1. 16.3-16.4 Buffer Effectiveness, Titrations, and pH Curves

2. 16.3: Buffer Effectiveness: Buffer Range and Buffer Capacity an effective buffer neutralizes small to moderate amounts of added acid or base in other words, the whole point of a buffer is to have enough acid and conjugate base to consume all of the added ions and not change the pH *buffers can be destroyed if you add too much acid or too much base Factors that influence buffer effectiveness relative amounts of acid and its conjugate base (or base to its conjugate acid) absolute concentrations of acid and conjugate base capacity of a buffer= How much added acid or base the buffer can effectively neutralize range of a buffer= The pH range over which a particular acid and its conjugate base can be effective buffers only effective if they keep within 1 pH unit of change

3. Relative Amounts of Acid and Base *Buffers are most effective when the concentrations of the acid and its conjugate base are equal ex.1.) Given a generic buffer composed of HA (acid) and A - (conjugate base) for which pK a = 5.00, calculate the percent change in pH upon addition of 0.010 mol of NaOH for two different 1.0 L solutions of this buffer system. Both solutions have 0.20 mol of total acid and conjugate base. Solution I has an initial pH of 5.00 and equal amounts of acid and conjugate base (0.10 mol of each). Solution II has an initial pH of 4.05 and has much more acid than conjugate base (0.18 mol HA and 0.020 mol A - )

4. Example 1 continued: Solution I 0.10 mol HA; 0.10 mol A - ; initial pH= 5.00 OH - (aq) + HA (aq) → H 2 O (l) + A - (aq) [OH - ] M[HA] M[A - ] M Before Addition≈0.00 mol1.00 mol Addition0.010 mol-0.010 mol+0.010 mol After Addition≈0.00 mol0.090 mol0.110 mol pH = pK a + log ([base] / [acid]) =5.00 + log (0.110 / 0.4090) = 5.09 % change in pH = ((5.09 - 5.00) / 5.00) x 100 = 1.8 % NaOH is a base because it is a group 1 hydroxide NaOH → Na + + OH - ionizes completely in water, releasing a large amounts of OH - ions BUT, in buffered solution with a base, the excess OH - ions combine with the cations present in the buffer, therefore lessening the effect of the added base.

5. Example 1 continued: Solution II 0.18 mol HA; 0.020 mol A - ; initial pH= 4.05 OH - (aq) + HA (aq) → H 2 O (l) + A - (aq) [OH - ] M[HA] M[A - ] M Before Addition≈0.00 mol0.18 mol0.020 mol Addition0.010 mol-0.010 mol+0.010 mol After Addition≈0.00 mol0.17 mol0.030 mol pH = pK a + log ([base] / [acid]) =5.00 + log (0.030 / 0.17) = 4.25 % change in pH = ((4.25 - 4.05) / 4.05) x 100 = 5.0 %

6. Example 1 continued: Answer/Significance: Solution I is a much better buffer because it resist pH change more effectively (has a lower % of change). This example proves that buffers are most effective when the concentrations of the acid and its conjugate base are equal. Guideline to Evaluate the Effectiveness of a Buffer The concentration of base to concentration of acid ratio should range from 0.10 to 10.0. Same as a difference no greater than 1 pH value *This is why pH is so useful!! [base] / [acid] = 0.10 → 10.0

7. Example 2 Given the same acids as in example 1, which acid would you use to create a buffer with the pH = 7.35? HClO pH = pK a + log ([base] / [acid]) 7.35 = 7.54 + log ([base] / [acid]) 10 -.19 = [base] / [acid] [base] / [acid] = 0.645654229

8. Example 3 Calculate the ratio of NaF to HF required to create a buffer with a pH= 4. (K a = 3.5 x 10 -4 ) pH = pK a + log ([base] / [acid]) 4 = -log (3.5 x 10 -4 ) + log ([base] / [acid]) 4 + log (3.5 x 10 -4 ) = log ([base] / [acid]) 10 (4 + log (3.5 x 10^ -4) = ([base] / [acid]) ([base] / [acid]) = 3.5

