1 / 21

Determining Ka for a Monoprotic Weak Acid through pH Measurements

This experiment aims to calculate the acid dissociation constant (Ka) for a monoprotic weak acid using pH measurements from solutions prepared with varying amounts of NaOH. By measuring the pH and determining the concentrations of both the weak acid (HA) and its conjugate base (A−), we establish a relationship that allows us to determine Ka. The experiment involves calculating the concentration of H3O+ from pH values and using stoichiometric principles to analyze molar concentrations, leading to the determination of the weak acid’s dissociation constant.

linda-hicks
Télécharger la présentation

Determining Ka for a Monoprotic Weak Acid through pH Measurements

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Equilibrium constant, Ka By Soo Jeon

  2. background • Most weak acids dissociate 5% or less, that is, 95% or more of the acid remains as HA. The smaller the value of Ka, the weaker the acid.

  3. Background .. • Monoprotic acids are those acids that are able to donate one proton per molecule during the process of dissociation(sometimes called ionization) as shown below (symbolized by HA): HA(aq)  +  H2O(l)  ->     H3O+(aq)  +  A-(aq)

  4. scenario • A student prepared three solutions by adding 12.00, 18.5, and 35.00ml of 7.5x10-2M NaOH solution to 50.00ml of a 0.100M solution of a weak monoprotic acid, HA. The solutions were labeled X,Y,and Z respectively. Each of the solutions were diluted to a total volume of 100.00ml with distilled water. The pH readings of these solutions, obtained through the use of a calibrated pH meter, were X,Y, Z. We are going to determine Ka for a monoprotic weak acid.

  5. Purpose • This experiment will determine Ka for a monoprotic weak acid using pH value and A-ion concentration.

  6. Materials • 7.5x10-2M NaOH solution • 0.100M solution of a weak monoprotic acid • distilled water • calibrated pH meter • 100 ml of Three beakers

  7. Using the calibrated pH meter, find pH of each solution. X : 6.50 Y : 6.70 Z : 7.10 pH of each solution

  8. Calculations • Convert pH to equivalent H30+ concentration. 1) pH = -log [H+] [H30+] = 10-pH • X : 10-6.50 = 3.16 x 10-7M H30+ • Y : 10-6.70 = 2.00 x 10-7M H30+ • Z : 10-7.10 = 7.94 x 10-8M H30+

  9. 2. Calculate the number of moles of acid added • The moles of acid was constant for each solution. 50.00ml x 1L x 0.100 moles HA 1 1000ml 1 L = 5.00 x 10-3 mol HA

  10. 3. Calculate the number of moles of OH- added • X : 12.00ml x 1L x 7.50 x 10-2 mol OH- 1 1000ml 1 L = 9.00 x 10-4 mol OH- • Y : 18.50ml x 1L x 7.50 x 10-2mol OH- 1 1000ml 1L = 1.39 x 10-3 mol OH- • Z : 35.00ml x 1L x 7.50 x 10-2mol OH- 1 1000ml 1L = 2.69 x 10-3 mol OH-

  11. 4. Determine the concentration of HA and A- ion X : [HA] = initial moles HA – moles OH- added volume of solution = 5.00 x 10-3 - 9.00 x 10-4 0.100 L = 4.10 x 10-2 Molarity Moles A- formed = moles OH- added [ A-] = 9.00 x 10-4 mol A- = 9.00 x 10-3Molarity 0.100L

  12. Y : [HA] = 5.00 x 10-3 – 1.39 x 10-3 0.1000L = 3.61 x 10-2 Molarity [A-] = 1.39 x 10-3 mol A- 0.1000L = 1.39 x 10-2 Molarity

  13. Z : [HA] = 5.00 x 10-3 - 2.63 x 10-3 0.10000L = 2.37 x 10-2 Molarity [A-] = 2.63 x 10-3 mol A- 0.10000L = 2.63 x 10-2 Molarity

  14. 5. Reciprocal of A- ion concentration • To show that the concentration is increasing, we need to flip the A- ion concentration. • X : 1/(9.00 x 10-3) = 1.11 x 10-2 M-1 Y : 1/(1.39 x 10-2) = 7.19 x 101 M-1 Z : 1/(2.63 x 10-2) = 3.80 x 101 M-1

  15. 6. Reciprocal of A- ion concentration against H3O+ ion concentration • If it is not reciprocal of A- ion concentration, then the slope (Ka) would be negative value.

  16. 7. Determine the slope of the line • Slope = ΔY = Δ[1/A-] ΔX = Δ[H30+] =1.11 x 102 – 3.80 x 101= 3.08 x 108 (3.16 x 10-7) – (7.94 x 10-8)

  17. 8. Determine the initial concentration of the acid, HA. = 5.00 x 10-3 mol HA = 5.00 x 10-2 M 0.10000L

  18. 9. Determine the Ka for the weak acid. 1 = Δ[H30+] = (H30+) · (A-) Slope Δ[1/A-] Ka = 1 = 1___________ HA x slope (5.00 x 10-2)·(3.08 x 101) = 6.49 x 10-8 M

  19. Conclusion • In this lab, I measured pH values of each solution and then determined the concentrations of HA and A- ion. I calculated the slope of the reciprocal of A- ion concentration on the ordinate against the H3O+ ion concentration and got the Ka for the weak acid, 6.49 x 10-8 M.

  20. Citation • http://answers.yahoo.com/question/index?qid=20090201045717AAjsY1g • http://wb7.itrademarket.com/pdimage/15/400615_cyberscanpc300.jpg

  21. Image citations • Slide #6 : http://www.micglobal.co.uk/images/beakers_14000BGYR.jpg • Slide # 7 : http://wb6.itrademarket.com/pdimage/15/400615_cyberscanpc300.jpg • Slide # 8 :http://www.flickr.com/photos/inthe_aeroplane/526874125/

More Related