Chemistry 222 Acids and Bases Acid-Base Equilibria (Chapter 10.) Problems: 2,3,6,13,16,18,21,30,31,33

Review acid-base theory and titrations. For all titrations, at the equivalence point, the two reactants have completely reacted with one another according to the stoichiometry of the equation. For acids and bases with a 1:1 mole ratio, at the equivalence point of a titration, moles acid = moles base

How do we find moles for a solutions? For a solid?

Acids and bases are classified as strong (ionize completely) or weak (ionize only slightly.)

An acid is a substance that can transfer a proton to another substance. A base is a substance that can accept a proton. A proton is a hydrogen ion, H+. Proton transfer is hydrogen ion transfer.

+ - Acid-base reactions: HCl + NH3 → NH4 + Cl

*In every acid-base reaction, equilibrium favors transfer of a proton from the stronger acid to the stronger base.

Water auto-ionizes to give equal, small concentrations of H+ and OH-. + - H2O(l) ⇔ H (aq) + OH (aq) + - 2 H2O(l) ⇔ H3O (aq) + OH (aq)

+ - Acid ionization in water: strong acid: HCl(aq) + H2O(l) → H3O (aq) + Cl (aq) Weak acid: HF(aq) + H2O(l) ⇔

How to write an equilibrium expression:

What does the value of the equilibrium constant tell us about the reaction?

+ - Write the equilibrium expression and constant for water: 2 H2O(l) ⇔ H3O (aq) + OH (aq) Kw =

Possible solutions: neutral acidic basic − In terms of [H+] and [OH ]

The pH scale is a way to measure hydronium ion concentration: pH = pOH = The pH is a measure of acidity. When pH = 7, the solution is neutral. When pH > 7, the solution is basic. When pH < 7, the solution is acidic

1 Chemistry 222 Acids and Bases A solution is a homogeneous mixture of solute and solvent. The amount of solute in a given volume is called the concentration of that solution. Molarity is a concentration; it tells us the number of moles of solute in 1 liter of solution. The symbols [ ] mean molarity. Thus, [HCl] = 0.100 M means we have a solution of HCl that is 0.100 M in concentration.

+ − pH relationships: pH = − log [H3O ] pOH = − log [OH ] pH + pOH = 14 + − pH - − pOH + - −14 [H3O ] = 10 [OH ]. = 10 [H3O ][OH ]. = 1×10

Strong Acids and Bases. Strong acids and bases ionize completely. + Monoprotic strong acids completely ionize their H. If [HCl] = 0.250 M, then {H3O ] = 0.250 M. The diprotic strong acid, H2SO4, completely ionizes the first H. The second H is weakly ionized. Each successive H-ionization in a polyprotic acid is progressively weaker.

− Strong bases (group I metals + Ca, Sr, or Ba) containing hydroxide ion will ionize all of the OH . − If [NaOH] = 0.098 M, then [OH ] = 0.098 M. − If [Ba(OH)2] = 0.205 M, then [OH ] = 2× 0.205 M = 0.410 M + - 1) A solution is 0.125 M Ba(OH)2. Calculate the [H3O ] and [OH ].

− 2) How many moles of OH are there in a mixture of 25.0 mL of a 0.264 M NaOH solution together with 35.0 mL of 0.998 M KOH solution? Determine pOH of the resulting mixture.

3) What is the molarity of an NaOH solution formed by adding 165.0 mL water to 200.0 mL of 0.692 M NaOH?

Ionization of a weak acid Equilibrium expression: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Ka =

+ + Given the Ka and the molarity of HA, we can always solve for [H3O ]; if we have [H3O ], we can find pH.

1. What is the pH of a 1.000 M solution of acetic acid, CH3COOH? + − −5 CH3COOH(aq) + H2O(l) ↔ H3O (aq) + CH3COO (aq) Ka = 1.7×10

2 Chemistry 222 Acids and Bases

Ionization of a conjugate base - + + − [][AHO3 ] For any general weak acid, HA, HA(aq) + H2O(l) ↔H3O (aq)+ A (aq) and Ka = [HA] a) What is the conjugate base of HA? b) Complete the equation for the ionization of the conjugate base in water: − A (aq) + H2O(l)↔

c) Write the equilibrium expression for this reaction. Is this the same as the Ka expression? What is different?

d) Ka values are ionization constants for weak acids in water. We need a value for Kb when we have a conjugate base in water. Ka × Kb = Kw, so Kb =

− Ex. Find the pH of a 0.010 M solution of F .

