1. Chemistry 21-30 Chem Olympiad Mini Quiz Buffer Notes Pancake-Ice Cream Sandwich Treat (throughout the class period)

2. AP Chemistry Study Session #1 02.23.13

3. 1. Acetic acid is a weak acid with a K a = 1.8 x 10 -5. (a) Write the dissociation reaction of acetic acid in water. CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + (b) Write the equilibrium expression for this reaction. K a = [CH 3 COO - ][ H 3 O + ] / [CH 3 COOH] (c) What is the equation to solve for pH? pH = -log [H 3 O + ]

4. (d) If you prepared 20.mL of a 0.10M solution of acetic acid, what is the pH? CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + [I] 0.10 0 0 [C] -x +x +x [E] 0.10-x x x K a = [CH 3 COO - ][ H 3 O + ] / [CH 3 COOH] 1.8 x 10 -5 = x 2 / 0.10x = 0.0013M pH = -log (0.0013) pH = 2.87

5. 2. Sodium hydroxide is a strong base. (a) Write an equation for this neutralization reaction. NaOH + CH 3 COOH  CH 3 COONa + H 2 O (b) Write it as a net ionic equation. OH - + CH 3 COOH  CH 3 COO - + H 2 O (c) If equal moles of weak acid and strong base were reacted the resulting pH would be basic. How many moles of NaOH must be added to reach this so called equivalence point? 0.020L of 0.10M AA is 0.0020 moles AA, therefore we need 0.0020moles of NaOH. If the concentration of our base is 0.20M, then we would need 10.mL of the base to react the equivalence point.

6. How much acetate ion would be produced? 0.0020moles acetate ion What is the concentration of the acetate ion? 0.0020 moles acetate ion in 0.030L total solution equals 0.067M AI. Is the acetate ion and acidic or basic ion? Since it pulls H + away from water in solution, it creates more OH - in the solution therefore the acetate ion is a basic ion.

7. What is the equation for the acetate ion in water? CH 3 COO - + H 2 O  CH 3 COOH + OH - What is the equilibrium constant expression for this reaction? K b = [CH 3 COOH][ OH - ]/[ CH 3 COO - ] Determine the [OH - ]. CH 3 COO - + H 2 O  CH 3 COOH + OH - [OH - ] = 6.1 x 10 -6 M

8. Determine the [OH-]. 5.6 x 10 -10 = x 2 / 0.067x = [OH-] = 6.1 x 10 -6 M What is the pOH? pOH = 5.21 What is the pH? pH = 8.79

9. 3. If you added 0.0010 moles NaOH, we would have excess acetic acid and the solution would be acidic. When you add half the required base needed to neutralize all the acid, we have reached what is called the half- equivalence point. (a) How much acetic acid remains at the half-equivalence point? There are 0.0010 moles of acetic acid in 0.025L (20mL AA + 5mL NaOH) total or 0.040M AA remaining, with some acetate (common ion) as well.

10. (b) How much common ion remains? We have 0.0010 moles in the 0.025L with a concentration of 0.040M as well. We have an equal amount of HA and A- at the half-equivalence point. (c) What is the pH now? HA + H2O  A- + H3O+ 0.040 0.040 0 -x +x +x 0.040 – x0.040 + x x 1.8 x 10 -5 = 0.040x / 0.040 pH = 4.74

11. (d) What is the pK a of the acid? pK a = 4.74 So, pH = pK a when the concentrations of the acid and its conjugate base are the same…is there an equation for this? pH = pKa + log[A-]/[HA] When [A-] = [HA] we have log 1 and log 1 = 0 so pH = pKa! This is called the Henderson-Hasselbalch equation and it is used for BUFFER solutions.

12. A buffer solution is a solution of a weak acid and its conjugate base that allows only small changes in the pH when small amounts of strong bases (like NaOH) are added. Also a buffer is a solution of a weak base and its conjugate acid that allows only small changes in the pH when small amounts of strong acids (like HCl) are added. In general, a buffer tries to keep the pH from changing.

13. 4. So, let’s add a little more base to our solution above. Let’s add 1.0 mL more of NaOH (or 0.00020 moles). This means that 0.00020moles of AA acid react and 0.00020moles of AI are produced. We now have 0.00080 moles of AA remaining and we have 0.0012 moles of AI in the solution. (a) What are the concentrations of AA and AI now? [AA] = 0.00080 moles / 0.0260L = 0.031M [AI] = 0.0012 moles / 0.0260L = 0.046M (b) Use HH to determine pH of this solution. Henderson says… pH = pKa + log[A-]/[HA] pH = 4.74 + log[0.046]/[0.026] = 4.99 So, adding a little base does not change its pH too much.

14. (c) What is the pH if we add another 1.0mL of NaOH? [AA] = 0.00060 moles / 0.0270L = 0.022M [AI] = 0.0014 moles / 0.0270L = 0.052M Henderson says… pH = pKa + log[A-]/[HA] pH = 4.74 + log[0.052]/[0.022] = 5.11 It will continue to creep up until we get to the equivalence point where the pH jumps to 8.79 almost at once…i.e. it jumps fast at the end…think about doing a titration and what 1 drop of base does…BOOM! Clear to bright pink!

15. 5. So, what happens on the other side of the equivalence point? If the equivalence point was adding 0.0020moles of base and the half- equivalence point was adding 0.0010moles of base…let’s see where we are if we add 0.0030moles of base. All of the acid reacts and we have 0.0010moles of OH- remaining and 0.0020moles of the AI. The amount of OH contributed by the weak base, AI, is insignificant in comparison to the amount of OH- the remaining NaOH will contribute so we will negate the AI.

16. (a) What is the pH of this remaining solution? For us to add 0.0030 moles of NaOH, we would have added 15mL of the base. This makes the total volume of solution 40mL of 0.040L for a concentration of 0.0010mole / 0.040L or 0.075M NaOH making the pH = 12.40.

17. (b) What is the pH if we added 12.5mL of the base instead of the 15mL? We get a pH = 12.12 So, we see that the slope is small when we add these amounts after the equivalence point.

18. 2.8 7 4.7 4 8.7 9 Half Equivalence Point 1010 5 1515 12.4 0 12.1 2