1. Strong base neutralizes weak acid Strong acid neutralizes weak base

2. How do we calculate the pH of a buffer ? 2 ingredient problem 0.10 M CH 3 COOH + 0.20 M CH 3 COO - pH = 5.05

3. Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] Remember pX = -logX so pK a = -log K a Henderson-Hasselbalch equation pH = pK a + log [conjugate base] [acid] Take log Assumption: Changes in [A - ] & [HA] will be negligible (within 5%) if K a < 0.01 & K a < 0.01 [base] [acid] use initial conc. of acid and base to calculate pH

4. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = 3.77 + log [0.52] [0.30] = 4.01 16.2 Mixture of weak acid and conjugate base!

5. A buffer solution is a solution of: 1.A weak acid or a weak base and 2.The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. 16.3 Add strong acid H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa

6. Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution 16.3

7. Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 K a = 5.6 x 10 -10 => pOH = 14 - pH = 4.75 [OH - ] = 1.8 x 10 -5 M

8. = 9.20 Calculate moles of all components. Remember we have 80.0 mL of the buffer solution @ 0.30 M NH 3 /0.36 M NH 4 Cl and 20.0 mL of 0.050 M NaOH. NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) 0.0290.0010 0.024 0.0280.00.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH 3 ] = 0.025 0.10 Moles NH 4 + = (0.080 L)(0.36 M) = 0.029 moles Moles NH 3 = (0.080 L)(0.30 M) = 0.024 moles Moles NaOH = (0.020 L)(0.050 M) = 0.0010 moles Moles/Volume(L) ΔpH = 0.03

9. Maintaining the pH of Blood 16.3

10. How do we build a better buffer? Add approximately equal quantities of acid and base Have relatively high concentrations of acid and base => the larger the [acid] & [base] the greater the buffer capacity How do we prepare a buffer at a given pH? Choose acid/base conjugate pair from table Check to be sure that they are unreactive in the system used pK a  pH typical rule of thumb pH + 1 = pK a

11. Create buffer with pH = 7.50. pH + 1 = pK a Look for pK a ‘s 6.5 => 8.5 => K a ‘s 3 x 10 -7 => 3 x 10 -7 HOCl / OCl K a = 3.5 x 10 -8 H 2 PO 4 - / HPO 4 -2 K a2 = 6.2 x 10 -8 H 2 AsO 4 - / HAsO 4 -2 K a2 = 8 x 10 -8 H 2 CO 3 / HCO 3 - K a1 = 4.3 x 10 -7 Calculate quantities of acid and base. pH = pK a + log [conjugate base] [acid]

12. H 2 CO 3 / HCO 3 - K a1 = 4.3 x 10 -7 pK a1 = 6.37 pH - pK a = log [HCO 3 - ] [H 2 CO 3 ] 7.50 - 6.37 = 1.13 10 -1.13 = 13.6 = [HCO 3 - ] [H 2 CO 3 ] Set [H 2 CO 3 ] = 0.0100 M (NOTE: This is a judgement call.) then [HCO 3 - ] = 13.6 [H 2 CO 3 ] = 0.136 M