1 For example: what is the molarity of a solution that contains 0.53 moles of HCl dissolved in mL of aqueous solution? Concentration of acids and bases is usually described in Molarity For example: How many moles of H  ion are present in 36.7 mL of M aqueous HNO 3 ?

2 A conjugate acid is a substance that will behave like an acid in the “reverse” reaction of an acid-base reaction. A conjugate base is a substance that will behave like a base in the “reverse” reaction of an acid-base reaction. NH 3 (aq) + H 2 O (l) ↔ NH 4  (aq) + OH  (aq) Identify the acid, base, conjugate acid, and conjugate base in the reaction above. BACACB Note: Conjugates always appear on the right side of an acid-base equation. Note: Technically, all acid-base reactions can be written as equilibrium reactions. CH 3 CO 2 H (aq) + NaHCO 3 (aq) ↔ NaCH 3 CO 2 (aq) + H 2 CO 3 (aq) Identify the acid, base, conjugate acid, and conjugate base in the reaction above. Conjugate Acids and Bases can be recognized by how they behave in a reaction.

3 Polyprotic Acids: Acids that have more than one acidic hydrogen atom. These acids undergo successive ionization steps when they are dissolved in water. H 2 SO 4 (aq) + H 2 O (l) ↔ H 3 O  (aq) + HSO 4  (aq) HSO 4  (aq) + H 2 O (l) ↔ H 3 O  (aq) + SO 4  (aq) First Ionization: Second Ionization: Overall Reaction (steps 1 and 2 added together) H 2 SO 4 (aq) + 2 H 2 O (l) ↔ 2 H 3 O  (aq) + SO 4  (aq) What is the meaning of: monoprotic, diprotic and triprotic?

4 Notice that the HSO 4  ion can act either as an acid or as a base. It acts like an acid when mixed with water, but will act like a base when mixed with a strong acid. Amphoteric: substances that can act like an acid or like a base. Water is an example of a common substance that is amphoteric. H 2 SO 4 (aq) + H 2 O (l) ↔ H 3 O  (aq) + HSO 4  (aq) NH 3  (aq) + H 2 O (l) ↔ HO  (aq) + NH 4  (aq) Water acting as an acid: Water acting as a base: Notice that when water acts as an acid, it makes the conjugate base hydroxide ion. When water acts as a base, it makes the conjugate acid hydronium ion. Water will act the way it does based on what is put into it!

5 Lewis definition of acids and bases: An acid is any substance that accepts an electron pair to form a covalent bond. A base is any substance that donates a pair of electrons to form a covalent bond. BF 3 (aq) + F  (aq) ↔ BF 4  (aq) BF 3 (aq) + H 2 O (l) ↔ BF   OH 2 Draw Lewis structures of the reactants and products of these reactions to clearly see the definition of Lewis Acid and Lewis Base. HCl (aq) + NaOH (aq) ↔ H 2 O (l) + NaCl (aq) Explain how the reaction above also fits with the Lewis definition of Acid-Base.

6 The strength of a conjugate is related to the strength of the substance that created it. A strong acid will create a conjugate base that is essentially not basic (extremely weak base). A weak acid will create a conjugate base that is a weak base (moderate). A strong base will create a conjugate acid that is essentially not acidic (extremely weak acid). A weak base will create a conjugate acid that is a weak acid (moderate). Acid Base Resulting Solution Strong Strong Neutral Weak Strong Basic Strong Weak Acidic The information above allows us to make the following table:

7 Acids undergo characteristic reactions with active metals (like alkalis and alkaline earths) to produce hydrogen gas (this is a single replacement reaction which is also a Redox reaction-see chapter 20). M (s) + HA (aq) → MA (aq) + H 2 (g) M is an active metal and HA is a generic form for any acid. MA will be an ionic compound made from the positive metal ion being paired with the negative ion that is left over when the acid loses an H + ion. Na (s) + HCl (aq) → NaCl (aq) + H 2 (g) Examples: Mg (s) + HBr (aq) → MgBr 2 (aq) + H 2 (g) Al (s) + H 2 SO 4 (aq) → Al 2 (SO 4 ) 3 (aq) + H 2 (g)

8 K W = [H 3 O  ][OH  ] = 1.0X10  14 K W is the ionization constant for water [H 3 O  ] = 1.0X10  14 [OH  ] [OH  ] = 1.0X10  14 [H 3 O  ] If the [OH  ] is known, the [H 3 O  ] can be found: If the [H 3 O  ] is known, the [OH  ] can be found: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH  (aq) K eq = [H 3 O  ][OH  ] [H 2 O][H 2 O] K eq [H 2 O] 2 = K W =[H 3 O  ][OH  ] Water self-ionizes (auto-ionization) Use these equations to complete practice problems 1a-d on page 474.

9 If [H 3 O + ] is greater than [OH  ], then a solution is acidic. If [OH  ] is greater than [H 3 O + ], then a solution is basic. If [H 3 O + ] is equal to [OH  ], then a solution is neutral. Applying Common Sense: Look at your work for practice problems 1a-d on page 474 and determine whether each solution is acidic, basic, or neutral.

10 pH HA (aq) + H 2 O (l) H 3 O + (aq) + A  (aq) What happens when an acid is dissolved in water? pH =  log([H 3 O + ]) The definition of pH is: If the [H 3 O + ] is known, the pH can be calculated. Acids produce H 3 O +, and it’s the amount of hydronium ion that determines what the pH of a solution is. If the pH is known, the hydronium ion concentration can be found: [H 3 O + ] = 10 (  pH) Example: what is the pH of a M H 3 O + solution? pH =  log([H 3 O + ]) =  log( M) = = 2.49 Example: what is the [H 3 O + ] of a solution with a pH = 4.70? [H 3 O + ] = 10 (  pH) = 10 (  4.70) = X10  5 = 2.0X10  5 M Look at your work for practice problems 1a-d on page 474 and determine the pH for each solution.

11 pOH B (aq) + H 2 O (l) HB + (aq) + OH  (aq) What happens when a base is dissolved in water? pOH =  log([OH  ]) The definition of pOH is: If the [OH  ] is known, the pOH can be calculated. Bases produce OH , and it’s the amount of hydroxide ion that determines what the pOH of a solution is. If the pOH is known, the hydroxide ion concentration can be found: [OH  ] = 10 (  pOH) Example: what is the pOH of a M OH  solution? pOH =  log([OH  ]) =  log( M) = = 3.12 Example: what is the [OH  ] of a solution with a pOH = 3.12? [OH  ] = 10 (  pOH) = 10 (  3.12) = X10  4 = 7.6X10  4 M Notice that the rounding we did to get 3.12 caused the molarity round off to a different number than what we started with!

12 H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH  (aq) K eq = [H 3 O  ][OH  ] [H 2 O][H 2 O] K eq [H 2 O] 2 = K W =[H 3 O  ][OH  ] For pure water at 25 o C, K W = 1.0X10  14 Since [H 3 O + ] = [OH  ], K W = [H 3 O + ] 2 = 1.0X10  14 and [OH  ] = 1.0X10  7 pH =  log(H 3 O + ) =  log(1.0X10 -7 ) = 7.00 pOH =  log(OH  ) =  log(1.0X10 -7 ) = 7.00 so [H 3 O + ] = √1.0X10  14 = 1.0X10  7 We know that pure water is neutral, but why is neutral called pH = 7?

13 Therefore a neutral solution has a pH of 7 and a pOH of 7. A neutral solution  neither acid nor basic  has equal amounts of H 3 O + and OH . pK W =  log(K W ) =  log(1.00X10  14 ) = but K W =[H 3 O  ][OH  ] so  log(K W ) =  log([H 3 O  ][OH  ]) and  log([H 3 O  ][OH  ]) =  log([H 3 O  ]) +  log([OH  ]) but pH =  log([H 3 O  ]) and pOH =  log([OH  ]) therefore, pK W = pH + pOH = Therefore, if pH is known, pOH can be found using this method instead of the other method of first finding [H 3 O  ] and then finding [OH  ] before finally calculating the pOH value. Find the pH and the pOH of each solution described in practice problems 1a-d on page 477.

14 Practice: find the pH; pOH; [H 3 O  ]; and [OH  ] for each of the following solutions: a) M HBr b)2.43X10  M LiOH c)0.65 M H 2 SO 4 d)7.8X10  M Ba(OH) 2 Answers: a)pH = 3.284; pOH = ; [H 3 O  ] = M; [OH  ] = 1.92X10  M b) pH = ; pOH = 2.614; [H 3 O  ] = 4.11X10  M; [OH  ] = 2.43X10  M c) pH =  0.11; pOH = 14.11; [H 3 O  ] = 1.3 M; [OH  ] = 7.7X10  M d) pH = 10.20; pOH = 3.80; [H 3 O  ] = 6.4X10  M; [OH  ] = 1.6X10  M pH =  log([H 3 O  ]) pOH =  log([OH  ]) [H 3 O + ] = 10 (  pH) [OH  ] = 10 (  pOH) [H 3 O  ][OH  ] = 1.0X10  14 pH + pOH = Useful equations:

15 Acid-Base Titration Titration is a process where the concentration of an acidic solution can be determined by reacting the acid with a measured volume of a basic solution of known concentration. Titrations make use of indicators. An indicator is a substance that changes color when the pH of the solution changes. A common indicator is Phenolphthalein. Colorless in acid and Pink in base.

