Aqueous Equilibria: Acids & Bases McMurry and Fay ch. 14

Acid-base definition (Arrhenius) Acid: increases H+ ions in solution Base: increases OH- ions in solution

Arrhenius Acids & Bases Acids: HCl  H+ + Cl- H2SO4  2 H+ + SO42- Bases: KOH  K+ + OH- NaOH  Na+ + OH-

Ammonia as a base Ammonia (NH3) neutralizes acids. How can it make OH- in solution?

An important equilibrium constant H2O H+ + OH- (self-ionization of water) [H+][OH-] = Kw

Ammonia as a base NH3 + H2O  NH3 + H+ + OH-  NH4+ + OH-

Acid-base definition (Brønsted-Lowry) Acid: proton donor Base: proton acceptor (the word proton refers to H+)

Brønsted-Lowry acid-base reactions are reversible NH3 + H2O NH4+ + OH- accepts proton BASE donates proton ACID donates proton ACID accepts proton BASE

Brønsted-Lowry acid-base reactions are reversible NH3 + H2O NH4+ + OH- CONJUGATE ACID CONJUGATE BASE accepts proton BASE donates proton ACID donates proton ACID accepts proton BASE

Conjugate acid-base pairs Two substances that differ by only one proton The acid has one more proton than the base The base has one less proton than the acid

In the following reaction, which is the conjugate base of CH3COOH In the following reaction, which is the conjugate base of CH3COOH? CH3COOH + H2O CH3COO- + H3O+ CH3COOH H2O CH3COO- H3O+

Hydronium ion H+ is not very stable by itself Charge on proton in solution is stabilized by IMFs H+ ion is frequently written as H3O+ H+ O H In this class we will use H+ and H3O+ interchangeably

Acid strength: old version The stronger the acid, the more H+ it generates Acid strength: old version HCl (aq) H+ (aq) + Cl- (aq) (strong acid because no HCl remains; all turns into H+) CH3COOH (aq) CH3COO-(aq) + H+ (aq) (weak acid because little H+ is made; most remains as CH3COOH)

Acid strength: Equilibrium Version Strength depends on equilibrium positions HCl + H2O H3O+ + Cl- CH3COOH + H2O CH3COO- + H3O+ In strong acids, the conjugate base form is favored at equilibrium In weak acids, the acid form is favored at equilibrium

Characteristics of a Strong Acid Weak bond (easy to break) between H & anion is weak Anion is stable in solution

Acid strength: Binary acids in the same group of the periodic table As valence shell increases, bond strength decreases Easier to break bonds between H & larger atoms H F Cl H Acid strength increases Br H I H

Acid strength: Binary acids in the same row of the periodic table - H H H H N N - As you move to the right, electronegativity increases More electronegative atoms are more happy as ions More electronegative elements are stronger acids H H H H H H H H O - O - F H F

Acid strength: Oxoacids Oxoacids are acids composed of hydrogen, oxygen, and some other element (examples: H2SO4, H3PO4, HNO3) General rule: as central atom electronegativity increases, strength of the O-H bond decreases  Higher electronegativity = stronger acid strength

Acid strength: Oxoacids For a given element, there can be oxoacids with different numbers of oxygen atoms bound to the central atom H2SO4 H2SO3 O O O S O O S O O

Acid strength: Oxoacids More resonance structures = more stable conjugate base General rule: The larger the number of lone oxygen atoms, the stronger the acid

Lewis Acids and Bases Explain acid-base chemistry that occurs outside of aqueous solutions Lewis acid: electron pair acceptor Lewis base: electron pair donor

Lewis Base examples: H N H H O H H .. .. ..

Lewis acid-base example NH3 + BF3

An important equilibrium constant H2O H+ + OH- (self-ionization of water) [H+][OH-] = Kw = 10-14 (at 25 oC) What does the Kw value tell us about this equilibrium?

Self-ionization of water H2O H+ + OH- [H+][OH-] = Kw = 10-14 (at 25 oC) Any aqueous solution will have both H+ ions and OH- ions. Kw will always be 10-14 at 25oC. You can’t have lots of acid AND lots of base

If you know [H+], you can calculate [OH-] (and vice versa) [H+][OH-] = Kw so Kw 10-14 [OH-] [OH-] [H+] = = Kw 10-14 [H+] [H+] [OH-] = =

If [H+] = 0.010 M, what is [OH-]? Kw [OH-] [H+] =

If [H+] = 6.24 x 10-13 M, what is [OH-]?

