Concentration of acids and bases is usually described in Molarity For example: what is the molarity of a solution that contains 0.53 moles of HCl dissolved in 423.5 mL of aqueous solution? For example: How many moles of H+1 ion are present in 36.7 mL of 0.867 M aqueous HNO3?

Conjugate Acids and Bases can be recognized by how they behave in a reaction. 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. NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq) B A CA CB Identify the acid, base, conjugate acid, and conjugate base in the reaction above. 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. CH3CO2H (aq) + NaHCO3 (aq) ↔ NaCH3CO2 (aq) + H2CO3 (aq) Identify the acid, base, conjugate acid, and conjugate base in the reaction above.

Polyprotic Acids: Acids that have more than one acidic hydrogen atom. These acids undergo successive ionization steps when they are dissolved in water. First Ionization: H2SO4 (aq) + H2O (l) ↔ H3O+ (aq) + HSO4- (aq) Second Ionization: HSO4- (aq) + H2O (l) ↔ H3O+ (aq) + SO4-2 (aq) Overall Reaction (steps 1 and 2 added together) H2SO4 (aq) + 2 H2O (l) ↔ 2 H3O+ (aq) + SO4-2 (aq) What is the meaning of: monoprotic, diprotic and triprotic?

Amphoteric: substances that can act like an acid or like a base. Notice that the HSO4- 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. Water acting as an acid: NH3 (aq) + H2O (l) ↔ HO- (aq) + NH4+ (aq) Water acting as a base: H2SO4 (aq) + H2O (l) ↔ H3O+ (aq) + HSO4- (aq) 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!

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. BF3 (aq) + F-1 (aq) ↔ BF4 -1 (aq) BF3 (aq) + H2O (l) ↔ BF3-OH2 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) ↔ H2O (l) + NaCl (aq) Explain how the reaction above also fits with the Lewis definition of Acid-Base.

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). The information above allows us to make the following table: Acid Base Resulting Solution Strong Strong Neutral Weak Strong Basic Strong Weak Acidic

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) + H2 (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. Examples: Na (s) + HCl (aq) → NaCl (aq) + H2 (g) Mg (s) + HBr (aq) → MgBr2 (aq) + H2 (g) Al (s) + H2SO4 (aq) → Al2(SO4)3 (aq) + H2 (g)

Water self-ionizes (auto-ionization) H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) Keq = [H3O+][OH-] [H2O][H2O] Keq[H2O]2 = KW = [H3O+][OH-] KW = [H3O+][OH-] = 1.0X10-14 KW is the ionization constant for water If the [OH-] is known, the [H3O+] can be found: [H3O+] = 1.0X10-14 [OH-] Use these equations to complete practice problems 1a-d on page 474. If the [H3O+] is known, the [OH-] can be found: [OH-] = 1.0X10-14 [H3O+]

Applying Common Sense: If [H3O+] is greater than [OH-], then a solution is acidic. If [OH-] is greater than [H3O+], then a solution is basic. If [H3O+] is equal to [OH-], then a solution is neutral. Look at your work for practice problems 1a-d on page 474 and determine whether each solution is acidic, basic, or neutral.

pH HA (aq) + H2O (l) H3O+ (aq) + A- (aq) pH = -log([H3O+]) What happens when an acid is dissolved in water? HA (aq) + H2O (l) H3O+ (aq) + A- (aq) pH = -log([H3O+]) The definition of pH is: If the [H3O+] is known, the pH can be calculated. Acids produce H3O+ , and it’s the amount of hydronium ion that determines what the pH of a solution is. Example: what is the pH of a 0.0032 M H3O+ solution? pH = -log([H3O+]) = -log(0.0032 M) = 2.49485 = 2.49 If the pH is known, the hydronium ion concentration can be found: [H3O+] = 10(-pH) Example: what is the [H3O+] of a solution with a pH = 4.70? [H3O+] = 10(-pH) = 10 (-4.70) = 1.99526X10-5 = 2.0X10-5 M Look at your work for practice problems 1a-d on page 474 and determine the pH for each solution.

