1. 1 Additional Aqueous Equilibria Chapter 17 Lawrence J. Henderson 1878-1942. Discovered how acid-base equilibria are maintained in nature by carbonic acid/ bicarbonate system in the blood. Developed buffer equation. Karl A. Hasselbalch 1874-1962 Developed logarithmic form of buffer equation.

2. 2 Use this “decision tree” to calculate pH values of solutions of specific solutions. 1.Is it pure water? If yes, pH = 7.00. 2.Is it a strong acid? If yes, pH = -log[HZ] 3.Is it a strong base? If yes, pOH = -log[MOH] or pOH = -log (2 x [M(OH) 2 ]) 4.Is it a weak acid? If yes, use the relationship K a = x 2 /(HZ – x), where x = [H +] 5.Is it a weak base? If yes, use the relationship K b = x 2 /(base – x), where x = [OH - ] 6.Is it a salt (MZ)? If yes, then decide if it is neutral, acid, or base; calculate its K value by the relationship K a K b = K w, where K a and K b are for a conjugate system; then treat it as a weak acid or base. 7.Is it a mixture of a weak acid and its weak conjugate base? It is a buffer; use the Buffer Equation.

3. 3 Common Ion Effect So far, we’ve looked at solutions of weak acids and solutions of weak bases. Weak acid equilibrium: HX  H + + X - (described by K a ) Weak base equilibrium: X - + H 2 O  HX+ OH - (described by K b ) What if you had both HX and X - in the same solution? This could be obtained by adding some Na + X - salt to a solution of HX. The result is called a buffered solution. A buffered solution resists changes to pH when an acid or base is added.

4. 4 Buffered Solutions A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X - ). An example of preparing a buffered solution: Add 0.10 mole of lactic acid (H-Z) and 0.14 mole of sodium lactate (Na-Z) to a liter of solution. A buffer resists a change in pH when a small amount of OH - or H + is added. The reason the buffered solution resists pH change becomes clear when we remember LeChâtlier’s principle. LeChâtlier says the above equation will (1) shift to the left if we add HCl or (2) shift to the right if we add NaOH -- thus resisting a change in [H + ]

5. 5 Buffered Solutions ] [ [X - ] [HX] KH a   We can predict the pH of a buffered solution: Taking –logs of both sides and rearranging: or Henderson-Hasselbalch equation or Buffer equation [HX] [H + ][X - ] KaKa  Rearranging,

6. 6 Buffered Solutions What is the pH of a buffered solution of 0.2 M HF and 0.1 M NaF? (buffer is weak acid HF and its conj. base (salt) F - ) Use: For HF, pK a = -log K a = -log(6.8 x 10 -4 ) = 3.16 So,

7. 7 One often wants a buffer to have a specific [base]/[acid] ratio. The H 2 CO 3 /HCO 3 - buffer in blood should be at a pH=7.40. What must the [base]/[acid] ratio be to obtain this? For H 2 CO 3  H + + HCO 3 - K a = 4.3x10 -7 pK a = 6.37 Henderson-Hasselbalch: Rearrange and solve for [base]/[acid]:

8. 8 Adding strong acids and strong bases to buffers Assume 1 L of a lactic acid/lactate buffer, [HLac] = 0.061 M; and [Lac -1 ] = 0.079 M. The pK a for lactic acid is 3.85. The pH is pK a + log[(0.079)/(0.061)] = 3.96 Suppose you add 0.020 mol H + (strong acid) to this buffer with no change in volume. Then, there will be a neutralization reaction between H + and Lac - H + + Lac -  HLac init. change-0.020M -0.020M +0.020 M after rxn 0 0.059 M 0.081 M Henderson-Hasselbalch: 0.020M 0.079M 0.061M

9. 9 Adding strong acids and strong bases to buffers Assume 1 L of the lactic acid/lactate buffer from earlier slide. [HLac] = 0.061 M; and [Lac -1 ] = 0.079 M The pK a for lactic acid is 3.85. Suppose you add 0.020 mol OH - (strong base) to this buffer with no change in volume. Then, there will be a neutralization reaction between OH - and HLac OH - + HLac  Lac - init. change after rxn 0.020M -0.020M -0.020M +0.020 M 0.061M 0.079M 0 0.041 M 0.099 M Henderson-Hasselbalch: 23. 4 ) +0.38 (85.3 041.0 099.0 log85.3  The pH is pK a + log[(0.079)/(0.061)] = 3.96

10. 10 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations buret (with 0.10 M OH - (aq) beaker (with 0.10 M H + (aq) & indicator) V=50.0 mL Titration curve HCl + NaOH  NaCl

11. 11 Acid-Base Titrations Strong Base – Weak Acid Titration Half-way through the titration, the pH = pK a = 4.74 At the end of the titration, the pH is determined by the concentration of [NaAc] = 0.050 M. The pH = 8.72 HAc + NaOH  NaAc

