1. Acids and Bases Chapter 8

2. Polyprotic acids However, the most ionization occurs in the first step.  K a1 >> K a2 > K a3.... Consequently, the [H + ] is predominantly established in the first ionization with the K a1 value. Subsequent ionizations (K a2 & K a3 ) only add minimal amounts of [H + ]. Use K a1 to determine the pH of the solution at equilibrium.

3. Acid-Base properties of Salt Solutions A salt is a solid that when dissolved in water dissociated into ions. Some salts may contain ions that alter the [H + ] or pH of a solution. The acidic or basic properties of a salt solution arise from the reaction of the dissociated ions of the salt with water.

4. Salts that form neutral solutions The salt of a strong acid/strong base dissolves in water to form neutral solutions.  Strong acid supplies the anion of the salt HCl  H + + Cl -  Strong base supplies the cation of the salt NaOH  Na + + OH -  When the water is removed (evaporation) a salt remains HCl + NaOH  H + + OH - + Cl - + Na +  NaCl Includes groups 1 & 2, but not Be 2+ The solution has a pH of 7

5. Salts that form acidic solutions The salt of a weak base (cation) and strong acid (anion) dissolves in water to form acidic solutions. The cation reacts with water to liberate H + The solution has a pH less than 7 NH 4 Cl  dissociates  NH 4 + & Cl - NH 4 + + H 2 O  H 3 O + (aq) + NH 3(aq)

6. Salts that form basic solutions The salt of a strong base (cation) and weak acid (anion) dissolves in water to form basic solutions. The anion reacts with water to liberate OH - The solution has a pH greater than 7 NaC 2 H 3 O 2  dissociates  Na + & C 2 H 3 O 2 - C 2 H 3 O 2 - (aq) + H 2 O  HC 2 H 3 O 2(aq) + OH - (aq)

7. Weak base/weak acid salts The salt of a weak base (cation) and weak acid (anion) dissolves in water and both ions react with water. If K a >K b, the solution is acidic. If K b >K a, the solution is basic. When an ion reacts with water to produce an acidic or basic solution (break water into its ions) it is called hydrolysis.

8. Weak base/weak acid salts NH 4 HCO 3 is the salt of a weak base (NH 3 ) and a weak acid (H 2 CO 3 ). Both ions react with water. NH 4 + + H 2 O  NH 3 + H 3 O + HCO 3 + H 2 O  H 2 CO 3 + OH - K a for NH 4 + = 5.6 x 10 -10 K b for HCO 3 - = 2.2 x 10 -8 Therefore, the solution would have more OH - in solution (K b >K a ) and have a large pH. It is basic!

9. Anion from Strong acid (Large K a ) Anion from Weak acid (Small K a ) Cation from Strong base (Large K b ) Neither ions reacts with water so the pH is neutral. Anion reacts with water to produce OH - so the pH is basic. Very large K b. Cation from Weak base (Small K b ) Cation reacts with water to produce H + so the pH is acidic. Very large K a. Both ions react with water. K a = K b  neutral K a > K b  acidic K b > K a  basic

10. Acid-Base Reactions Neutralization reaction  Double displacement reaction between an acid and a base to produce a salt and water (solvent).  The acidic and basic properties are destroyed – neutralized.

11. Acid-Base Reactions Titration  Analytical lab technique used to determine the concentration of a solution.  Quantitative neutralization reaction  Titrant – the solution of known concentration in buret.  Sample – the solution of unknown concentration, but known volume (in flask).

12. Acid-Base Reactions Titration  Standard solution – a stock solution that is of known concentration that is used to create the titrant.  Primary standard – a chemical that is available in a pure and stable form that can be used to produce an accurate concentration.  Standardization – a titration used to find the concentration of the titrant using a a primary standard.