pOH B (aq) + H2O (l) HB+ (aq) + OH- (aq) pOH = -log([OH-]) What happens when a base is dissolved in water? B (aq) + H2O (l) HB+ (aq) + OH- (aq) 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. Example: what is the pOH of a 0.00075 M OH- solution? pOH = -log([OH-]) = -log(0.00075 M) = 3.124938 = 3.12 If the pOH is known, the hydroxide ion concentration can be found: [OH-] = 10(-pOH) Example: what is the [OH-] of a solution with a pOH = 3.12? [OH-] = 10(-pOH) = 10 (-3.12) = 7.58577X10-4 = 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!

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) We know that pure water is neutral, but why is neutral called pH = 7? H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) Keq = [H3O+][OH-] [H2O][H2O] Keq[H2O]2 = KW = [H3O+][OH-] For pure water at 25oC, KW = 1.0X10-14 Since [H3O+] = [OH-], KW = [H3O+] 2 = 1.0X10-14 so [H3O+] = √1.0X10-14 = 1.0X10-7 and [OH-] = 1.0X10-7 pH = -log(H3O+) = -log(1.0X10-7) = 7.00 pOH = -log(OH-) = -log(1.0X10-7) = 7.00

pKW = -log(KW) = -log(1.00X10-14) = 14.00 A neutral solution-neither acid nor basic-has equal amounts of H3O+ and OH-. Therefore a neutral solution has a pH of 7 and a pOH of 7. pKW = -log(KW) = -log(1.00X10-14) = 14.00 but KW = [H3O+][OH-] so -log(KW) = -log([H3O+][OH-]) and -log([H3O+][OH-]) = -log([H3O+]) + -log([OH-]) but pH = -log([H3O+]) and pOH = -log([OH-]) therefore, pKW = pH + pOH = 14.00 Therefore, if pH is known, pOH can be found using this method instead of the other method of first finding [H3O+] 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.

Practice: Useful equations: find the pH; pOH; [H3O+]; and [OH-] for each of the following solutions: 0.000520 M HBr 2.43X10-3 M LiOH 0.65 M H2SO4 7.8X10-5 M Ba(OH)2 pH = -log([H3O+]) pOH = -log([OH-]) [H3O+] = 10(-pH) [OH-] = 10(-pOH) [H3O+][OH-] = 1.0X10-14 pH + pOH = 14.00 Useful equations: Answers: pH = 3.284; pOH = 10.716; [H3O+] = 0.000520 M; [OH-] = 1.92X10-11 M b) pH = 11.386; pOH = 2.614; [H3O+] = 4.11X10-12 M; [OH-] = 2.43X10-3 M c) pH = -0.11; pOH = 14.11; [H3O+] = 1.3 M; [OH-] = 7.7X10-15 M d) pH = 10.20; pOH = 3.80; [H3O+] = 6.4X10-11 M; [OH-] = 1.6X10-4 M

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.

The point during a titration where the moles of OH- added is equal to the moles of H3O+ originally present is called the equivalence point. The point during a titration where the indicator changes color is called the end 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 8-10.

Titration Calculations: we use titrations to find the concentration of an acid or base solution. 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 (Vacid). We start to add a base solution whose concentration we know (Mbase). We add just enough of the base to cause the color to change and we accurately measure the volume of base (Vbase) needed to make the color change. From this information we can calculate the concentration of the acid solution (Macid) that we started with. The mathematics is based on the idea that the color change occurs very near the equivalence point. Remember that at the Equivalence point: nacid = nbase MacidVacid = MbaseVbase 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.

