O+] = 1 x 10-2 • Can calculate pOH from [OH-] pOH = - log[OH-] • Also, since Kw = [H3O+][OH-] then pKw = - log Kw = - log (1.0 x 10-14) = 14.00 • And, pKw = pH + pOH = 14.00 • So, if pH = 3.00, then pOH = 14.00 - 3.00 = 11.00

  13. Reactions of Acids and Bases • Acids and bases are involved in a variety of chemical reactions (we’ll study 3 types here) • Acids react with certain metals to produce metal salts and H2 gas, for example: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) • Acids react with carbonates and bicarbonates to produce salts, H2O and CO2 gas, for example: NaHCO3(aq) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g) • Acids react with bases (neutralization reactions) to form salts and H2O, for example: HBr(aq) + LiOH(aq)  LiBr (aq) + H2O(l) • Neutralization reactions are balanced with respect to moles of H+ and moles of OH-, for example: H2SO4 + 2NaOH  Na2SO4 + 2H2O

  14. Acidity of Salt Solutions • Salts dissolved in water can affect the pH • When salts dissolve, they dissociate into their ions NaCl  Na+ + Cl- • If one of those ions can donate a proton to H2O, or accept one from H2O, the pH will change: Na2S  2Na+ + S2- S2- + H2O  HS- + OH- S2- is a weak base that can accept an H+ from H2O Since [OH-] is increased, the solution is basic

  15. Salts that form Neutral Solutions • When a strong acid dissolves in water, a weak conjugate base is formed that can’t remove a proton from water • When a strong base dissolves in water, the metal that dissociates can’t form H3O+ • So, salts containing ions that come from strong acids and bases do not affect the pH of the solution • Example: KBr  K+ + Br- (KOH = strong base, HBr = strong acid) (KOH + HBr  KBr + H2O) K+ has no proton to donate, so can’t form H3O+ Br- is too weak of a base to pull a proton off of H2O, so can’t form OH- So, the solution remains neutral

  16. Salts that form Basic Solutions • When a weak acid dissolves in water, the conjugate base formed is usually strong enough to remove a proton from H2O to form OH- • So, salts that contain ions that come from a weak acid and a strong base form basic solutions • Example: NaCN  Na+ + CN- (HCN is a weak acid) CN- + H2O  HCN + OH- (HCN + NaOH  NaCN + H2O) (Na+ doesn’t affect the pH, it’s from a strong base)

  17. Salts that form Acidic Solutions • When a weak base dissolves in water, the conjugate acid formed is usually strong enough to donate a proton to H2O to form H3O+ • So, salts that contain ions from a weak base and a strong acid form acidic solutions • Example: NH4Br  NH4+ + Br- (from NH3 and HBr) (NH3 + HBr  NH4Br) NH4+ + H2O  NH3 + H3O+ (Br- doesn’t affect the pH)

  18. Buffer solutions • A small amount of strong acid or base added to pure water will cause a very large change in pH • A buffer is a solution that can resist changes in pH upon addition of small amounts of strong acid or base • Body fluids, such as blood, are buffered to maintain a fairly constant pH • Buffers are made from conjugate acid/base pairs (either a weak acid and a salt of its conjugate base or a weak base and a salt of its conjugate acid) • Thus, they contain an acid to neutralize any added base, and a base to neutralize any added acid • Buffers can’t be made from strong acids or bases and the salts of their conjugates since they completely ionize in H2O

  19. How to Make a Buffer Solution • An acetate buffer is made from acetic acid and a salt of its conjugate base: CH3CO2H and CH3CO2Na • The salt is used to increase the concentration of CH3CO2- in the buffer solution • Recall: CH3CO2H + H2O  CH3CO2- + H3O+ (the equilibrium favors reactants, so the concentration of CH3CO2- is low) But, CH3CO2Na  CH3CO2- + Na+ CH3CO2H + CH3CO2Na + H2O  2 CH3CO2- + H3O+ + Na+ • If acid is added: CH3CO2- + H3O+  CH3CO2H + H2O • If base is added: CH3CO2H + OH-  CH3CO2- + H2O • Buffer capacity = how much acid or base can be added and still maintain pH (depends on buffer type and concentration)

  20. Calculating pH of a Buffer • The pH of a buffer solution can be calculated from the acid dissociation constant (Ka) • Example (for acetate buffer): CH3CO2H + H2O  CH3CO2- + H3O+ Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5 [H3O+] = Ka x [CH3CO2H]/[CH3CO2-] • What is the pH of an acetate buffer that is 1.0 M CH3CO2H and 0.50 M CH3CO2Na? [H3O+] = 1.8 x 10-5 x 1.0 M/0.50 M = 3.6 x 10-5 M pH = - log[H3O+] = - log(3.6 x 10-5) = 4.44

  21. Dilutions • Often solutions are obtained and stored as highly concentrated stock solutions that are diluted for use (i.e. cleaning products, frozen juices) • When a solution is diluted by adding solvent, the volume increases, but amount of solute stays the same, so the concentration decreases: Mol solute = concentration (mol/L) x V (L) = constant So, C1V1 = C2V2 • For molarity, it becomes: M1V1 = M2V2

  22. Dilution Calculations • Example 1: What volume (in mL) of 8.0 M HCl is needed to prepare 1.0 L of 0.50 M HCl? M1V1 = M2V2 V1 = M2V2/ M1 V1 = 0.50 M x 1.0 L/ 8.0 M = 0.0625 L = 63 mL • Example 2: How many L of water do you need to add to dilute 0.50 L of a 10.0 M NaOH solution to 1.0 M ? V2 = M1V1/ M2 V2 = 10.0 M x 0.50 L/ 1.0 M = 5.0 L volume of water needed = 5.0 L - 0.50 L = 4.5 L

  23. Acid-Base Titration • Molarity of an acid or base solution of unknown concentration can be determined by titration: • A measured volume of the unknown acid or base is placed in a flask and a few drops of indicator dye (such as phenolpthalein) are added • A buret is filled with a measured molarity of known base or acid (the “titrant”) and small amounts are added until the solution changes color (neutralization endpoint) • At neutralization endpoint [H3O+] = [OH-] • Molarity of unknown is calculated from moles of titrant added (mole ratio comes from balanced chemical equation)

  24. Example: Titration of H2SO4 with NaOH • What is the molarity of a 10.0 mL sample of H2SO4 if the neutralization endpoint is reached after adding 15.0 mL of 1.00 M NaOH? • Calculate moles NaOH added: • 15.0 mL x (1 L/ 1000 mL) x (1.00 mol/ 1 L) = 0.0150 mol NaOH • Write the balanced chemical equation: H2SO4 + 2NaOH  Na2SO4 + 2H2O • Calculate moles H2SO4 neutralized: 0.0150 mol NaOH x 1 mol H2SO4/ 2 mol NaOH = 0.00750 mol H2SO4 • Calculate molarity of H2SO4: 0.00750 mol H2SO4/ 0.0100 L = 0.750 M H2SO4

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