9. Absolute Concentrations of the Acid and Conjugate Base *Buffers are most effective when concentrations of acid and conjugate base are high ex.1.) Given a generic buffer composed of HA and A - for which pK a = 5.00, calculate the percent change in pH upon addition of 0.010 mol of NaOH for two different 1.0 L solutions of this buffer system. In solution I, the acid and the base are 10 times more concentrated than the acid and base of solution II. Both solutions have equal relative amounts of acid and conjugate base and therefore have the same initial pH of 5.00.

10. Example 1 continued: Solution I 0.50 mol HA; 0.50 mol A - ; initial pH= 5.00 OH - (aq) + HA (aq) → H 2 O (l) + A - (aq) [OH - ] M[HA] M[A - ] M Before Addition≈0.00 mol0.50 mol Addition0.010 mol-0.010 mol+0.010 mol After Addition≈0.00 mol0.49 mol0.51 mol pH = pK a + log ([base] / [acid]) =5.00 + log (0.51 / 0.49) = 5.02 % change in pH = ((5.02 - 5.00) / 5.00) x 100 = 0.4%

11. Example 1 continued: Solution II 0.050 mol HA; 0.050 mol A - ; initial pH= 5.00 OH - (aq) + HA (aq) → H 2 O (l) + A - (aq) [OH - ] M[HA] M[A - ] M Before Addition≈0.00 mol0.050 mol Addition0.010 mol-0.010 mol+0.010 mol After Addition≈0.00 mol0.040 mol0.060 mol pH = pK a + log ([base] / [acid]) =5.00 + log (0.060 / 0.040) = 5.18 % change in pH = ((5.18 - 5.00) / 5.00) x 100 = 3.6 %

12. Example 1 continued: Answer/Significance: Solution I is a much better buffer because it resist pH change more effectively (has a lower % of change). This example proves that buffers are most effective when concentrations of acid and conjugate base are high.

13. Example 1 ex.1.) Which acid would you choose to combine with its sodium salt to make a solution buffered at pH 4.25? For the best choice, calculate the ratio of the conjugate base to the acid required to attain the desired pH. a.) chlorous acid (HClO 2 )pK a = 1.95 b.) nitrous acid (HCHO 2 )pK a = 3.34 c.) formic acid (HCHO 2 )pK a = 3.74 d.) hypochlorous acid (HClO)pK a = 7.54

14. Buffer Range recall: concentration of acid to concentration of base ratio should not differ by greater than a factor of 10 (1 pH level). To determine the effectiveness of a buffer, find the effective pH range. Hasselbalch equation allows you to find outermost points of effective range *the effective range for a buffer is one pH unit on either side of pK a Highest pH for effective buffer (upper limit): when base is 10 times more concentrated than acid pH = pK a + log ([base] / [acid]) = pK a + log10 = pK a + 1 Lowest pH for effective buffer (lower limit): when base is 1/10 as concentrated as acid pH = pK a + log ([base] / [acid]) = pK a + log 0.10 = pK a - 1

15. Example 1 continued: How would you solve this? pH = pK a + log ([base] / [acid])set up Hasselbalch equation 4.25 = 1.95 + log ([base] / [acid])plug in the values you have (pH, pK a ) log ([base] / [acid]) = 4.25 -1.95isolate the ratio and simplify log ([base] / [acid]) = 2.30 10 2.30 = ([base] / [acid])solve the log 199.5262315 = ([base] / [acid])Analyze the results conclusion: This is NOT right! 200 is nowhere near a pH of 4.25. This means that ratio of [base] to [acid] is increased by a factor larger than 10.

16. Example 1 continued: Correct answer: C. formic acid pH = pK a + log ([base] / [acid]) 4.25 = 3.74 + log ([base] / [acid]) log ([base] / [acid]) = 4.25 - 3.74 log ([base] / [acid]) = 0.51 10 0.51 = ([base] / [acid]) 3.24 = ([base] / [acid]) explanation: This value is within the effective range which is one pH unit on either side or different by a factor no greater than 10.