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)

Find the concentration of an unknown acid solution if 10 Find the concentration of an unknown acid solution if 10.45 mL of the acid solution required 38.65 mL of 0.1087 M NaOH solution to reach a phenolphthalein end point. Vacid = 10.45 mL Vbase = 38.65 mL Mbase = 0.1087 M Macid = ? Remember that at the End point (Equivalence point): nacid = nbase MacidVacid = MbaseVbase 𝑴𝒂𝒄𝒊𝒅= 𝑴𝒃𝒂𝒔𝒆𝑽𝒃𝒂𝒔𝒆 𝑽𝒂𝒄𝒊𝒅 𝑴𝒂𝒄𝒊𝒅= (𝟎.𝟏𝟎𝟖𝟕 𝑴)(𝟑𝟖.𝟔𝟓 𝒎𝑳) (𝟏𝟎.𝟒𝟓 𝒎𝑳) = 0.4020 M How would this problem change if Sr(OH)2 had been used instead of NaOH?

Find the concentration of an unknown acid solution if 10 Find the concentration of an unknown acid solution if 10.45 mL of the acid solution required 38.65 mL of 0.1087 M Sr(OH)2 solution to reach a phenolphthalein end point. Vacid = 10.45 mL Vbase = 38.65 mL Mbase = (0.1087 mol/L Sr(OH)2)(2 mol OH-/1 mol Sr(OH)2) Mbase = 0.2174 M OH- Macid = ? 𝑴𝒂𝒄𝒊𝒅= 𝑴𝒃𝒂𝒔𝒆𝑽𝒃𝒂𝒔𝒆 𝑽𝒂𝒄𝒊𝒅 𝑴𝒂𝒄𝒊𝒅= (𝟎.𝟐𝟏𝟕𝟒 𝑴)(𝟑𝟖.𝟔𝟓 𝒎𝑳) (𝟏𝟎.𝟒𝟓 𝒎𝑳) = 0.8041 M Using Sr(OH)2 instead of NaOH caused the calculated concentration of the acid to double.

Practice: Find the concentration of an unknown acid solution if 25.35 mL of the acid solution required 17.25 mL of 0.0876 M NaOH solution to reach a phenolphthalein end point.

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 0.200 M NaOH: At the start: pH = 0.70 and mole HCl = (0.200 mol/L)(0.0500 L) = 0.0100 mol After 10.0 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0100 L) = 0.00200 mol Mole of HCl remaining = (0.0100 mol – 0.00200 mol) = 0.0080 mol Concentration of HCl now = (0.0080 mol/(0.0500 L + 0.0100 L)) = 0.13 M pH now = -log(0.13) = 0.89

After 30.0 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0300 L) = 0.00600 mol Mole of HCl remaining = (0.0100 mol – 0.00600 mol) = 0.0040 mol Concentration of HCl now = (0.0040 mol/(0.0500 L + 0.0300 L)) = 0.050 M pH now = -log(0.050) = 1.30 After 49.5 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0495 L) = 0.00990 mol Mole of HCl remaining = (0.0100 mol – 0.00990 mol) = 0.0001 mol Concentration of HCl now = (0.0001 mol/(0.0500 L + 0.0495 L)) = 0.001 M pH now = -log(0.001) = 3.0

After 50.5 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0505 L) = 0.0101 mol Mole of HCl remaining = (0.0100 mol – 0.0101 mol) = no meaning for negative numbers Concentration of NaOH now = (0.0101 mol - 0.0100 mol) = 0.0001 mol = (0.0001 mol/(0.0500 L + 0.0505 L) = 0.001 mol/L pOH now = -log(0.001) = 3.0 pH = 14.0 – 3.0 = 11.0

The pH Curve for the Titration of 100. 0 mL of 0. 10 M of HCl with 0 The pH Curve for the Titration of 100.0 mL of 0.10 M of HCl with 0.10 M NaOH

The pH Curve for the Titration of 50 mL of 0. 1 M HC2H3O2 with 0 The pH Curve for the Titration of 50 mL of 0.1 M HC2H3O2 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?

