1. Drawing titration curves

2. 20.0 mL of 0.0750 mol L –1 HCOOH is titrated against 0.100 mol L –1 NaOH. Draw the titration curve for this titration up to a total of 25.0 mL of NaOH added. pK a (HCOOH) = 3.74 Before we start graphing, there are specific points on the graph that we need to calculate. the pH before any NaOH is added the volume of NaOH at the equivalence point the volume of NaOH when pH = pK a the pH at the equivalence point the pH after 25.0 mL of NaOH is added.

3. 1 Find the pH of 0.0750 mol L –1 HCOOH, pK a = 3.74. HCOOH(aq) + H 2 O(l) HCOO – (aq) + H 3 O + (aq) Assume 1 [HCOO – ] = [H 3 O + ] 2 [HCOOH] = 0.0750 mol L –1

5. 2 Calculate the volume of NaOH at the endpoint. Titration reaction is: HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H 2 O(l)

6. 3 Calculate the volume of NaOH when pH = pK a. The volume of NaOH at the equivalence point is 15.0 mL. 15.0 ÷ 2 = 7.5 mL The pH = pK a when 7.5 mL of NaOH have been added.

7. 4 Calculate the pH at the equivalence point. At the equivalence point we will have 1.50 mol L –1 of HCOONa in (20.0 mL + 15.0 mL = 35.0 mL) of solution. We need to calculate the pH of a solution of HCOONa with this concentration.

8. HCOO – (aq) + H 2 O HCOOH(aq) + OH – (aq) Assume 1 [HCOOH] = [OH – ] 2 [HCOO – ] = 0.0429 mol L –1

10. 5 Calculate the pH after 25 mL of NaOH has been added. Since the equivalence point is at 15.0 mL of NaOH, this results in an excess of 10.0 mL of NaOH. Although the HCOONa formed hydrolyses slightly in water, the [OH – ] from this reaction is very small compared to the [OH – ] from the NaOH. So we will assume all the OH – comes from the NaOH. Total volume of solution = 20.0 mL + 25.0 mL = 45.0 mL

11. NaOH is a strong base. Assume [OH – ] = c(NaOH) = 0.0222 mol L –1

12. Now we have the key data points: the pH before any NaOH is added = 2.43 the volume of NaOH at the equivalence point = 15.0 mL the volume of NaOH when pH = pK a = 7.5 mL the pH at the equivalence point = 8.19 the pH after 25.0 mL of NaOH is added. = 12.3

13. * Plot these points on the graph: (0.0 mL, 2.43) (7.5 mL, 3.74) (15.0 mL, 8.19) (25.0 mL, 12.3) * * *

14. * * * * Draw a vertical line centred on the equivalence point, about 3 pH units long. Draw a line, 7.5 mL wide, centred on the pH = pKa point. Tilt it so it has a slope of 1 pH unit. Draw a short horizontal line at the end of the titration. Join these lines with curves. The very start will be quite steep.

15. Note: sometimes you are asked to sketch a titration curve and told that further pH calculations are not required. (Or, you’ve run out of time to do the calculations.) In these cases, estimate the pH of the various points. Most weak acid solutions you will deal with have a pH of about 3 and weak base solutions will be about 11. A weak acid-strong base titration usually has an equivalence point between pH = 8 and 9. For weak base–strong acid titrations it will be between 5 and 6. The final pH is about half a pH point below the pH of the base added (or half a pH point above the pH of the strong acid added).