Chemistry 115B
	Hydrolysis/Buffers
	An acid is any substance which can donate a proton, and a base is any substance which can accept a proton, according to the Bronsted-Lowry definition of acids and bases. For example, when HCl(g) is dissolved in water an acid-base reaction occurs in which the strong acid HCl donates a proton to the base, water, which accepts the proton. CHl(g) + H2O(l) <--> H3O+(aq) + Cl-(aq) acid base acid base In the reaction a new acid and a new base are generated. The two bases (H2O and Cl-) can be considered to be in competition for the proton. The fact that this particular reaction goes essentially to H3O+ and Cl- simply means that H2O is a much stronger base than Cl-. Since HCl and H3O+ are the conjugate acids of Cl- and H2O, respectively, HCl is a much stronger acid than H3O+. If a weaker acid such as acetic acid, CH3COOH, (Ka = 1.8 x 10-5) is dissolved in water, ionization occurs to a lesser extent than for a strong acid. CH3COOH + H2O <---> H3O+ + CH3COO- In this case, CH3COO- is a stronger base than water or H3O+ is a stronger acid than CH3COOH. HYDROLYSIS If the salt of a weak acid and a strong base, such as sodium acetate, is dissolved in water, an acid-base reaction involving the anion of the weak acid will occur to generate the acid and hydroxide ion. This type of reaction is referred to as hydrolysis. H2O + CH3COO- <--> CH3COOH + OH- acid base acid base   As in any other acid-base reaction the extent to which this reaction will occur is determined by the relative strength of the two bases (or the two acids). Since OH- is a strong base (H2O is a very weak acid), hydrolysis does not proceed far to the right unless the salt of a very weak acid is used. Nevertheless, a solution of sodium acetate is slightly basic. Hydrolysis can also occur when the salt of a strong acid and a weak base such as ammonium chloride is dissolved in water. The resulting solution is acidic. NH4+ + H2O <--> H3O+ + NH3 acid base acid base   Hydrolysis of the acetate ion involves a competition for the proton from water by the two bases CH3COO- and OH-. The position of equilibrium depends upon the relative values of the constants Kw and Ka: Kw Ka (H2O) <--> H+ = OH-) + CH3COO- <--> CH3COOH For the hydrolysis reaction: Kh CH3COO- + H2O <--> CH3COOH + OH- an equilibrium expression can be written which is numerically equal to the hydrolysis constant Kh: [CH3COOH][OH-]/[CH3COO] = Kh Since this hydrolysis amounts to a competition for a proton between the acid anion and hydroxide ion it can be shown that the hydrolysis expression is equal to Kw/Ka as shown below: Kw/Ka = [H+][OH-]/[H+][CH3COO-][CH3COOH] = [CH3COOH][OH-]/[CH3COO-] = Kh The equation for the hydrolysis of the ammonium ion is NH4+ + H2O <--> H3O+ + NH3 The equilibrium constant expression for hydrolysis of the ammonium ion is equal to Kw/Kb and the magnitude of Kh depends upon the relative strengths of H3O+ and NH4+ as acids. Kw/Kb = [H3O+][OH-]/[NH4+][OH-][NH3] = [NH3][H3O+]/[NH4+] = Kh Hydrolysis can also occur with polyvalent metal ions to generate acidic solutions. Cations in aqueous solutions are bonded to water molecules forming complex ions such as Al(H2O)63+. The higher the charge on these ions, the greater the tendency to hydrolyze to form acidic solutions. Al(H2O)63+ + H2O <--> H3O+ Al(H2O)5OH2+ acid base acid base The percent of hydrolysis is an indication of the relative strength of the two acids or bases involved in the hydrolysis reaction. It is calculated by dividing the hydroxide concentration by the initial concentration of the anion of the weak acid and multiplying this quotient by 100. HYDROLYSIS Experimental Procedure 1. Using a 30 mL sample in a 100 mL beaker, test the pH of each of the following salt solutions with a pH meter and record your observed pH for each solution. Calculate the concentration of H+ and OH- in each solution. 