VIEWS: 51 PAGES: 13 POSTED ON: 10/10/2010
IONIC EQUILIBRIUM INVOLVING WEAK ELECTROLYTES SECTION I KCl K+ + Cl- HCl + H2O H3O+ + Cl- HCl H+ + Cl- 1. H2A + H2O H3O+ + HA- 2. HA- + H2O H3O + A= 1a. H2A H+ + HA- 2a. HA- H+ + A= 3. H2A 2H+ + A= 3a. H2O + HOAc H3O+ + OAc- 4. K= H O OAc 3 H 2 O HOAc 5. K[H2O] = H O OAc 3 HOAc K[H2O] = Ka 6. Ka = H O OAc 3 HOAc 7. Ka = H OAc HOAc Example 1 Calculate the Ka for HOAc if .1 M solution of HOAc is 1.34% ionizable. Solution HOAc H+ + OAc- Since HOAc is 1.34% ionizable 1.34 Concentration of [H+] which is equal to the concentration of [OAc-] = .1 = 1.34 10-3 m.l. 100 1.34 The concentration of HOAc = .1 - .1 = .099 100 Ka = 134 10 134 10 = 1.8 10 . 3 . 3 -5 .099 Example 2 Kb for NH4OH is 1.8 10-5. Calculate the concentration of NH4+ and the OH- and the % ionization if .5 molar NH4OH solution. Solution NH4OH NH4+ + OH- The concentration of NH4+ = the concentration of OH-. Let “x” be the concentration of NH4+. Then “x” is the concentration of OH- and the concentration of NH4OH is .5 - x. Kb = NH OH 4 NH 4 OH x. x 1.8 10-5 = .5 x x is very small compared to .5. It can be deleted from the denominator: x2 1.8 10-5 = .5 .9 10-5 = x2 9 10-6= x2 9 10 6 x 3 10-3 = x = NH4+ = OH- concentration of NH 4 + or concentration of OH - % ionization = 100 concentration of original base 3 10 3 % ionization = 100 = .6% .5 SECTION II 1. Calculate the concentration of each ion in the following solutions. (a) .05 M HCl (b) 1.2 M NaOH (c) .1 M NaOAc 2. Calculate the H+ and the % ionization in a .1 M HCN (Ka HCN 4 10-10). 3. .1 M NH4OH solution is 1.34% ionizable. Calculate the Kb for NH4OH. 4. Calculate the H+ and the % ionization in a 1 M HOAc (Ka 1.8 10-5). ANSWERS FOR SECTION II: 1. (a) [H+] = .05 [Cl-] = .05 (b) [Na+] = 1.2 [OH-] = 1.2 (c) [Na+] = .1 M [OAc-] = .1 2. [H+] = 6.3 10-6 % ionization = 6.3 10-3% 3. 1.85 10-5 4. [H+] = 4.24 10-3 % ionization = .424% SECTION III (Common Ion) PbCl2 Pb+2 + 2Cl- HCl H+ + Cl- If HCl is added, the solubility of PbCl2 will be reduced. HOAc H+ + OAc- Example 1 Calculate the [H+] in a .1 M HOAc which contains .1 M NaOAc (Ka HOAc 1.8 10-5). HOAc H+ + OAc- Solution HOAc is a weak electrolyte. The Ka expression is given by Equation 7 of Section 1. H OAc Ki = HOAc Assume the concentration of H+ = x. The concentration of OAc- from OHAc = x. NaOAc is a strong electrolyte; thus the concentration of the OAc- = .1, and the total concentration of OAc- is .1 + x. 1.8 10-5 = x .1 x .1 The value of x compared to .1 is very small. It can be neglected. Hence: 1.8 10-5 = x .1 .1 H+ = x = 1.8 10-5 Reaction 1: HOAc H+ + OAc- Reaction 2: NaOAc Na+ + OAc- H+ + OH H2O Example 2 Calculate the [OH-] in a .2 M NH3 solution which also contains .5 M NH4Cl. Kb NH3 is 1.8 10-5. Solution NH4OH is a weak electrolyte. NH3 + H2O NH4 + + OH- Kb = NH OH 4 NH 3 NH4Cl NH4+ + Cl- Assume the concentration of OH- = x. Concentration of NH4+ from NH4OH is also x. But in addition to NH4+ from NH4OH, we have + NH4 from NH4Cl which is .5 M. The total NH4+ = .5 + x. Substitute the values obtained from the ionization expression for NH4OH 1.8 10-5 = . 5 x x . 2 x is very small compared to .5; thus it can be neglected. 1.8 10-5 = . 5 x .2 .36 10-5 = .5x .72 10-5 = x 7.2 10-6 = x x = 7.2 10-6 Section IV 1. Calculate the concentration of [H+] in a .1 M HCN solution which also contains .2 M NaCN (Ka HCN 4 10-10). 2. Calculate the [H+] in a .2 M HOAc solution which also contains .2 M NaOAc (Ka HOAc 1.8 10-5). 3. How many m/l of hydrogen ions are contained in a .1 M HOAc solution which contains .2 Mole of NaOAc? (Ka HOAc 1.8 10-5 ). ANSWERS TO SECTION IV: 1. [H+] = 2 10-10 mole/liter 2. [H+] = 1.8 10-5 mole/liter 3. [H+] = 9 10-6 mole/liter Section V (Ionization of Water) Reaction 3: H2O + H2O H3O+ + OH- Reaction 4: H2O H+ + OH- Eq. 8 Ke = H OH H 2 O Ke[H2O] = [H+][OH-] Eq. 9 Kw = [H+][OH-] Kw is the ion product constant for water. Kw is a constant at 25C and has a value of 1 10-14. Eq. 10 [H+][OH-] = 1 10-14 It follows that in an aqueous solution at 25C. Eq. 11 pH + pOH = 14 In pure water the [H+] = [OH-] = 1 10-7. For example, calculate the [OH-] in a .01 molar HCl. Solution [H+][OH-] = 1 10-14 [1 10-2][OH-] = 1 10-14 1 10 14 OH- = 1 10 2 = 1 10-12 Section VI (pH) Eq. 12 pH = -log[H+] Eq. 13 pOH = -log[OH-] Example 1 Calculate the pH of a solution in which [H+] is 1 10-3 m/l. Solution Substitute the value given in Equation 12. pH = -log[H+] = -log[1 10-3] = -[log 1 + log 10-3] = -[-3 log 10] = -[-3] = 3 Example 2 Calculate the pH of .0055 molar HCl. Solution HCl is completely ionzied; thus the H+ is .0055 or 5.5 10-3. Substitute the value in Equation 12. pH = -log[5.5 10-3 = [log 5.5 + log 10-3] = -[.74 - 3] = -[-2.26] = 2.26 If the pOH is required, remember that: pH + pOH = 14 pOH = 14 - 2.26 pOH = 11.74 Example 3 Calculate the pH of .1 M HOAc. Ka 1.8 10-5. Solution HOAc is a weak electrolyte. Ki expression is given by Equation 7. Ka = H OAc HOAc H+ = x OAc- = x x. x 1.8 10-5 = .1 1.8 10-6 = x2 1.34 10-3 = x therefore, H+ = 1.34 10-3 therefore, pH = -log[1.34 10-3] = -[.11 - 3] = -[-2.89] = 2.89 IMPORTANT POINTS TO REMEMBER 1. The pH of a neutral solution is 7; at this point the [H+] = [OH-] = 1 10-7 m/l. 2. If the pH of solution is less than 7, then the solution is acidic and the [H +] is larger than 1 10-7. 3. If the pH of a solution is more than 7, then the solution is basic and the [H +] is less than 1 10-7. Section VIII 1. Calculate the pH of .1 M HCl. 2. Calculate the pH of .1 M HOAc (Ka HOAc 1.8 10-5) 3. Calculate the pH, pOH of .0005 M HCl. 4. Calculate the pH of .05 M HCN (Ka 4 10-10) 5. Calculate the pOH of .5 M HOA (Ka HOAc 1.8 10-5) ANSWERS TO SECTION VIII: 1. pH = 1 2. pH = 2.89 3. pH = 3.3 pH = 10.7 4. pH = 5.35 5. pOH = 11.48 Section IX (Hydrolysis) Reaction 5: NH4Cl + H2O NH4OH + HCl Reaction 6: NaOAc + H2O NaOH + HOAc Reaction 7: NH4+ + HOH NH4OH + H+ NH 4 OH H NH 4 HOH Eq. 14 Ka = Ka[HOH] = Kh NH 4 OH H NH Eq. 15 Kh = 4 Kw Eq. 16 Kh = Kb Eq. 9 [H+][OH-] = Kw From the hydrolysis of the NH4+ NH4+ + HOH NH4OH + H+ NH 4 OH H NH Eq. 17 Therefore, Kh = 4 Substitute the value of [H+] from Eq. 9 into Eq. 17 and obtain Eq. 18. NH 4 OH Kw NH OH Eq. 18 Kh = 4 NH 4 OH 1 NH OH but k1 4 Kw i.e. = Kh Kb Example 1 Calculate the pH of a solution which is .1 M NaOAc. (Ka HOAc 1.8 10-5) Solution OAc- + HOH HOAc + OH- The hydrolysis of OAc- indicates an increase in the OH-; hence, the solution should be basic. Assume the concentration of HOAc = x. Then the concentration of OH- = x. Kw HOAc OH Kh = Ka OAc 1 10 14 x. x 5 18 10 . .1 10 10 16 x2 18 10 5 . 5.5 10-11 = x2 55 10-12 = x 7.5 10-6 = x pOH = -log[OH-] = -log[7.5 10-6] = [.88 - 6] = 5.12 therefore, pH = 14 - 5.12 = 8.88 Example 2 Calculate the pH of a .1 M NaCN solution. Ka HCN 4 10-10. Solution CN- + HOH HCN + OH- From the reaction between CN- and water, it can be seen that there is an increase in the OH-, and therefore the solution will be basic. Let x be the concentration of HCN; then the concentration of OH - is x. That is, [HCN] = [OH-] = x. Kw HCN OH Eq. 19 Kh = Ka CN 10 10 15 x. x 4 10 10 .1 2.5 10-6 = x2 1.58 10-3 = x pOH = -log[1.58 10-3] = [.2 - 3] = 2.8 pH = 14 - 2.8 = 11.2 Section X 1. Calculate the [OH-], [H+], and the pH in .2 M (NaCN). (Ka HCN = 4 10-10) 2. Calculate the [H+] and the pOH in .2 M NH4Cl. (Kb NH4OH = 1.8 10-5) 3. Calculate the [H+] and the pH of .5 M NaOAc. (Ka HOAc 1.8 10-5) ANSWERS FOR SECTION X: 1. [OH-] = 2.24 10-3 2. [H+] = 1.05 10-5 3. [H+] = 6 10-10 [H+] = 4.46 10-12 pOH = 9.02 pH = 9.22 pH = 11.35