10.5 Buffer Solutions What is a buffer?? … an acid with its conjugate base

How do we recognize a buffer system? 1) Obvious buffer systems: + a mixture of NH3 and NH4 ; a mixture of HF and NaF; a mixture of HCOOH and HCOONa. 2) Not so obvious buffer systems: a mixture of HF and NaOH a mixture of HCOOH and KOH Henderson-Hasselbalch equation.

Example 1: Suppose that you make up a buffer solution by mixing 60.0 mL of 0.100 M NH3 with 60.0 mL of a 0.090 M NH4Cl. What is the pH of this solution? + + −10 NH4 (aq) + H2O(l) ↔ NH3(aq) + H3O (aq) Ka = 5.6×10

3 Chemistry 222 Acids and Bases

Example 2: Suppose we add 2.00 mL of 0.088 M NaOH to 25.0 mL of 0.0800 M H2CO3. Calculate the pH of the resulting solution. + − −7 H2CO3 (aq) + H2O(l) ↔ H3O (aq) + HCO3 (aq) Ka = 4.3×10

What is the function of a buffer? What does a buffer do?

− Example: Suppose we have a buffer system made by mixing 0.250 mol HN3 with 0.130 mol N3 so that the total volume of the mixture is 1-liter. − + −5 HN3(aq + H2O(l) ↔ N3 (aq) + H3O (aq) Ka =1.9×10

+ a) Calculate [H3O ] in this buffer solution.

+ b) Calculate [H3O ] after adding an additional 3.00 mL of 0.0600 M HCl to the buffer solution.

+ c) Calculate [H3O ] after adding an additional 3.00 mL of 0.0600 M NaOH to the buffer solution.

− Table 10.1 For every 10:1 ratio of [A ] / [HA], pH changes by 1 unit

− Ex. NaOCl was dissolved in a solution buffered to 6.20. Find the ratio [OCl ] / [HOCl}.

Ex. Find the pH of a solution prepared by dissolving 12.43 g of tris (121.136 g/mole) plus 4.67 g of tris hydrochloride (157.597 g/mole) in 1.000 L. Tris is the acid and tris hydrochloride is the conjugate base.

Titrations of weak acids with strong bases give buffer solutions that, at the equivalence point, convert to the opposite conjugate. A titration of a weak acids with a strong base will turn basic at the equivalence point as the weak acid is converted to the weak conjugate base. A titration of a weak base with a strong acid will turn acidic at the equivalence point as the initial weak base is converted to the weak acid..

At the half-equivalence volume in the titration of a weak acid with a strong base, pH = pKa

4 Chemistry 222 Acids and Bases The Leveling Effect (p 245) + H3O is the strongest acid that can exist in water. This means that all strong acids in water will result in + H3O , effectively acting as if they have the same acid strength. − − If a base is stronger than OH , it deprotonates H2O to make OH . This is called the leveling effect. To combat this effect, we have to switch away from water solutions. Example:. Acetic acid solvent is less basic than water, so

+ − −5 HClO4 + CH3COOH ↔ CH3CO2H2 + ClO4 K = 1.3×10 + − −9 HCl + CH3CO2H ↔ CH3CO2H2 + Cl K = 2.8×10 A. Equilibria and LeChatelier’s Principle: Qualitative Understanding

+ − −4 HF(aq) + H2O(l) ↔ H3O (aq) + F (aq) Ka = 6.6×10

− 1) What are three salts that contain F ? − 2) In which direction will the equilibrium shift when additional F is added? (To left? To right?) + 3) How does this affect [H3O ]? (increase, decrease, remain the same) 4) How does this affect the pH? (increase, decrease, remain the same) 5) How does this affect the ionization of HA? (increase, decrease, remain the same)

+ − −4 HF(aq) + H2O(l) ↔ H3O (aq) + F (aq) Ka = 6.6×10

Suppose we add a small amount of HCl(aq) to HF(aq) . 6) What ion does HCl add to the equilibrium mixture? 7) In which direction will the equilibrium shift? (To left? To right?) 8) Will the solution become more acidic or more basic? 9) Will the pH increase or decrease?

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