Understanding Acid-Base Titrations: Experimentation and Analysis

Introduction

Acid-base titrations are used to determine the unknown concentrations of basic or acidic substances through acid base reactions1. A reaction between an acid and a base are known as neutralization reactions, which result in the formation of water and an ionic compound (salt)1. By observing the net ionic equation in Figure 1, it is evident that the essential feature of the neutralization process in a solution, is when hydrogen ions combine with hydrogen ions for water to be formed.

The analyte is the solution with an unknown molarity1.

The reagent is the solution that has a known molarity that reacts with the analyte1. The substance being studied is dissolved into the solution in order to prepare the analyte1. For a titration, a small amount of indicator is then added into the flask along with the analyte1. To titrate, one solution of a known concentration is slowly added into a (known) volume of another solution of unknown concentration, until neutralization is reached2.

To know when a reaction has neutralized, there will be an indication of color change2.

In this specific lab, Phenolphthalein was the indicator used to help indicate the end point. An end point is the moment at which the color change can finally be observed. The equivalence point is when the [H+] = [OH-]. The half-equivalence point is the required volume needed to reach the exact halfway mark to the equivalence point. At the half-way equivalence point, this is where the pH and pKa are equal to each other.

Titrations are extremely relevant to everyday life.

For example, titrations are important in laboratory medicine to help find the unknown concentrations of chemicals in urine and blood3. Titrations are also utilized by pharmacists to help develop new pharaceuticals3. Water content, fat content and concentrations of vitamins can also be determined using titrations3.

For the first week of experimentation, an acid-base reaction was performed to determine the concentration of a sodium hydroxide (NaOH) solution through a standardization. The standardization was executed by being titrated against the unknown acid #1. The acid in the experiment was potassium hydrogen phthalate (KHC8H4O4) and the mass was measured for each trial. The average concentration of the NaOH found in week one was 0.08244 M. The data collected was precise because there were four trials performed, and the right number of significant figures were recorded when finding the mass of the KHP and the volume of the solution in the burette.

𝐻𝐶8𝐻4𝑂4(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) →𝐻2𝑂(𝑙) + 𝐾𝑁𝑎𝐶8𝐻4𝑂4 (𝑎𝑞) (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛1)

Reaction 1 supported that KHP had the same number of moles as the NaOH did, and this supports the claim that the moles of NaOH will be equal to the number of moles of KHP at the equivalence point. By using this knowledge and by understanding that molarity is calculated by taking the number of moles divided by the volume (L), the concentration of NaOH was calculated as follows;

In week two of experimentation, the goal was two-fold. The concentration of an unknown weak acid #1 was determined by measuring the volume of standardized NaOH that was required to neutralize the unknown acid that was present. The ionization constant, Ka, of the weak acid was also determined by creating a titration curve in excel.

To find the ionization constant, Ka:

It was hypothesized that the average concentration of NaOH would be about 0.1 M since the 1 L of H2O had 10 mL of the 10 M NaOH solution added to it. It was hypothesized that the average ionization constant, Ka, can be calculated by using the anti-log of the pKa value from the half-equivalence point (which is the same value as the pH at this point) of both trial’s titration curves. The Ka value would be significantly small because weak acids were utilized as the unknowns in the lab.

Experimental Methods

Reference the Principles of Chemistry II Laboratory Manual Emmanuel College 2019, “Titration of Acids and Bases” for the materials that were used and the for the procedures that were followed. During the week one experiment, there were four trials performed instead of three trials. After each reading, the burette was filled back up to 0 mL (as seen in Table 1). There were 25 mL of the unknown solution #1 that was obtained before the experiment. There were no other changes made to the procedure (Emmanuel College, 2019).

Results/Data

Trial 1 2 3 4

H2O 1 L

10 M NaOH Stock Solution 10.0 mL

Mass of KHP (g) 0.5132 0.5134 0.5136 0.5138

Initial Burette Reading (mL) 0 0 0 0

Final Burette Reading (mL) 24.81 32.19 33.21 32.74

Volume of NaOH Used (mL) 24.81 32.19 33.21 32.74

Volume of NaOH Used (L) 0.02481 0.03219 0.03321 0.03274

Concentration of NaOH (M) 0.1013 0.07812 0.07574 0.07458

Average Concentration of NaOH: 0.0824M

Calculations for Table 1:

Volume of NaOH Used 1 mL*1L/(1000 mL)

Sample 1:

= (final buret reading) – (initial buret reading) = (24.81 – 0) mL = 24.81mL = 0.02481 L

Sample 2

= (final buret reading) – (initial buret reading) = (32.19– 0) mL = 32.19 mL = 0.03219 L

Sample 3

= (final buret reading) – (initial buret reading) = (32.21 – 0) mL = 32.21 mL = 0.03221 L

Sample 4

= (final buret reading) – (initial buret reading) = (32.74 – 0) mL = 32.74 mL = 0.03274 L

Concentration of NaOH for Table 1:

Sample 1:

= mass of KHP*(1 molKHP)/(molar mass KHP)*(1 molNaOH)/(1 molKHP)*1/(volume of NaOH used)

= 0.5132 g of KHP*(1 molKHP)/(204.2 g KHP)*(1 molNaOH)/(1 molKHP)*1/(0.02481 L of NaOH used)

= 0.101298 ~ 0.1013 M NaOH

Sample 2:

= mass of KHP*(1 molKHP)/(molar mass KHP)*(1 molNaOH)/(1 molKHP)*1/(volume of NaOH used)

= 0.5134 g of KHP*(1 molKHP)/(204.2 g KHP)*(1 molNaOH)/(1 molKHP)*1/(0.03219 L of NaOH used)

= 0.078105 ~ 0.07812 M NaOH

Sample 3:

= mass of KHP*(1 molKHP)/(molar mass KHP)*(1 molNaOH)/(1 molKHP)*1/(volume of NaOH used)

= 0.5136 g of KHP*(1 molKHP)/(204.2 g KHP)*(1 molNaOH)/(1 molKHP)*1/(0.03321 L of NaOH used)

= 0.0757356 ~ 0.07574 M NaOH

Sample 4:

= mass of KHP*(1 molKHP)/(molar mass KHP)*(1 molNaOH)/(1 molKHP)*1/(volume of NaOH used)

= 0.5138 g of KHP*(1 molKHP)/(204.2 g KHP)*(1 molNaOH)/(1 molKHP)*1/(0.0274 L of NaOH used)

= 0.0745750 ~ 0.07458 M NaOH

Average Concentration of NaOH

(0.1013+0.07812+0.07574+0.07458)/4 = 0.08244 M = 8.24 x 10-2 M

Figure 1 – For the first trial, the pH of the NaOH solution was measured every mL until the end point was reached. The endpoint was at 21.15 mL and had a pH value of 9.41. A titration curve for trial 1 was generated to display the equivalence point and the half-equivalence point. The half-equivalence point was where the pH equaled the pKa. The pKa found for trial 1 was 4.01, and this helped determine the Ka and molarity.

Figure 2 – For the second trial, the pH of the NaOH solution was measured every mL until the end point was reached. The endpoint was at 21.01 mL and had a pH value of 9.72. A titration curve for trial 2 was generated to display the equivalence point and the half-equivalence point. The half-equivalence point was where the pH equaled the pKa. The pKa found for trial 2 was 4.12, and this helped determine the Ka and molarity.

Trial

1 2

Equivalence Point Volume (mL) 17.02

17.40

Half-Equivalence Point Volume (mL)

8.51

8.70

pH = pKa

at the Half- Equivalence Point 4.01 4.12

Trial 1 2

pH = pKa

4.01 4.12

Ka

9.77 x 10-5 7.59 x 10-5

Avg Ka

8.68 x 10-5

Vol (mL) of Base at Eq Point

17.02 17.40

Concentration of NaOH (M = mol/L )

0.0836M

Volume of Acid #1

25 mL = 0.025 L

Molarity (M) 5.69 x 10-2 5.82 x 10-2

Avg Concentration of Acid #1 = 5.76 x 10-2 M

Solving for Ka:

Trial 1:

= pKa = -logKa

= 4.01 = -logKa

= -4.01 = logKa

= 10-4.01 =Ka

Ka = 9.77 x 10-5

Trial 2:

= pKa = -logKa

= 4.12 = -logKa

= -4.12 = logKa

= 10-4.12 =Ka

Ka = 7.59 x 10-5

Average Ka:

= ((9.77*〖10〗^(-5 ))+(7.59*〖10〗^(-5)))/2 = 8.68 x 10-5

Molarity

Trial 1:

= 0.01702 L NaOH*(0.0824 mol)/(1 L) NaOH*(1 mol acid)/(1 molNaOH)*1/(0.025 Lacid)

= 0.05609 ~ 5.61 x 10-2 M acid

Trial 2:

= 0.01740 L NaOH*(0.0824 mol)/(1 L) NaOH*(1 mol acid)/(1 molNaOH)*1/(0.025 Lacid)

= 0.05735 ~ 5.74 x 10-2 M acid

Average Concentration of the Acid

((5.61x10^(-2)+5.74x10^(-2)))/2 = 5.68 x 10-2 M acid

Discussion & Conclusion

The purpose of this lab was to determine the concentration of a NaOH solution through a standardization and to find the pH and the ionization constant, Ka, of the unknown weak acid #1. This experiment was successful because it was found that the average concentration of the NaOH solution was 8.24 x 10-2 M, the Ka was 8.68 x 10-5, and the average concentration of the unknown acid was 5.68 x 10-2 M.

The hypothesis predicted that the concentration of the NaOH would be about 0.1 M, and this hypothesis was proven correct because according to Table 1, the average concentration of NaOH was 0.0824M, which is close to 0.1 M. The hypothesis was also proven correctly because the titration curves for trial 1 & 2 as seen in Figure 1 & 2, accurately displayed the half-equivalence point and equivalence point which helped determined the pKa (pKa was 4.01 and 4.12).

There was minimal error in this experiment, but there may have been minor systematic or random errors that may have occurred. For example, there may have been a random error when reading the burette. The meniscus may have not been observed properly, or the incorrect number of significant figures may have been noted. These random errors can greatly affect the precision of the results, so to avoid these errors in the future it is important to read the meniscus correctly at eye level and report the burette measurements with four significant figures to ensure precise results.

As for systematic error, there may have been and error in the visual detection/observation of when the solution reached its end point. The observable end point is at a volume that is larger than the equivalence point, therefore it has a higher result than the true value of the concentration. Another possible systematic error could have been that the scale for measuring the KHP could have read too high or too low or the calibration was off. Systematic error is important to avoid because these types of errors affect the accuracy of the results. To avoid these errors, it’s important to perform multiple trials to ensure the endpoint is being observed accurately and to check to that the scales and tools are calibrated and function properly.

Sources

1. Libretexts. Acid-Base Titrations. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Demos,_Techniques,_and_Experiments/General_Lab_Techniques/Titration/Acid-Base_Titrations (accessed Dec 9, 2019).
2. Libretexts. Titration. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Demos,_Techniques,_and_Experiments/General_Lab_Techniques/Titration (accessed Dec 9, 2019).
3. Bandos, G. Titration. http://chemteacher.chemeddl.org/joomla/index.php?option=com_content&view=article&id=56 (accessed Dec 9, 2019).