1 M NaCH3COO, 1 M NaCl, 1 M NH4Cl, 1 M NH4CH3COO, 1 M Na2CO3 2. Place approximately 3 g of aluminum sulfate in a 250 mL beaker and add approximately 50 mL of distilled water. Test the pH of the resulting solution with pH paper to estimate its pH. Record your data. To this aluminum sulfate solution add about 3 g of sodium hydrogen carbonate and mix well. Note the results and determine the pH of the final mixture using pH paper. Is the solution more or less acidic than the aluminum sulfate solution? Results HYDROLYSIS 1. a. SALT pH of 1M Solution H+ Conc. OH- Conc. Net Ionic Equation for Hydrolysis, if any NaCH3COO NaCl NH4Cl NH4CH3COO Na3CO3   b. Calculate the percent of hydrolysis of 1 M NaCH3COO and 1 M Na2CO3 based on your experimental data of the concentration of OH- in each solution. % hydrolysis of 1 M NaCH3COO _________ % hydrolysis of 1 M Na2CO3 __________ c. Calculate the theoretical percent of hydrolysis for these solutions given that Ka for CH3COOH is 1.8 x 10-5 and Ka for NaHCO3 is 4.7 x 10-11. Theoretical % hydrolysis of 1 M NaCH3COO ____ Theoretical % hydrolysis of 1 M Na2CO3 ______   2. a. The observed pH of your aluminum sulfate solution is _________________.   b. The observed pH of the solution of aluminum sulfate to which sodium hydrogen carbonate has been added is _____________.   c. Write an overall equation which explains the results from the addition of sodium hydrogen carbonate to an aluminum sulfate solution. BUFFERS A mixture of relatively large amounts of weak acid and its salt or a weak base and its salt is known as a buffer. Buffers have the property of maintaining a relatively constant pH even when considerable acid or base has been added. For example, a solution which contained 1.00 M acetic acid and 1.00 M sodium acetate would be a buffer whose H+ concentration would be 1.8 x 10-5 M and have a pH of 4.74. CH3COOH <--> CH3COO- + H+ Ka = [CH3COO-][H+]/[CH3COOH] = 1.8 x 10-5 If 0.05 mole of HCl were added per liter of solution, the CH3COOH concentration would become 1.05 M and the CH3COO- concentration would become 0.95 M. The new H+ concentration can be calculated using the Ka expression: [H+] = 1.05/0.95 1.8 x 10-5 M = 2.0 x 10-5 M The pH of this solution is 4.70. In a similar manner, if 0.05 mole of NaOH were added to the original solution, the H+ concentration would be: [H+] = 0.95/1.05 1.8 x 10-5 M = 1.6 x 10-5 M and the pH would be 4.78. In other words, the pH would only change by .04 in the buffer solution in each case. This relative pH stability would be true until one had exceeded the capacity of the buffer to absorb H+ or OH-. For comparison, if 0.05 mole of HCl were added to one liter of pure water, the [H+] of the resulting solution would be 5.0 x 10-1 M and the pH would be 1.3. Thus, for the buffered solution the pH changed by .04 units whereas for water it changed by 5.7 units. BUFFERS Experimental Procedure 1. Record the pH of a 0.10 M acetic acid solution. Calculate Ka for acetic acid and compare your value with that given above. 2. In a beaker, mix together 25 mL of 0.10 M acetic acid and 25 mL of 0.050 M NaOH. The resulting solution will be a half-neutralized acetic acid solution. Record the pH of the solution. Calculate the Ka and compare your value to that calculated in step 1 above. 3. Add 50 mL of a 0.001 M NaOH solution to the solution in step 2. Record the pH of the resulting solution. Compare this value with that recorded for step 2. 4. In another beaker mix together 50 mL of the 0.001 M NaOH solution with 50 mL of deionized water. Record the pH of this solution and compare it to that for step 3. 5. Dilute the solution in step 3 by a factor of 2 with deionized water. Record the pH and compare this pH with that recorded in step 3. Explain the pH results in steps 3-5 with ionic equilibrium calculations. Results BUFFERS pH [H+] [CH3COO-] [CH3COOH] Ka(exp) Ka(literature) 1 2 pH 1 2 3   DBT/DSR:clm 12/83
	Les Brooks | Lecture Outline | Lab Outline | 115B Home Page
	kh 7/29/99