Summary for predicting pH of solutions at the equivalence point of a titration. pH at Equivalence Point Strong Acid + Strong Base 7.00 Strong Acid + Weak Base Acidic Weak Conjugate Acid Weak Acid + Strong Base Basic Weak Conjugate Base Actual pH depends upon Ka and Kb values of the conjugates.

Equivalence Points for the Titrations of Weak and Strong Acids

The pH Curves for the Titrations of 50. 0-mL Samples pf 0 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

The pH Curve for the Titration of 100. 0 mL of 0. 050 M NH3 with 0 The pH Curve for the Titration of 100.0 mL of 0.050 M NH3 with 0.10 M HCl. The pH at the Equivalence Point is Less than 7, Since the Solution Contains the Weak Acid NH4+

The Useful pH Ranges for Several Common Indicators

Finding the pH of a weak acid solution What is the pH of a 0.53 M HF solution? HF (aq) + H2O (l) H3O+ (aq) + F- (aq) Ka = [H3O+][F-] [HF] Ka = (x)*(x) (0.53 – x) = 7.2X10-4 Create an “ICE” chart HF H3O+ F- Initial 0.53 M 0 0 Change - X + X + X Equilibrium 0.53 – x 0 + x 0 + x Now use the ICE chart to plug into the Ka equation, then solve for [H3O+]

What is the pH of a 0.53 M HF solution? Continued HF (aq) + H2O (l) H3O+ (aq) + F- (aq) Ka = [H3O+][F-] [HF] Ka = (x)*(x) (0.53 – x) = 7.2X10-4 If “x” is small, then the problem simplifies to: x2 = (0.53)*(7.2X10-4) and x = 0.019534584 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(1.953X10-2) = 1.71

HA (aq) + H2O (l) H3O+ (aq) + A- (aq) Acid Strength Any acid that does not dissociate 100% into ions when dissolved in water is called a weak acid. HA (aq) + H2O (l) H3O+ (aq) + A- (aq) The equilibrium constant equation for this dissociation is used to determine acid strength for weak acids. Keq = [H3O+][A-] [HA][H2O] Ka = [H3O+][A-] [HA] Base Strength Any base that does not dissociate 100% into ions when dissolved in water is called a weak base. B (aq) + H2O (l) HB+ (aq) + OH- (aq) The equilibrium constant equation for this dissociation is used to determine base strength for weak bases. Keq = [HB+][OH-] [B][H2O] Kb = [HB+][OH-] [B]

HClO (aq) + H2O (l) H3O+ (aq) + ClO- (aq) Finding Ka from pH of a weak acid solution. HClO (aq) + H2O (l) H3O+ (aq) + ClO- (aq) [H3O+][ClO-] Ka = [HClO] If 0.028 M HClO solution has pH of 3.68, what is Ka for HClO? [H3O+] = 10(-pH) = 10 (-3.68) = 2.089296X10-4 = 2.1X10-4 From the stoichiometry, [H3O+] = [ClO-], and [HClO] = [HClO]0 – [H3O+] Therefore, Ka = (2.1X10-4)(2.1X10-4) (0.028 – 0.00021) = 1.5869X10-6 = 1.6X10-6

Many oxides of Metals and Nonmetals will become bases and acids respectively in water. Example of metal oxide: 4 Na + O2 → 2 Na2O Na2O + H2O → 2 NaOH Example of nonmetal oxide: 2 S + 3 O2 → 2 SO3 SO3 + H2O → H2SO4 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.

[H3O+][F-] Ka = = [H3O+] for this buffer [HF] 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 pKa of the weak acid (or pKb of the weak base). pKa = -log(Ka) Example: What is the pH of a buffer made with 0.10 M HF and 0.10 M NaF? [H3O+][F-] Ka = = [H3O+] for this buffer [HF] -log(6.3X10-4) = 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.