12. 12 Acid-Base Titrations Titrations of Polyprotic Acids Titrations of Polyprotic Acids In polyprotic acids, each ionizable proton dissociates in steps. In the titration of H 2 CO 3 with NaOH there are two equivalence points: –one for the formation of HCO 3 - H 2 CO 3 + OH - → HCO 3 - + H 2 O –one for the formation of CO 3 2- HCO 3 - + OH - → CO 3 2- + H 2 O

13. 13 Solubility Equilibria Solubility-Product Constant, K sp Solubility-Product Constant, K sp Consider = K sp K sp is the solubility product. (Remember, BaSO 4 is ignored because it is a pure solid so its concentration is constant.) The larger the solubility product, the more soluble the salt. for which

14. 14 Solubility Equilibria Solubility-Product Constant, K sp Solubility-Product Constant, K sp In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers. Examples: for Ag 2 CO 3 K sp = [Ag + ] 2 [[CO 3 -2 ] for Al 2 (SO 4 ) 3 is K sp = [Al +3 ] 2 [SO 4 -2 ] 3 Solubility is the amount (grams) of substance that dissolves to form a saturated solution. Molar solubility (s) is the number of moles of solute dissolving to form a liter of saturated solution.

15. 15 Solubility and K sp Problem solving – always “let s = molar solubility” Two types of problems: (1)Calculate K sp from solubility (2)Calculate solubility from K sp Problem examples: (a) Calculate K sp for PbS, if the solubility = 1.73 x 10 -14 M Let s = molar solubility = [PbS(aq)] = 1.73 x 10 -14 [Pb +2 ] = s = 1.73 x 10 -14 [S -2 ] = s = 1.73 x 10 -14 K sp = [Pb +2 ][S -2 ] = s 2 = (1.73 x 10 -14 ) 2 = 2.95 x 10 -28

16. 16 (b) Calculate K sp for Co(OH) 2 ; solubility = 6.88 x 10 -6 M Let s = molar solubility = [Co(OH) 2 (aq)] = 6.88 x 10 -6 [Co +2 ] = s = 6.88 x 10 -6 [OH-] = 2s = 2(6.88 x 10 -6) = 1.38 x 10 -5 K sp = [Co +2 ][OH - ] 2 = (6.88 x 10 -6 )(1.38 x 10 -5 ) 2 = 1.31 x 10 -15 [Alternatively, K sp = 4s 3 = 4(6.88 x 10 -6 ) 3 = 1.30 x 10 -15 ] (c) Calculate solubility of CdS; K sp = 8 x 10 -28 Let s = molar solubility (unknown) = [CdS(aq)] K sp = [Cd +2 ][S -2 ] = (s)(s) = s 2 = 8 x 10 -28 s = 3 x 10 -14 (d) Calculate solubility of CaF 2 ; K sp = 3.9 x 10 -11 Let s = molar solubility (unknown) = [CaF 2 (aq)] K sp = [Ca +2 ][F - ] 2 = (s)(2s) 2 = 4s 3 = 3.9 x 10 -11 s = 2.1 x 10 -4

17. 17 Factors That Affect Solubility Common-Ion Effect Common-Ion Effect Solubility is decreased when a common ion is added. This is an application of Le Châtelier’s principle: As F - (from NaF, say) is added, the equilibrium shifts to the left. Therefore, CaF 2 (s) is formed and precipitation occurs. As NaF is added to the system, the solubility of CaF 2 decreases.

18. 18 Common-Ion Effect What is solubility of AgCl in a 0.1 M solution of NaCl (which contains Cl -, a common ion)? Example problem. What is solubility of AgCl with and without NaCl? Given: K sp = 1.8 x 10 -10 for AgCl Let s = solubility of AgCl (unknown) = [AgCl(aq)] K sp = 1.8 x 10 -10 = s 2 Solubility = s = (1.8 x 10 -10 ) 1/2 = 1.35 x 10 -5 M (without NaCl) AgCl (s)  Ag + (aq) + Cl - (aq) x x+0.1 K sp = [Ag + ][Cl - ] = (s)(s+0.1)  (s)(0.1) = 1.8 x 10 -10 s = 1.8 x 10 -9 M (with NaCl)

19. 19 Solubility and pH How would we increase the solubility of a sparingly soluble salt such as CaF 2 ? In water, the equilibrium is: CaF 2 (s)  Ca 2+ (aq) + 2 F - (aq) LeChatelier says that to increase solubility, we must remove F - ions. If we add a strong acid, H +, this will cause the reaction: H + (aq) + F - (aq) HF (aq) (Remember, HF is a weak acid) This reaction removes F - ions thus driving equilibrium in the top equation to the right which means more CaF 2 dissolves. As acid is added, pH decreases and solubility of CaF 2 increases.