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Acid-base titrations: results

titration curve of sodium carbonate

Figure 2: titration curve of sodium carbonate against HCl

Figure 2 shows the titration curve of sodium carbonate with HCl. There are two abrupt pH changes in the curve. These correspond to the following successive reactions:

Na2CO3 + HCl -> NaHCO3 + NaCl    (conversion of carbonate into bicarbonate)

NaHCO3 + HCl -> CO2 + H2O + NaCl

The calculated molarity of HCl in this experiment is 0.95 mol/l.

titration curve of NaOH

Figure 3: titration curve of NaOH against HCl

Figure 3 shows the titration curve of the reaction:

NaOH + HCl -> NaCl + H2O

This reaction involves strong acid (HCl) and strong base (NaOH). You can notice how the pH changes from a very high to very low pH value. In such reactions, the pH at the equivalence point is 7. Move the cursor on the screen and see that the steepest trace occurs at pH 7.

The calculated molarity of NaOH in this experiment is 1.01 mol/l.

titration curve of baking powder

Figure 4: titration curve of baking powder against HCl

Figure 4 shows the pH changes during the titration of baking powder with HCl. In contrast to the carbonate experiment, we can see here only one abrupt pH step, which corresponds to the conversion of bicarbonate into carbon dioxide. It is interesting to note that Figure 4 is similar to the second portion only of Figure 2.

titration curve of Vinegar

Figure 5: titration curve of Vinegar against NaOH

Figure 5 shows the titration curve of vinegar against sodium hydroxide. Note that the pH of the solution increases during the titration due to the addition of NaOH.

The reaction involved is:

CH3COOH + NaOH -> CH3COO-Na+ + H2O

The calculated concentration of acetic acid in vinegar in this experiment is 5.85 % (w/w)

titration curve of HCl

Figure 6: titration curve of HCl against NaOH

Figure 6 is the opposite of Figure 3 where HCl is being titrated with NaOH. It is clear that the pH jump is larger in the case of titration of strong acids (e.g. HCl) than that in the titration of weak acids (e.g. acetic) with an alkali.


  1. Explain why the pH at equivalence point in Figures 3 and 6 is 7 whereas in Figure 5 is 8.86
  2. Calculate the pKa of acetic acid from Figure 5.
  3. Predict the titration curve if you titrate a mixture of 0.1 mol/l sodium carbonate and 0.1 mol/l sodium bicarbonate with HCl.
  4. What would happen if you did not calibrate the flow rate?
The first derivative of figure 2

Figure 7: the first derivative of figure 2, for more accurate end point location.

Further study

  1. Repeat part 3 with the concentrations of HCl and NaOH reduced 10 and 100 times. Study the effect of concentration on the pH change.
  2. Repeat part 6 with H3PO4 instead of HCl. (You should see a rising curve with two pH jumps corresponding to the first two hydrogen atoms of phosphoric acid)
  3. Plot the first derivative (dpH/dt) against time of all the previous experiments and locate the end point in each case. The end point in the first derivative curves are defined as the point at the maximum value. The first derivative curves can be obtained by graphing programs such as MicroCal Origin. (A sample is shown in Figure 7.)

Teachers’ notes

Target groups

Primarily, College students (General Chemistry, Introduction to Analytical Chemistry and Instrumental Analysis Courses) Partially, High school students.

Required knowledge

  1. Expression of solution concentration (e.g. Molarity and percentage concentration).
  2. Acid-Base equilibria.
  3. Basics of volumetric analysis.

Safety tips

Avoid contact of all chemicals with eyes or skin.


  1. Figure 3 and 6 show titrations involving a strong acid and a strong base. The solution at the equivalence point contains their salt (NaCl) which is neutral, i.e., pH 7. Whereas, Figure 5 shows the titration of a weak acid (acetic) against strong base (NaOH). At this equivalence point, the solution contains their salt (sodium acetate) which is a basic salt the pH of which is > 7 , 8.86 in this case.
  2. pKa of acetic acid can be calculated by determining the pH at half neutralization. At this point pH = pKa (theoretical value = 4.74, the experimental value 4.6).
  3. The second step will be as twice as the first step.
  4. You can still get the titration curve but you cannot tell the volume of the titrant required to reach the equivalence point, and of course cannot calculate the unknown concentration.