17. Example 2 Determine whether or not the mixing of each pair of solutions results in a buffer. 1.100.0 mL of 0.10 M NH 3 ; 100.0 mL pH 0.15 M NH 4 Cl 1.50.0 mL of 0.10 M HCl; 35.0 mL of 0.15 M NaOH 1.50.0 mL of 0.15 M HF; 20.0 mL of 0.15 M NaOH 1.175.0 mL of 0.10 M NH 3 ; 150.0 mL of 0.12 M NaOH 1.125.0 mL of 0.15 M NH 3 ; 150.0 mL of 0.20 M NaOH

18. Example 2 Determine whether or not the mixing of each pair of solutions results in a buffer. 1.100.0 mL of 0.10 M NH 3 ; 100.0 mL pH 0.15 M NH 4 Cl YES: weak base and its conjugate acid 1.50.0 mL of 0.10 M HCl; 35.0 mL of 0.15 M NaOH NO: strong acid cannot be part of buffer 1.50.0 mL of 0.15 M HF; 20.0 mL of 0.15 M NaOH YES: 1.175.0 mL of 0.10 M NH 3 ; 150.0 mL of 0.12 M NaOH NO: two bases 1.125.0 mL of 0.15 M NH 3 ; 150.0 mL of 0.20 M NaOH NO: two bases

19. 3.) explained OH - HFF-F- before addition3.07.50 addition-3.0 + 3.0 after addition04.53.0 0.15 M = (x / 0.05 L) x= 0.0075 mol HF 0.02 L NaOH x (0.15 mol NaOH / 1 L NaOH) x (1 mol OH - / 1 mol NaOH) = 0.003 mol OH - OH - + HF → H 2 O + F - Conclusion: Because the acid and its conjugate base are both present in this mixture, the buffer capacity has not been exceeded. This means that this mixture creates a buffer!

20. Buffer Capacity buffer capacity= amount of acid or base that you can add to a buffer without causing a large change in the pH absolute concentrations of buffer components ↑, buffer capacity ↑, the more concentrated the weak acid and conjugate base are that compose the buffer, the higher the capacity is relative concentrations of buffer components become more similar to each other, buffer capacity ↑ as the ratio of buffer components gets closer to 1, the overall capacity of the buffer becomes greater (the ability to neutralize added acid and added base) in some cases, a buffer that must neutralize a primarily added acid or base may be overweighted in one of the buffer components this can be applied to the buffer system in human blood (pg. 786)

21. 16.4: Titrations and pH Curves acid base titration= a basic or acidic solution of unknown concentration reacts with an acidic or basic solution of low concentration slowly add known solution to unknown while monitoring the pH pH can be measured by a pH meter or an indicator indicator= a substance that changes color based on pH the acid and base neutralize each other as the combine until the equivalence point is reached equivalence point= the point in a titration when the number of moles of base is stoichiometrically equal to number of moles of acid *indicates the end of a titration neither reactant is in excess

22. Titrations continued: titration curve/pH curve= a plot of the pH of the solution during a titration

23. Titration Curves Continued: ex. acid titrated with a base before you add any base, pH low as you slowly add the base, acid becomes less acidic (neutralization) when you reach middle of curve, the equivalence point, pH changes quickly if using a pH meter, you will see the rapid increase of pH value if using an indicator, the color of the solution with change quickly if you continue to add base after you reach the equivalence point, the solution will be basic Why? you already neutralized all of the acid present and now your are adding excessive base to the solution https://youtu.be/g8jdCWC10vQ