16 The point during a titration where the moles of OH  added is equal to the moles of H 3 O  originally present is called the equivalence point. In order for an indicator to be useful for a titration, the end point must occur at or very near to the equivalence point (usually just after) For strong acid-strong base titrations, the equivalence point will be at pH of 7. So indicators that change color at or very near pH 7 will work. Phenolphthalein is a common indicator that changes from colorless to pink at pH of The point during a titration where the indicator changes color is called the end point.

17 Titration Calculations: we use titrations to find the concentration of an acid or base solution. Remember that at the Equivalence point: n acid = n base M acid V acid = M base V base An acid-base titration usually starts with an acid solution of unknown concentration. We accurately measure the amount of acid solution we want to use in the titration (V acid ). We start to add a base solution whose concentration we know (M base ). We add just enough of the base to cause the color to change and we accurately measure the volume of base (V base ) needed to make the color change. From this information we can calculate the concentration of the acid solution (M acid ) that we started with. The mathematics is based on the idea that the color change occurs very near the equivalence point. The next slide shows a titration curve. A titration curve is a graph made by plotting pH of the solution versus the volume of base added during the titration.

18 A titration curve: a strong acid being titrated with NaOH. End Point for phenolphthalein is at pH of 8. Since the titration curve is almost vertical here, the volume of base added going from pH 7 to pH 8 is usually less than one or two drops (0.05 mL or less)

19 Remember that at the End point (Equivalence point): n acid = n base M acid V acid = M base V base Find the concentration of an unknown acid solution if mL of the acid solution required mL of M NaOH solution to reach a phenolphthalein end point. V acid = mL V base = mL M base = M M acid = ? How would this problem change if Sr(OH) 2 had been used instead of NaOH?

20 Find the concentration of an unknown acid solution if mL of the acid solution required mL of M Sr(OH) 2 solution to reach a phenolphthalein end point. V acid = mL V base = mL M base = ( mol/L Sr(OH) 2 )(2 mol OH  /1 mol Sr(OH) 2 ) M base = M OH  M acid = ? Using Sr(OH) 2 instead of NaOH caused the calculated concentration of the acid to double.

21 Practice: Find the concentration of an unknown acid solution if mL of the acid solution required mL of M NaOH solution to reach a phenolphthalein end point.

22 Why is Phenolphthalein a good indicator even though its color change occurs at a pH of about 8.0? One: Its acidic form is colorless, so that makes the color change easier to detect. Two: The titration curve is nearly vertical near the equivalence point, so adding one or two drops of base will cause a very large change in pH. The error associated with one or two drops of added base is small. Calculations for the titration of 50.0 mL of 0.20 M HCl with M NaOH: At the start: pH = 0.70 and mole HCl = (0.200 mol/L)( L) = mol After 10.0 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)( L) = mol Mole of HCl remaining = ( mol – mol) = mol Concentration of HCl now = ( mol/( L L)) = 0.13 M pH now =  log(0.13) = 0.89

23 After 30.0 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)( L) = mol Mole of HCl remaining = ( mol – mol) = mol Concentration of HCl now = ( mol/( L L)) = M pH now =  log(0.050) = 1.30 After 49.5 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)( L) = mol Mole of HCl remaining = ( mol – mol) = mol Concentration of HCl now = ( mol/( L L)) = M pH now =  log(0.001) = 3.0

24 After 50.5 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)( L) = mol Mole of HCl remaining = ( mol – mol) = no meaning for negative numbers Concentration of NaOH now = ( mol  mol) = mol = ( mol/( L L) = mol/L pOH now =  log(0.001) = 3.0 pH = 14.0 – 3.0 = 11.0

25 The pH Curve for the Titration of mL of 0.10 M of HCl with 0.10 M NaOH

26 The pH Curve for the Titration of 50 mL of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH; Phenolphthalein Will Give an End Point Very Close to the Equivalence Point of the Titration Why is the pH at the equivalence point close to 8.5?

27 Summary for predicting pH of solutions at the equivalence point of a titration. Strong Acid + Strong Base pH at Equivalence Point 7.00 Strong Acid + Weak BaseAcidic Weak Acid + Strong BaseBasic Actual pH depends upon K a and K b values of the conjugates. Weak Conjugate AcidWeak Conjugate Base

28 Equivalence Points for the Titrations of Weak and Strong Acids

29 The pH Curves for the Titrations of 50.0-mL Samples pf 0.10 M Acids with Various Ka Values with 0.10 M NaOH

30 The pH Curve for the Titration of mL of M NH 3 with 0.10 M HCl. The pH at the Equivalence Point is Less than 7, Since the Solution Contains the Weak Acid NH 4 +

31 The Useful pH Ranges for Several Common Indicators

32 Finding the pH of a weak acid solution HF (aq) + H 2 O (l) H 3 O + (aq) + F  (aq) K a = [H 3 O  ][F  ] [HF] What is the pH of a 0.53 M HF solution? Initial 0.53 M 0 0 Change  X  X  X Equilibrium 0.53 – x 0 + x 0 + x HF H 3 O  F  Create an “ICE” chart Now use the ICE chart to plug into the K a equation, then solve for [H 3 O + ] K a = (x)*(x) (0.53 – x) = 7.2X10 

HF (aq) + H 2 O (l) H 3 O + (aq) + F  (aq) K a = [H 3 O  ][F  ] [HF] What is the pH of a 0.53 M HF solution? Continued K a = (x)*(x) (0.53 – x) = 7.2X10  If “x” is small, then the problem simplifies to: x 2 = (0.53)*(7.2X10  ) and x = M If “x” is not small then the quadratic equation will need to be used to solve for “x”. “x” is considered to be small if “x” divided by the initial concentration of HF is less than 5%. In this case, (0.195/0.53)*100 is about 3.7%, so the assumption is valid. Therefore: pH =  log(1.953X10  

34 Acid Strength Any acid that does not dissociate 100% into ions when dissolved in water is called a weak acid. HA (aq) + H 2 O (l) H 3 O + (aq) + A  (aq) The equilibrium constant equation for this dissociation is used to determine acid strength for weak acids. K eq = [H 3 O  ][A  ] [HA][H 2 O] K a = [H 3 O  ][A  ] [HA] Base Strength Any base that does not dissociate 100% into ions when dissolved in water is called a weak base. B (aq) + H 2 O (l) HB + (aq) + OH  (aq) The equilibrium constant equation for this dissociation is used to determine base strength for weak bases. K eq = [HB  ][OH  ] [B][H 2 O] K b = [HB  ][OH  ] [B]

35

36 Finding K a from pH of a weak acid solution. HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO  (aq) K a = [H 3 O  ][ClO  ] [HClO] If M HClO solution has pH of 3.68, what is K a for HClO? [H 3 O + ] = 10 (  pH) = 10 (  3.68) = X10  4 = 2.1X10  4 From the stoichiometry, [H 3 O + ] = [ClO  ], and [HClO] = [HClO] 0 – [H 3 O + ] Therefore, K a = (2.1X10  4 )(2.1X10  4 ) (0.028 – ) = X10  6 = 1.6X10 

37 Many oxides of Metals and Nonmetals will become bases and acids respectively in water. 4 Na + O 2 → 2 Na 2 ONa 2 O + H 2 O → 2 NaOH Example of metal oxide: 2 S + 3 O 2 → 2 SO 3 SO 3 + H 2 O → H 2 SO 4 Example of nonmetal oxide: These oxides are called anhydrides. The feature that determines which will become an acid and which will become a base is the nature of the bond that each forms with oxygen. Metals form ionic bonds with oxygen and these can dissociate in water which means that they will form hydroxide ions (bases) in solution. Nonmetals form covalent bonds with oxygen, and these do not dissociate in water which means that they will form hydrogen ions (acids) in solution.

38 Buffers A solution that resists changes in pH when small amounts of acid or base are added are called buffers. A buffer is made whenever a weak acid is mixed with its conjugate base or when a weak base is mixed with its conjugate acid. If equal amounts of acid and conjugate base (or base and conjugate acid) are used, the pH of the buffer solution will be equal to the pK a of the weak acid (or pK b of the weak base). Example: What is the pH of a buffer made with 0.10 M HF and 0.10 M NaF? K a = [H 3 O  ][F  ] [HF] = [H 3 O  ] for this buffer pK a =  log(K a )  log(6.3X10  ) = 3.20 If the concentration of weak acid is not equal to the concentration of conjugate base, the problem of finding the pH becomes more complex.