1. Weak Acids & Weak Bases

2. Review Try the next two questions to see what you remember Try the next two questions to see what you remember

3. The pH of rainwater collected on a particular day was 4.82. What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = 10 -4.82 = 1.5 x 10 -5 M The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH - ]= -log (2.5 x 10 -7 )= 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

4. New Topic Weak Acids & Weak Bases Weak Acids & Weak Bases

5. Strong Electrolyte – 100% dissociation NaCl (s) Na + (aq) + Cl - (aq) Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Strong Acids are strong electrolytes HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq)

6. HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Weak Acids are weak electrolytes HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) HSO 4 - (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2- (aq) H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) Strong Bases are strong electrolytes NaOH (s) Na + (aq) + OH - (aq) KOH (s) K + (aq) + OH - (aq) Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq)

7. F - (aq) + H 2 O (l) OH - (aq) + HF (aq) Weak Bases are weak electrolytes NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Conjugate acid-base pairs: The conjugate base of a strong acid has no measurable strength. H 3 O + is the strongest acid that can exist in aqueous solution. The OH - ion is the strongest base that can exist in aqueous solution.

9. Strong AcidWeak Acid

10. What is the pH of a 2 x 10 -3 M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End 0.002 M 0.0 M What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End 0.018 M 0.036 M0.0 M pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6

11. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Weak Acids (HA) and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a is the acid ionization constant KaKa weak acid strength

13. Important Points When weak acids and bases are added to water the concentration of [H 3 O + ] and [OH - ] is not the same as the concentration of the dissolved acid. When weak acids and bases are added to water the concentration of [H 3 O + ] and [OH - ] is not the same as the concentration of the dissolved acid. The concentration can be determined using the equilibrium constant values for acids and bases termed K a and K b. The concentration can be determined using the equilibrium constant values for acids and bases termed K a and K b.

14. Important Points K a values are the equilibrium constant expressions for the dissolving of various acids in water. K a values are the equilibrium constant expressions for the dissolving of various acids in water. The larger the value of K a the stronger the acids are. The larger the value of K a the stronger the acids are. The same is true of bases and K b. The following link contains a table of acids and their K a values The same is true of bases and K b. The following link contains a table of acids and their K a valuesa table of acidsa table of acids

15. Calculating [H 3 O + ] and [OH - ] of Weak Acids & Bases We can calculate the concentration of the [H 3 O + ] or [OH - ] for a weak acid or base using K a and K b values and the following procedure. We can calculate the concentration of the [H 3 O + ] or [OH - ] for a weak acid or base using K a and K b values and the following procedure. The procedure assumes that the concentration of the dissolved acid or base does not change significantly when the weak acid dissolves. The procedure assumes that the concentration of the dissolved acid or base does not change significantly when the weak acid dissolves.

16. Procedure for Calculating [H 3 O + ] and [OH - ] of Weak Acids & Bases 1.Write the dissolving reaction for the weak acid. 2.These are typically monoprotic (acids or bases) that produce only one H 3 O + ion or OH - ion. 2.These are typically monoprotic (acids or bases) that produce only one H 3 O + ion or OH - ion. 3.Write the K a or K b expression for the acid or base. 3.Write the K a or K b expression for the acid or base. 4.Calculate the concentration of the acid or base in moles per liter.

17. 5.Using a table of acids and bases locate the value of K a or K b for the acid and base and then use the following formulas to calculate the [H 3 O + ] or [OH - ] [H 3 O + ] 2 = [Acid]* K a [OH - ] 2 = [Base] *K b 6.Using the K w formula calculate the other concentration or [OH - ] or [H 3 O + ]

18. Problem 1: Calculate the concentration of [H 3 O + ] and [OH - ] ions in a 0.0056 mol/l solution of HCN. Calculate the concentration of [H 3 O + ] and [OH - ] ions in a 0.0056 mol/l solution of HCN. (K a value = 6.2 X10 -10 ) Strategy: 1.Write the dissolving reaction: HCN (aq) + H 2 O (l) ↔ H 3 O + (aq) + CN - (aq) 2.Write the equilibrium expression K a : K a = [H 3 O + ]*[CN - ] / [HCN] 3.Determine the concentration of the weak acid, [HCN] = 0.0056 mol/l

19. 3.Using a table of acids determine that the K a value is = 6.2 X10 -10 a table of acids a table of acids 4.Calculate the [H 3 O + ] by substituting information into the formula to get [H 3 O + ] 2 = (0.0056 mol/L)(6.2 X10 -10 ) [H 3 O + ] = 1.86 X10 -6 mol/L 5.Using the K w formula calculate the [OH - ] ; remember K w = 1.0 X10 -14 [OH - ] = K w / [H 3 O + ] [OH - ] = 1.0 X10 -14 / 1.86 X 10 -6 mol/L [OH - ] = 5.37 X10 -9 mol/L

20. Important Note: The procedure and formula used above only work if the concentration of the weak acids and bases is not significantly changed when it dissolves. The procedure and formula used above only work if the concentration of the weak acids and bases is not significantly changed when it dissolves. This is true when the [Acid] ÷ K a > 500 This is true when the [Acid] ÷ K a > 500 Work through the following questions by creating ICE tables that support this same logic or strategy!

21. What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4  0.50 – x  0.50 The ratio is > 500, x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M

22. When can I use the approximation? 0.50 – x  0.50 K a << 1 When the ratio of initial concentration/K a is greater than 500. x = 0.5 0.50 M 7.1 x 10 -4 = 714 which is > 500 What is the pH of a 0.05 M HF solution (at 25 0 C)? 0.05 7.1 X 10 -4 = 71.4 which is < 500  We can not neglect the “x” value. We must solve for x exactly using quadratic equation or method of successive approximation.

23. Solving weak acid ionization problems: 1.Identify the major species that can affect the pH. In most cases, you can ignore the auto-ionization of water. Ignore [OH - ] because it is determined by [H + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4.Calculate concentrations of all species and/or pH of the solution.

24. What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 x 10 -4 ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx K a = x2x2 0.122 - x = 5.7 x 10 -4 Ka  Ka  x2x2 0.122 = 5.7 x 10 -4 0.122 – x  0.122 K a << 1 0.122 5.7 X 10 -4 = 214.04 This value is < 500  Approximation not valid.

25. K a = x2x2 0.122 - x = 5.7 x 10 -4 x 2 + 0.00057x – 6.95 x 10 -5 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = 0.0081x = - 0.0081 HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx [H + ] = x = 0.0081 M pH = -log[H + ] = 2.09

26. percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration

27. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak Bases and Base Ionization Constants K b = [NH 4 + ][OH - ] [NH 3 ] K b is the base ionization constant KbKb weak base strength Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].

29. Chemistry In Action: Antacids and the Stomach pH Balance NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) Mg(OH) 2 (s) + 2HCl (aq) MgCl 2 (aq) + 2H 2 O (l)