Sodium Hydroxide Dilution and pH Analysis Using McVc = MdVd Formula

Objective

To learn about the principles of dilution, understand different concentration units, and practice converting. We aim to determine the molar concentration of dilute sodium hydroxide solution by measuring pH.

Abstract

In the present work, three successive dilutions were carried out, each of which pH was measured. The results showed that the second and third dilutions were identical in measuring the concentration of the starting substance, unlike the first dilution, for which the pH was probably mismeasured.

Introduction

Will work with solutions, which are highly concentrated solutions of common reagents. In Reaction 1 of the copper cycle, a 6M nitric acid solution was prepared forma 15M nitric acid solution through dilution. Dilution is a fundamental concept in chemistry, and it is used to change the concentration of a solution. The change is calculated using the formula M_cV_c = M_dV_d, where:

• M_c is the concentration of the concentrated solution.
• V_c is the volume of the concentrated solution.
• M_d is the molarity of the dilute solution.
• V_d is the volume of the dilute solution.

Concentration can be determined using various methods. If a solution is colored, we can measure its absorbance at a specific wavelength using Beer’s law. Absorbance (A) is related to the molar concentration (C) of the sample by the equation A = abc, where:

• a is the molar absorptivity coefficient.
• b is the path length in centimeters.
• c is the molar concentration (M) of the sample.

Alternatively, pH can be used to determine the concentration of ????⁺ ions in a solution. Finally, we can use chemical reactions via titration to determine concentration.

Experimental Procedure

Preparing Dilute Sodium Hydroxide Solutions

1. Record the molarity of the concentration sodium hydroxide solution.
2. Calculate the amount of stock solution needed to create 25mL of 0.10M sodium hydroxide.
3. Determine the amount of water required to reach 25mL 0.10M sodium hydroxide.
4. Measure the concentrated sodium hydroxide and water with the appropriate graduated cylinders.
5. Mix the sodium hydroxide and water, and measure the pH (target pH ranger: 12.85-13.05)
6. If the pH falls within the range, transfer 20mL to a clean, dry beaker and dispose of the remainder properly.
7. With the guidance of your instructor, use glass pipettes to create 200mL of a 1.0mM solution recording the pH.
8. Transfer the 1.0mM solution to a clean, dry beaker.
9. Rinse the glass pipettes and graduated cylinders, then use them to prepare 20mL of a 10μ???? solution appropriately.

Preparing Dilute Potassium Permanganate Solution

This part included the preparation of dilute potassium permanganate solution from concentrated stock solution at different concentration variations, but this part was not done in the present laboratory work.

Results and Calculations

First Dilution

The results of the experiments performed are summarized in Table 1. To calculate the required number of ml for the sodium hydroxide stock solution, the original formula was used, viz:

M_cV_c = M_dV_d

(1.0M)(V_c) = (0.10M)(25mL)

V_c = (0.10M)(25mL) / (1.0M)

V_c = 2.5 mL

So, it took this much water to prepare 25mL of solution:

V_w = V_d – V_c = 25 mL – 2.50 mL = 22.50 mL

Second Dilution

The second dilution required calculating the amount of 0.10M solution needed to make 20 mL of 1mM solution:

M_cV_c = M_dV_d

(0.10M)(V_c) = (0.001M)(20mL)

V_c = (0.001M)(20mL) / (0.10M)

V_c = 0.20mL

Then, the amount of water:

V_w = V_d – V_c = 20 mL – 0.20 mL = 19.80 mL

Third Dilution

Finally, the third dilution required determining the amount of 1mM solution needed to produce 20 mL of 1µM solution:

M_cV_c = M_dV_d

(1mM)(V_c) = (10µM)(20mL)

V_c = (10µM)(20mL) / (1mM)

V_c = 0.20 mL

Then, the amount of water:

V_w = V_d – V_c = 20 mL – 0.20 mL = 19.80 mL

Table 1 – Experimental Findings.

Discussion

The concentration of sodium hydroxide could be calculated through the measured pH values for each of the three dilutions, viz.:

1. [NaOH] = [OH^–] = 10^{– (14 – pH)} = 10^{– (14 – 12.99)} = 10^{– 1.01} = 0.098M
2. [NaOH] = [OH^–] = 10^{– (14 – 11.20)} = 10^{– 2.8} = 0.002M
3. [NaOH] = [OH^–] = 10^{– (14 – 8.20)} = 10^{– 5.8} = 0.000002M

Once the concentrations have been calculated based on the pH algorithm, the dilution formula can be used again to determine the concentration of the stock solution:

1. [NaOH]_stock = 0.098M ∙ 10 = 0.98M
2. [NaOH]_stock = 0.002M ∙ 10³ = 1.58M
3. [NaOH]_stock = 0.000002M ∙ 10⁶ = 1.58M

As follows from the calculations, the second and third dilutions gave identical results, in contrast to the first dilution. The reason could be that the pH measurement for the first dilution was inaccurate and therefore the calculated concentration was not true.

Some sources of error in this work could include incorrect pH readings, for example due to a contaminated probe, and incorrect measurement of the amounts of alkali and water taken for the dilutions. These can be minimized through practicing laboratory skills and increasing the number of repetitions to reduce the error.

Conclusion

The present work was aimed at investigating the dilution technique of solutions which is numerically based on the rule McVc = MdVd. Three consecutive dilutions were performed and the last two of them showed identical results in calculating the effluent concentration. In addition, it was shown that the higher the dilution factor, and thus the lower the alkali concentration, the lower the pH of the final solution.

References

“Experiment 8. Dilution” in Chemistry 113.1 Introduction to Chemical Techniques Laboratory Manual, C. M. Evans, F. H. Watson, and G. L. Findley (Queens College, New York, 2012).