13. Acid-Base Reactions Titration  Indicator – an acid base indicator that will change colour at a known pH to signify a specific pH in the neutralization reaction.  End point – the point in the titration when the indicator changes colour.  Equivalence point – the point in the titration when chemically equivalent amounts of reactants have reacted. Usually an equilibrium has been established.

14. Titration of a strong acid with a strong base At the equivalence point [H + ]=[OH - ] or pH=7. Phenolphthalein is a popular indicator because it is colourless in acidic solutions and pink in basic. Remember to consider the reaction at the molar level. (convert to moles!)  C=n/V and C 1 V 1 = C 2 V 2

15. Titration of a strong acid with a strong base Titration curve – a plot of the pH vs. Volume of titrant added. pH Vol. of titrant 7 0 14 Equivalence point (pH=7.0)

16. Titration of a strong base with a strong acid pH Vol. of titrant 7 0 14 Equivalence point (pH=7.0)

17. Titration of a weak acid with a strong base In the titration of a weak acid with a strong base the neutralization occurs equilibrating the number of H + and OH - ions. In doing so, a salt ion is produced that adjusts the hydrolysis of water due to the ions “desire” to form a weak acid. The shift in the ionization of water produces a pH other than 7 at the equivalence point.

18. Titration of a weak acid with a strong base pH Vol. of titrant 7 0 14 Equivalence point (pH> 7.0) Buffer region ½ way to Equivalence point

19. Titration of a weak base with a strong acid pH Vol. of titrant 7 0 14 Equivalence point (pH<7.0) Buffer region

20. Titration of a polyprotic acid with a strong base pH Vol. of titrant 7 0 14 Second Equivalence point First Equivalence point

21. Buffer During the “buffer region” one has a combination of weak acid and its conjugate base. A buffer solution can “absorb” either H + or OH – ions added to the solution. A buffer is an equal mixture of a weak acid and its conjugate base.

22. Buffer solution of H 2 CO 3 & HCO 3 - Consumes H + H + (aq) + HCO 3 - (aq)  H 2 CO 3 (aq) + H 2 O (l) ConjugateWeak baseacid Consumes OH - OH - (aq) + H 2 CO 3 (aq)  HCO 3 - (aq) + H 2 O (l) Weak Conjugate acidbase

23. Acid-base calculations Determine [H + ] and pH Percent ionization Use k a to determine [H + ] and pH Neutralization reactions

24. Determine [H + ] and pH

25. The pH and pOH add up to 14 for a conjugate acid-base pair.

26. Determine pK a and pK b The pK a and pK b add up to 14 for a conjugate acid-base pair.

27. Percent Ionization

28. Use k a to determine [H + ] and pH Given the concentration and equilibrium constant for an acid one can determine the hydronium ion concentration and the pH of the solution.

29. Calculate the hydrogen ion concentration and the pH of a 0.15 mol/L solution of hydrofluoric acid (HF (aq) ) with a k a of 6.6 x 10 -4. 1. Produce a balanced chemical equation for the hydrolysis equilibrium. HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq) 2. Create an equilibrium expression.

30. Calculate the hydrogen ion concentration and the pH of a 0.15 mol/L solution of hydrofluoric acid (HF (aq) ) with a k a of 6.6 x 10 -4. 3. Generate and ICE table to account for concentration changes. HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq) Initial0.15 mol L -1 00 Change-x+x Equilibrium0.15 -x0 + x = x

31. Calculate the hydrogen ion concentration and the pH of a 0.15 mol/L solution of hydrofluoric acid (HF (aq) ) with a k a of 6.6 x 10 -4. 4. Use the equilibrium expression to generate a mathematical model for the equilibrium.