24. The Titration of a Strong Acid with a Strong Base ex.1.) Titrate 25 mL of 0.100 M HCl with 0.100 M NaOH. 1.Find Volume of NaOH Required to Reach the Equivalence Point during titration, added NaOH neutralizes HCl HCl (aq) + NaOH (aq) → H 2 O (l) + NaCl (aq) eq. point reached when moles of added base = initial moles of acid calculate amount of acid initially in soln: 0.0250 L x (0.100 mol / 1 L) = 0.00250 mol HCl initially now that we know that there were 0.00250 mol HCl initially, calculate volume of NaOH required: 0.00250 mol x (I L / 0.100 mol) = 0.0250 L = 2.50 mL NaOH added

25. 2. Find Initial pH (before adding any base) Because HCl is a strong acid, the [H 3 O + ] = [HCl] [H 3 O + ] = [HCl][HCl] = 0.100 M[H 3 O + ] = 0.100 M pH = -log[H 3 O + ] pH = -log(0.100) pH = 1.00 3. Find pH (after adding 5.00 mL NaOH) as you add NaOH, it neutralizes the H 3 O + OH - (aq) + H 3 O + (aq) → 2H 2 O (l) calculate the amount of H 3 O + at any given point before the equivalence point (refer to original equation: HCl (aq) + NaOH (aq) → H 2 O (l) + NaCl (aq)) *already have this step doe 1 mol NaOH neutralizes 1 mol H 3 O + [H 3 O + ] initial = 0.00250 mol calculate the number of moles of NaOH added at 5.00 mL: 0.005 L x (0.100 mol / 1 L) = 0.005 mol NaOH added

26. the addition of OH - causes amount H + to decrease Calculate hydronium concentration and then pH [H 3 O + ] = 0.002 mol H 3 O + / (0.025 L + 0.005 L) [H 3 O + ]= 0.0667 M pH = -log(0.0667) pH= 1.18 *You can repeat this step over and over with different volumes up to the equivalence point of added NaOH to form a complete titration curve (more work on this shown on pg. 771) At the eq. point, the pH will always be 7 because the strong acid is completely neutralized. Water is the only thing left that contributes to the ionization, and its concentration is negligible as we know. OH - H3O+H3O+ Before Addition≈0 mol0.0025 mol Addition0.0005 mol-0.0005 mol After Addition≈0 mol0.002 mol

27. 4. Find pH after equivalence point (after adding 30.00 mL NaOH) because NaOH is being added after the equivalence point, it is excessive (“excessive reagent”) Calculate the number of moles of OH - added at 30.00 mL initial H 3 O + = 0.0025 mol 0.0300 L x (0.100 mol / 1 L) = 0.003 mol OH - added Calculate OH - concentration by dividing by total volume [OH - ] = 0.0005 mol / (0.025 L + 0.03 L)[OH - ] = 0.00909 M Calc [H 3 O + ] and pH [H 3 O + ] [OH - ]= 10 -14 OH - H3O+H3O+ Before Addition≈0 mol0.0025 mol Addition0.003 mol-0.003 mol After Addition≈0.0005 mol≈0 mol

28. [H 3 O + ] = (10 -14 / [OH - ]) = (10 -14 / 0.00909) = 1.10 x 10 -12 M pH = -log (1.10 x 10 -12 ) pH = 11.96 *continue for other NaOH volumes after the equivalence point

29. Example 2 Graphs a and b show titration curves for two equal-volume samples of monoprotic acids, one weak and one strong. Both titrations were carried out with the same concentration of strong base.

30. Example 2 explained Graphs a and b show titration curves for two equal-volume samples of monoprotic acids, one weak and one strong. Both titrations were carried out with the same concentration of strong base i.a.) pH= 8 i.b.) pH= 7 ii.a.) Weak Acid: because the initial pH is higher, pH rapid rise is shorter, and the eq. point is higher ii.b.) Strong Acid: because the initial pH is lower, pH rapid rise is longer, and the eq. point is lower

31. College Board Any 6C learning objective pertains to acid base chemistry: pg. 108 http://media.collegeboard.com/digitalServices/pdf/ap/IN120085263_ChemistryCED_Effective_Fall_2013_l kd.pdf these are the most important to this presentation: