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 H 2 O + NH 3 NH 4 + + OH -

2.  In the solution, the NH 3 + ion reacts with water (called hydrolysis) according to the equation: NH 3 + + H 2 O NH 4 + OH -.  The acidity constant can be derived from K w and K b. Ammonia K b =1.75e-5  K w K a = ---------------- K b = 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 - + H 2 O HCLO + Na + OH -   step 2 determine initial, change and equilibrium [M] of reactants and products CLO - HCLOOH - Initial.1 M00 ChangexM equilibrium.1M -xXM

4.  What is the pH 0f a.1M NaCLO solution if Ka for HCLO 1s 3.0e-8.  step 3 Find the Kb value Kw K b = ---------------- ------ Ka 1.0e-14 3.3e-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 - ] K b = ---------------- ------ [CLO] [X 2 ] 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 pOH to 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) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) General

11.  This is a strong acid and is completely dissociated (in other words, 100 percent dissociated) in dilute aqueous solution. Consequently, the [H 3 O + ] concentration of a 0.1 M HCl solution is 0.1 M. Thus HCl is a stronger acid than water and completely donates a proton to water to form H 3 O +. Lab 25

12.  This is a strong acid and is completely dissociated (in other words, 100 percent dissociated) in dilute aqueous solution. Consequently, the [H 3 O + ] concentration of a 0.1 M HCl solution is 0.1 M. Thus HCl is a stronger acid than water and completely donates a proton to water to form H 3 O +. 

13.  By contrast, acetic acid, HC 2 H 3 O 2 (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 K a = = 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 V e ).  pH = pK a @ 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, K a, of the weak acid

15. . C. Determination of pK a of Unknown  Pipet a 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 = pK a @ 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