32. Calculate the hydrogen ion concentration and the pH of a 0.15 mol/L solution of hydrofluoric acid (HF (aq) ) with a k a of 6.6 x 10 -4. 5. Resolve for “x” and relate to [H + ]. x = [H + ] = 3.3 x10 -4 mol/L

33. Calculate the hydrogen ion concentration and the pH of a 0.15 mol/L solution of hydrofluoric acid (HF (aq) ) with a k a of 6.6 x 10 -4. 6. Calculate the pH. [H + ] = 3.3 x10 -4 mol/L has two significant digits (3.3) while the exponent is a certain value. The pH value must reflect the certainty of the exponent and significance of the measured coefficient. (3 – indicates the exponent,.48 – the two SD of the coefficient) 3.48  2SD The [H + ] = 3.3 x10 -4 mol/L and the pH= 3.48.

34. Neutralization of a weak/strong combination Stoichiometric calculation of quantities during neutralization. Determine adjusted specie concentration due to modified volumes Look at the common ion affect on the hydrolysis of water equilibrium.

35. Calculate the pH of a sample when 25.00 mL of 0.200 mol/l HClO (aq) is titrated with 10.00 mL of 0.200 mol/L KOH (aq). There are two reactions to consider in this situation: The neutralization reaction which is a quantitative reaction that is governed by stoichiometry. HClO (aq) + KOH (aq) → H 2 O (l) + KClO (aq) The neutralization produces a salt that adjusts an acid- base equilibrium system for the conjugate acid-base pair. HClO (aq) + H 2 O (l) ⇌ H 3 O + (aq) + ClO - (aq) This might be considered as a common ion effect where the neutralization generates and ion that affects an equilibrium.

36. Calculate the pH of a sample when 25.00 mL of 0.200 mol/l HClO (aq) is titrated with 10.00 mL of 0.200 mol/L KOH (aq). 1. Use the neutralization reaction to determine the quantity of acid and base that react. HClO (aq) + KOH (aq) → H 2 O (l) + KClO (aq) HClO (aq) KOH (aq) Concentration (C)0.200 mol/L Volume (V)0.02500 L0.01000L Moles (n)

37. 2. Use the neutralization reaction to determine the quantity of “salt” produced in the reaction. HClO (aq) + KOH (aq) → H 2 O (l) + KClO (aq) 2.00 x 10 -3 mol of KOH (aq) is used up completely and produces 2.00 x 10 -3 mol of KClO (aq) or ClO - (aq). In doing so there is a quantity of HClO (aq) that is left unreacted in the neutralization reaction. The unreacted HClO (aq) is 5.00 x 10 -3 mol – 2.00 x 10 -3 mol = 3.00 x 10 -3 mol Calculate the pH of a sample when 25.00 mL of 0.200 mol/l HClO (aq) is titrated with 10.00 mL of 0.200 mol/L KOH (aq).

39. 3. Using the quantities of weak acid and conjugate base determine the affects on the hydrolysis of water. HClO (aq) + H 2 O (l) ⇌ H 3 O + (aq) + ClO - (aq) Calculate the pH of a sample when 25.00 mL of 0.200 mol/l HClO (aq) is titrated with 10.00 mL of 0.200 mol/L KOH (aq). HClO (aq) H 3 O + (aq) ClO - (aq) [Initial] = 3.00 x 10 -3 mol/ 0.035 L = 0.0857 mol/L 0 = 2.00 x 10 -3 mol/ 0.035 L = 0.0571 mol/L [Change in]-x+x [Equilibrium]0.0857 - xx0.0571 + x

40. 4. Use the k a value and equilibrium expression to solve for [H + ] and pH HClO (aq) + H 2 O (l) ⇌ H 3 O + (aq) + ClO - (aq) Calculate the pH of a sample when 25.00 mL of 0.200 mol/l HClO (aq) is titrated with 10.00 mL of 0.200 mol/L KOH (aq).

41. 4. Use the k a value and equilibrium expression to solve for [H + ] and pH HClO (aq) + H 2 O (l) ⇌ H 3 O + (aq) + ClO - (aq) Calculate the pH of a sample when 25.00 mL of 0.200 mol/l HClO (aq) is titrated with 10.00 mL of 0.200 mol/L KOH (aq).