Determining the Equivalent Mass and Dissociation Constant of an Unknown Weak Acid by Titrimetry Essay

DETERMINING THE EQUIVALENT MASS AND DISSOCIATION CONSTANT OF AN UNKNOWN WEAK ACID BY TITRIMETRY INTRODUCTION Acids are substances that donate hydrogen ions and bases are substances that accept hydrogen ions. Acids and bases react with each other by transferring hydrogen ions. One way to distinguish an acid is by its equivalent mass, which is the number of grams of the acid needed to transfer one mole of hydrogen ion to a base. For a monoprotic acid, which only transfers one hydrogen ion, its equivalent mass equals its molar mass.

For a diprotic acid, which transfers two hydrogen ions, its equivalent mass equals half its molar mass.

We will write a custom sample essay on
Determining the Equivalent Mass and Dissociation Constant of an Unknown Weak Acid by Titrimetry
specifically for you for only $13.9/page
Order now

The equivalent mass of a base is simply the number of grams required to accept one mole of hydrogen ion. The equivalent mass of an acid or base is also equal to the mass of the acid or base titrated divided by the number of equivalents of the acid or base. Acid strength is measured by how much it dissociates.

This is determined by the amount of hydronium ion formed in water. Most often, it is expressed by pH, which equals the –log[H3O+]. The equilibrium constant for an acid is represented by Ka = [H3O+] [A-] / [HA], where HA is the acid and A- is the dissociated acid. When comparing Ka’s, most commonly pKa is used, which is equal to –log Ka.

In this experiment, the pKa of an unknown acid is determined by titrating it with NaOH and graphing its pH levels versus volume of NaOH titrated. The inflection point found by graphing is the equivalence point, and at half that volume is the half-equivalence point. At half equivalence, the [A-] = [HA], so they cancel out in the equation Ka = [H3O+] [A-] / [HA], leaving Ka = [H3O+]. With pH being known, Ka is found by [H3O+] equaling 10-pH, and pKa = pH. EXPERIMENTAL First, about 2 g of an unknown weak acid was obtained. Then about 0. 10 to 0. 12 g of the acid was weighed out on an analytical scale. This mass was recorded to 4 decimal places.

The acid was then transferred to a clean 125 mL Erlenmeyer flask. Distilled water was added to dissolve the acid, and then 3 drops of phenolphthalein indicator was added to the solution. Next, a buret was filled with NaOH solution and its initial reading was recorded to 2 decimal places. The acid solution was titrated with the NaOH solution until the acid solution turned and stayed pink. This final reading was recorded to 2 decimal places and the volume of NaOH used was calculated. Another sample of unknown acid was weighed out using an analytical scale requiring about 25 mL of standardized NaOH solution.

This mass was recorded to 4 decimal places. The acid was transferred into a clean 150 mL beaker, and was again dissolved using distilled water. A pH meter and electrode was then obtained. The electrode was rinsed with distilled water and put into the acid solution. Standardized NaOH solution was again used to titrate the acid. The initial buret reading was recorded to 2 decimal places and NaOH was added 0. 5 mL at a time. After every portion added, the buret reading and pH meter reading were recorded to 2 decimal places. This was done until the pH spiked and stopped increasing significantly.

