AP Chemistry · Free-response skill

Finding Ka and concentration from a titration

The move. Use titration data to find the equivalence volume, the concentration of an unknown weak acid and its pKa, and explain the shape of the curve with the species present.

How to do it

  1. Find the equivalence point in the middle of the steepest rise in pH; read its volume from the data.
  2. At equivalence, moles of base = moles of a monoprotic acid. Divide those moles by the sample volume in liters for the concentration, and keep the significant figures of the data.
  3. Use the equivalence volume to find the point where the acid is half used up, read the pH there, and say what the mixture holds at that point.
  4. To explain the shape of a stretch of the curve, say which species are in the flask there and what happens to each drop of added base.

Practice prompt Try it

A student dissolves an unknown monoprotic weak acid, HA, in water to make 25.00 mL of solution. She titrates it with 0.1000 M NaOH, recording the pH with a calibrated pH meter.

pH during the titration of 25.00 mL of HA with 0.1000 M NaOH

Volume of NaOH added (mL)pH
0.002.71
5.003.83
10.004.26
12.004.40
15.004.62
20.005.10
23.005.76
23.806.48
24.008.55
24.2010.61
25.0011.30
30.0012.04

Work through parts a to e with the student’s data, showing each step of your arithmetic.

  1. The equivalence point falls on one of the volumes in the table. Which volume is it, and what change in the pH readings around it tells you so?
  2. How many moles of HA were in the sample, and what was its molar concentration?
  3. At what volume of NaOH has exactly half of the HA reacted? Read the pH there and give the pKa\text{p}K_a of HA.
  4. Tell why the pH at that half-way volume is equal to the pKa\text{p}K_a.
  5. Between 5.00 mL and 20.00 mL of NaOH, the pH rises by only about 1.3 units. Say what happens, as a balanced equation, to the hydroxide ions added in this stretch.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 8 points

Model answer, annotated

(a) The equivalence point is at 24.00 mL of NaOH. Around it the pH jumps by more than 4 units over only 0.40 mL, from 6.48 at 23.80 mL to 10.61 at 24.20 mL.

  • Point 1 Names the volume at the middle of the jump.
  • Point 2 Backs the choice with readings on both sides of it.

(b) Moles of NaOH at equivalence = (0.02400 L)(0.1000 M)=2.400×10−3(0.02400 \text{ L})(0.1000 \text{ M}) = 2.400 \times 10^{-3} mol, which equals the moles of HA (1:1). Concentration: 2.400×10−3 mol/0.02500 L=0.09602.400 \times 10^{-3} \text{ mol} / 0.02500 \text{ L} = 0.0960 M.

  • Point 3 Uses the 1:1 ratio of NaOH to HA at equivalence.
  • Point 4 Divides by the 25.00 mL sample volume in liters.

(c) Half of the HA has reacted at half the equivalence volume, 12.00 mL. The pH there is 4.40, so pKa=4.40\text{p}K_a = 4.40.

  • Point 5 Halves 24.00 mL to get the half-way volume.
  • Point 6 Reads the pH at 12.00 mL from the table.

(d) At 12.00 mL, half of the HA has become A−^-, so [HA] = [A−^-]. Then pH=pKa+log⁡1=pKa\text{pH} = \text{p}K_a + \log 1 = \text{p}K_a.

  • Point 7 States that the acid and its conjugate base are equal and why that makes the pH equal the pKa.

(e) Each portion of added hydroxide reacts with HA: HA+OH−→A−+H2O\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}. The hydroxide is used up and only the ratio of A−^- to HA changes, so the pH rises slowly.

  • Point 8 Gives the reaction that uses up the added base.