1 / 17

Lab 24 - Hydrolysis

Lab 24 - Hydrolysis. A salt formed between a strong acid and a weak base is an acid salt . Ammonia is a weak base, and its salt with any strong acid gives a solution with a pH lower than 7. For example, let us consider the reaction. H 2 O + NH 3 NH 4 + + OH -.

hovan
Download Presentation

Lab 24 - Hydrolysis

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. Lab 24 - Hydrolysis A salt formed between a strong acid and a weak base is an acid salt. Ammonia is a weak base, and its salt with any strong acid gives a solution with a pH lower than 7. For example, let us consider the reaction • H2O + NH3 NH4+ + OH-

  2. In the solution, the NH3+ion reacts with water (called hydrolysis) according to the equation: NH3++ H2O NH4+ OH-. • The acidity constant can be derived from Kw and Kb. • Ammonia Kb=1.75e-5 • KwKa = ---------------- Kb    = 1.00e-14 / 1.75e-5 = 5.7e-10.

  3. What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8. • step 1 write the net ionic formula • Na + + CLO- + H2O HCLO + Na + OH- • step 2 determine initial, change and equilibrium [M] of reactants and products

  4. What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8. • step 3 Find the Kb value KwKb= ---------------- ------ Ka 1.0e-143.3e-7.= ---------------- ------3.0e-8.

  5. What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8. • step 3 Find the M of [OH}           [HCLO] [OH-]Kb= ---------------- ------ [CLO] [X2] 3.3e-7= ---------------- [0.1] = 1.8 e -4

  6. What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8. • step 4 Convert [OH] to pOH -log 1.8 e -4 = 3.74 step 5 Convert pOHto pH 14 -3.74 = 10.26

  7. A. Hydrolysis of salts 1. Add 5ml of Di water into 6 test tubes 2. Add 3 drops of indicator (as listed on page 270) to each test tube 3. Using the indicators chart on page 262 determine the pH of the water 4. Repeat steps 1-3 using boiled water 5. Repeat steps 1-3 using .1M solutions of the salts listed on page 269

  8. A. Data 1. Fill out chart on pg 269 based on the hydrolysis behavior on page 260 2. Determine (H +) for each pH 3. Determine (OH -) for each pH 4. Using data from pg 269fill out chart on page 271

  9. calculations • [H+] = 10-pH ( anti log of –pH) • [OH-] = Kw / [H+] • [OH-] = [1.0 x 10-14] / [H+] • Ka or Kb = [M] Products over [M] reactants • Omit water and assume both products have the same molarity

  10. Lab 25 – dissociation of a weak acid General Theory According to the Brønsted-Lowry acid-base theory, the strength of an acid is related to its ability to donate protons. All acid-base reactions are then competitions between bases of various strengths for these protons. For example, the strong acid HCl reacts with water according to Equation [1]: HCl(aq) + H2O(l)H3O+(aq) + Cl-(aq) General

  11. Lab 25 • This is a strong acid and is completely dissociated (in other words, 100 percent dissociated) in dilute aqueous solution. Consequently, the [H3O+] concentration of a 0.1 MHCl solution is 0.1 M. Thus HCl is a stronger acid than water and completely donates a proton to water to form H3O+.

  12. This is a strong acid and is completely dissociated (in other words, 100 percent dissociated) in dilute aqueous solution. Consequently, the [H3O+] concentration of a 0.1 MHCl solution is 0.1 M. Thus HCl is a stronger acid than water and completely donates a proton to water to form H3O+.

  13. By contrast, acetic acid, HC2H3O2 (abbreviated HOAc), is a weak acid and is only slightly dissociated, as shown in Equation [2]: • Its acid dissociation constant, as shown by Equation [3], is therefore small • Ka= = 1.8 x 10-5

  14. titrate the weak acid • If we titrate the weak acid HA with a base, there will be a point in the titration at which the number of equivalents of base is just one-half the number of equivalents of acid present (at ½ Ve). • pH = pKa@ ½ [H+] • By titrating a weak acid with a strong base and recording the pH versus the volume of base added, one can determine the dissociation constant, Ka, of the weak acid

  15. . C. Determination of pKa of Unknown • Pipeta 25.00 mL aliquot of your unknown acid solution into a 250-mL beaker and carefully immerse the previously rinsed electrode into this solution. • Measure the pH of this solution. Record the pH in your notebook. Begin your titration by adding 1 mL of your standardized base from a buret and record the volume of titrant and pH. • Repeat with successive additions of 1 mL of base until you approach the end point. • equivalence point--where the graph is steepest • See pg 277 • Add 0.1-mL increments of base and record the pH and milliliters of NaOH added untill the pH no longer changes • plot graph to determine ½ equivalence point-

  16. Formulas • pH = pKa @ ½ equivalence point- • ½ equivalence point = 4.3 • pKa = 4.3 • Ka = antilog - 4.3 = 5 e-5

  17. Due • Pg 264,265 • Questions 1-3

More Related