RESULTS Identification code of weak unknown acidT Mass of unknown acid, g0. 1167 Final buret reading, mL14. 37 Initial buret reading, mL0. 75 Volume of NaOH solution required, mL13. 62 Concentration of NaOH solution used, M1. 00 x 10-1 Mass of unknown acid requiring 25 mL of NaOH solution for titration, g0. 2142 Mass of unknown acid used, g0. 2152 Buret Reading, mLpH 0. 502. 95 0. 993. 07 1. 503. 22 2. 003. 38 2. 493. 50 3. 003. 60 3. 493. 70 4. 003. 78 4. 503. 86 4. 993. 92 5. 493. 99 6. 014. 04 6. 504. 09 6. 994. 14 7. 504. 19 7. 994. 23 8. 504. 28 8. 984. 32 9. 504. 36 10. 004. 40 10. 84. 43 11. 004. 47 11. 484. 51 12. 004. 55 12. 504. 59 13. 004. 62 13. 494. 66 13. 984. 69 14. 494. 73 15. 004. 77 15. 504. 80 15. 994. 84 16. 504. 88 17. 004. 92 17. 494. 96 17. 995. 00 18. 505. 05 19. 005. 09 19. 505. 14 19. 995. 19 20. 485. 23 20. 995. 30 21. 505. 36 22. 005. 43 22. 505. 50 23. 005. 59 23. 505. 70 24. 005. 83 24. 506. 02 25. 006. 30 25. 506. 88 25. 9910. 18 26. 5010. 73 27. 0010. 98 27. 4811. 13 27. 9911. 23 28. 5011. 32 28. 9911. 38 29. 5011. 44 30. 0011. 49 30. 5011. 53 31. 0011. 57 32. 0211. 63 33. 0111. 68 Equivalence Point, mL25. 75 Half-Equivalence Point, mL12. 8 Half-Equivalence pH4. 62 Ka2. 40 x 10-5 pKa4. 62 pKa of trans-crotonic4. 69 Percent Error1. 49% Number of equivalents of NaOH, equiv. 2. 58 x 10-3 Number of equivalents of acid titrated, equiv. 2. 58 x 10-3 Equivalent mass of acid, g/equiv. 83. 41 Equivalent mass of trans-crotonic, g/equiv. 86. 09 Percent Error3. 11% CALCULATIONS In order to find the volume of NaOH used, I subtracted the initial buret reading from the final reading (14. 37 mL – 0. 75 mL = 13. 62 mL). To find the mass of unknown acid needed using 25 mL of NaOH solution, I used a proportion. The proportion was 0. 167 g / 13. 62 mL = x / 25. 00 mL, with x equaling the mass of unknown acid needed. To find the half-equivalence point, I divided the equivalence point volume by 2. I then just looked at the graph to find the pH at the half-equivalence point. The pKa is equal to the pH, which was 4. 62. The Ka is equal to the concentration of [H3O+]. [H3O+] is equal to 10-pH, so using my numbers, 10-4. 62 = 2. 40 x 10-5. To find percent error, I took the theoretical pKa – my found pKa, divided by the theoretical pKa­, and multiplied by 100%. Using my numbers, (4. 69 – 4. 62) / 4. 69 x 100% = 1. 49%.

In order to find the number of equivalents of NaOH, I took the volume at the equivalence point (25. 75 x 10-3 L) and multiplied by the molarity (1. 00 x 10-1 M). The number of equivalents of acid titrated was equal to the equivalents of NaOH. Finally, to find the equivalent mass of the acid, I divided the mass of the acid used by the number of equivalents of acid. Using my numbers, 0. 2152 g / 2. 58 x 10-3 equiv. = 83. 41 g/equiv. DISCUSSION/CONCLUSIONS Looking at the table in the lab book, the acid with the closest pKa to my calculated pKa is trans-crotonic. It has a pKa of 4. 69 while my calculated pKa was 4. 2. I feel that these numbers are very close, as the percent error is only 1. 49%. The percent error for my calculated equivalent mass of the acid is also small at 3. 11%. Error could have come from reading the graph, as I had to estimate the equivalence point. I should have taken measurements at smaller intervals when approaching the inflection point. This would have allowed me to read the graph more accurately. Error could have also come from the pH meter, which may not have been calibrated precisely. Lastly, error could have resulted if I inaccurately read the buret while recording the readings.

Haven’t Found A Paper?

Let us create the best one for you! What is your topic?

By clicking "SEND", you agree to our terms of service and privacy policy. We'll occasionally send you account related and promo emails.

Eric from Graduateway Hi there, would you like to get an essay? What is your topic? Let me help you

logo

Haven't found the Essay You Want?

Get your custom essay sample

For Only $13.90/page