2. 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 - ) - Bases form OH - in water (NaOH  Na + + OH - ) Brønsted-Lowry (B-L): - Acids donate H + and Bases accept H + HCl + NaOH  H 2 O + NaCl HCl is the acid, it donates H + to OH - (the base)

3. B-L Acids and Formation of H 3 O + 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 H 3 O + from an acid in water HBr + H 2 O  H 3 O + + Br -

4. 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 + H 2 O  H 3 O + + Cl - Acid = HCl and Base = H 2 O (HCl gives H 2 O an H + ) NH 3 + H 2 O  NH 4 + + OH - Acid = H 2 O and Base = NH 3 (H 2 O gives NH 3 an H + ) CH 3 CO 2 H + NH 3  CH 3 CO 2 - + NH 4 + Acid = CH 3 CO 2 H and Base = NH 3 (CH 3 CO 2 H gives NH 3 an H + )

5. 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

6. Identifying Conjugate Acid/Base Pairs 1.Identify the acid and base for the reactants 2.Identify the acid and base for the products 3.Identify the conjugate acid/base pairs acid conjugate base + + base conjugate acid HF H2OH2O H3O+H3O+ F -

7. Acid and Base Strength Strong acids give up protons easily and completely ionize in water: HCl + H 2 O  H 3 O + + Cl - Weak acids give up protons less easily and only partially ionize in water: CH 3 CO 2 H + H 2 O  CH 3 CO 2 - + H 3 O + Strong bases have a strong attraction for H + and completely ionize in water: KOH(s)  K + (aq) + OH - (aq) NaNH 2 + H 2 O  NH 3 + NaOH Weak bases have a weak attraction for H + and only partially ionize in water: HS - + H 2 O  H 2 S + OH -

8. 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 + H 2 O ? H 3 O + + Br - stronger acid (HBr)  weaker acid (H 3 O + ) (equilibrium favors products) Example 2:NH 3 + H 2 O ? NH 4 + + OH - weaker acid (H 2 O)  stronger acid (NH 4 + ) (equilibrium favors reactants)

9. Dissociation Constants Since weak acids dissociate reversibly in water, we can write an equilibrium expression: HA + H 2 O  H 3 O + + A - K eq = [H 3 O + ][A - ]/[HA][H 2 O] But, since [H 2 O] remains essentially constant we can write: K a = K eq x [H 2 O] = [H 3 O + ][A - ]/[HA] The acid dissociation constant (K a ) is a measure of how much the acid dissociates (A higher K a = a stronger acid) Example:CH 3 CO 2 H + H 2 O  CH 3 CO 2 - + H 3 O + K a = [H 3 O + ][CH 3 CO 2 - ]/[CH 3 CO 2 H] = 1.8 x 10 -5 Can also write dissociation constants for weak bases: NH 3 + H 2 O  NH 4 + + OH - K b = [NH 4 + ][OH - ]/[NH 3 ] = 1.8 x 10 -5

10. Ionization of Water Since H 2 O can act as either a weak acid or a weak base, one H 2 O can transfer a proton to another H 2 O: H 2 O + H 2 O  H 3 O + + OH - K eq = [H 3 O + ][OH - ]/[H 2 O][H 2 O] Since [H 2 O] is essentially constant, we can write: K w = K eq x [H 2 O] 2 = [H 3 O + ][OH - ] (where K w = the ion-product constant for water) For pure water: [H 3 O + ] = [OH - ] = 1.0 x 10 -7 M So, K w = [H 3 O + ][OH - ] = (1.0 x 10 -7 M) 2 = 1.0 x 10 -14 (units are omitted for K w as for K eq and K a )

11. Using K w If acid is added to water, [H 3 O + ] goes up - for an acidic solution [H 3 O + ] > [OH - ] If base is added to water, [OH - ] goes up - for a basic solution [OH - ] > [H 3 O + ] K w is constant (1.0 x 10 -14 ) for all aqueous solutions Can use K w to calculate either [H 3 O + ] or [OH - ] if given the other concentration Example: if [H 3 O + ] = 1.0 x 10 -4 M, what is the [OH - ]? K w = [H 3 O + ][OH - ] [OH - ] = K w / [H 3 O + ] = 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 [H 3 O + ] > [OH - ], it’s an acidic solution

13. The pH Scale pH is a way to express [H 3 O + ] in numbers that are easy to work with [H 3 O + ] has a large range (1.0 M to 1.0 x 10 -14 M) so we use a log scale: pH = - log [H 3 O + ] The pH scale goes from 0 - 14 Each pH unit = a ten-fold change in [H 3 O + ] pH 7 = neutral, 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

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

16. 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 H 2 gas, for example: Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) Acids react with carbonates and bicarbonates to produce salts, H 2 O and CO 2 gas, for example: NaHCO 3 (aq) + HCl(aq)  NaCl(aq) + H 2 O(l) + CO 2 (g) Acids react with bases (neutralization reactions) to form salts and H 2 O, for example: HBr(aq) + LiOH(aq)  LiBr (aq) + H 2 O(l) Neutralization reactions are balanced with respect to moles of H + and moles of OH -, for example: H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O

17. 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 H 2 O, or accept one from H 2 O, the pH will change: Na 2 S  2Na + + S 2- S 2- + H 2 O  HS - + OH - S 2- is a weak base that can accept an H + from H 2 O Since [OH - ] is increased, the solution is basic

18. 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 H 3 O + 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 + H 2 O) K + has no proton to donate, so can’t form H 3 O + Br - is too weak of a base to pull a proton off of H 2 O, so can’t form OH - So, the solution remains neutral

19. 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 H 2 O 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 - + H 2 O  HCN + OH - (HCN + NaOH  NaCN + H 2 O) (Na + doesn’t affect the pH, it’s from a strong base)

20. 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 H 2 O to form H 3 O + So, salts that contain ions from a weak base and a strong acid form acidic solutions Example: NH 4 Br  NH 4 + + Br - (from NH 3 and HBr) (NH 3 + HBr  NH 4 Br) NH 4 + + H 2 O  NH 3 + H 3 O + (Br - doesn’t affect the pH)

22. 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 H 2 O

24. How to Make a Buffer Solution An acetate buffer is made from acetic acid and a salt of its conjugate base: CH 3 CO 2 HandCH 3 CO 2 Na The salt is used to increase the concentration of CH 3 CO 2 - in the buffer solution Recall: CH 3 CO 2 H + H 2 O  CH 3 CO 2 - + H 3 O + (the equilibrium favors reactants, so the concentration of CH 3 CO 2 - is low) But, CH 3 CO 2 Na  CH 3 CO 2 - + Na + CH 3 CO 2 H + CH 3 CO 2 Na + H 2 O  2 CH 3 CO 2 - + H 3 O + + Na + If acid is added: CH 3 CO 2 - + H 3 O +  CH 3 CO 2 H + H 2 O If base is added: CH 3 CO 2 H + OH -  CH 3 CO 2 - + H 2 O Buffer capacity = how much acid or base can be added and still maintain pH (depends on buffer type and concentration)

26. Calculating pH of a Buffer The pH of a buffer solution can be calculated from the acid dissociation constant (K a ) Example (for acetate buffer): CH 3 CO 2 H + H 2 O  CH 3 CO 2 - + H 3 O + K a = [H 3 O + ][CH 3 CO 2 - ]/[CH 3 CO 2 H] = 1.8 x 10 -5 [H 3 O + ] = K a x [CH 3 CO 2 H]/[CH 3 CO 2 - ] What is the pH of an acetate buffer that is 1.0 M CH 3 CO 2 H and 0.50 M CH 3 CO 2 Na? [H 3 O + ] = 1.8 x 10 -5 x 1.0 M/0.50 M = 3.6 x 10 -5 M pH = - log[H 3 O + ] = - log(3.6 x 10 -5 ) = 4.44

27. 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, C 1 V 1 = C 2 V 2 For molarity, it becomes:M 1 V 1 = M 2 V 2

29. 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? M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 / M 1 V 1 = 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 ? V 2 = M 1 V 1 / M 2 V 2 = 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

30. Acid-Base Titration Molarity of an acid or base solution of unknown concentration can be determined by titration: 1.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 2.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 [H 3 O + ] = [OH - ] 3.Molarity of unknown is calculated from moles of titrant added (mole ratio comes from balanced chemical equation)

32. Example: Titration of H 2 SO 4 with NaOH What is the molarity of a 10.0 mL sample of H 2 SO 4 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: H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O Calculate moles H 2 SO 4 neutralized: 0.0150 mol NaOH x 1 mol H 2 SO 4 / 2 mol NaOH = 0.00750 mol H 2 SO 4 Calculate molarity of H 2 SO 4 : 0.00750 mol H 2 SO 4 / 0.0100 L = 0.750 M H 2 SO 4