Definition of acidic & basic solutions Neutral solution: [H+] = [OH-] Acidic solution [H+] > [OH-] Basic solution [H+] < [OH-]

p Notation When talking about measurements that span several orders of magnitude, it is easiest to use a logarithmic scale pX = -log X

Easily measured using pH electrode pH = -log [H+] Easily measured using pH electrode

pH of strong acid solutions A strong acid is completely dissociated…. HCl  H+ + Cl- (no HCl remains) If you start with 0.200 M HCl, you end with 0.200 M H+ 0.200 M Cl-

To find the pH of a strong acid solution: 1. Find molarity of acid solution 2. Assume [H+] = molarity 3. Use equation pH = -log [H+]

What is the pH of a 0.0325 M solution of HNO3? 1. Find molarity of acid solution 2. Assume [H+] = molarity 3. Use equation pH = -log [H+]

Converting from pH to [H+] pH = -log [H+] Divide each side by -1 - pH = log [H+] Do the anti-log of both sides (raise 10 to the power of each side) 10-pH = [H+]

Converting from pH to [H+] 10-pH = [H+] pH 4.0 [H+] = 10-4 pH 8.0 [H+] = 10-8 pH 5.75 [H+] = 10-5.75 = 1.78 x 10-6

What is [H+] for an acid solution with pH 4.27?

pOH = -log [OH-] [H+][OH-] = 10-14 log[H+][OH-] = log (10-14) Can’t be measured directly; calculate from pH [H+][OH-] = 10-14 log[H+][OH-] = log (10-14) log[H+] + log [OH-] = -14 -log [H+] – log [OH-] = 14 pH + pOH = 14

pH of strong base solutions 100% of strong base molecules generate OH- ions pH + pOH = 14 pOH = 14 - pH

pH of strong base solutions 1. Find molarity of base solution 2. Assume [OH-] = molarity 3. Use equation pOH = -log [OH-] 4. Use equation pH = 14 - pOH

What is the pH of a 0.245 M solution of KOH? 1. Find molarity of base solution 2. Assume [OH-] = molarity 3. Use equation pOH = -log [OH-] 4. Use equation pH = 14 - pOH

What is the pH of 0.000581 M solution of NaOH? 1. Find molarity of base solution 2. Assume [OH-] = molarity 3. Use equation pOH = -log [OH-] 4. Use equation pH = 14 - pOH

Definition of weak acids and bases— not everything dissociates pH is not a simple calculation

Acid equilibria HA H+ + A- Generic equation for an acid dissociation reaction is: HA H+ + A- acid (protonated) conjugate base (deprotonated)

Acid equilibrium constant, Ka HA H+ + A- [H+][A-] [HA] Ka =

pKa pKa = - log Ka Ka = 5.7 x 10-3 pKa = 2.2 Ka = 9.8 x 10-7 pKa = 6.0

Determining a Ka value from initial concentration & equilibrium pH 1. Make an ICE table 2. Use equation [H+] = 10-pH 3. Use step 2 to calculate change in [H+] 4. Use step 3 and stoichiometric ratios to calculate other concentration changes 5. Calculate equilibrium concentrations 6. Use final concentrations to calculate Ka

A student prepared a 0.10 M solution of formic acid (HCHO2) and measured a pH of 2.38. Calculate Ka for formic acid.

1. Make an ICE table HCHO2 H+ + CHO2- [HCHO2] (M) [H+] (M) [CHO2-] (M) Initial Change Equilibrium A student prepared a 0.10 M solution of formic acid (HCHO2) and measured a pH of 2.38. Calculate Ka for formic acid.

2. Use equation [H+] = 10-pH HCHO2 H+ + CHO2- [HCHO2] (M) [H+] (M) [CHO2-] (M) Initial Change Equilibrium A student prepared a 0.10 M solution of formic acid (HCHO2) and measured a pH of 2.38. Calculate Ka for formic acid.

3. Use step 2 to calculate change in [H+] HCHO2 H+ + CHO2- [HCHO2] (M) [H+] (M) [CHO2-] (M) Initial Change Equilibrium

4. Use step 3 and stoichiometric ratios to calculate other concentration changes HCHO2 H+ + CHO2- [HCHO2] (M) [H+] (M) [CHO2-] (M) Initial Change Equilibrium

5. Calculate equilibrium concentrations HCHO2 H+ + CHO2- [HCHO2] (M) [H+] (M) [CHO2-] (M) Initial Change Equilibrium

6. Use final concentrations to calculate Ka HCHO2 H+ + CHO2- [HCHO2] (M) [H+] (M) [CHO2-] (M) Initial Change Equilibrium [H+][CHO2-] [HCHO2] Ka =

Calculating pH using Ka 1. Make an ICE table; set change in [H+] to +x 2. Use stoichiometric ratios to calculate other concentration changes 3. Calculate equilibrium concentrations 4. Substitute equilibrium concentrations in to the Ka expression 5. Rewrite equation to form ax2 + bx + c 6. Solve for x using the quadratic equation 7. Use value of x to calculate pH

Calculate the pH of a 0.30 M solution of acetic acid (HC2H3O2), whose Ka = 1.8 x 10-5

1. Make an ICE table; set change in [H+] to +x C2H3O2H H+ + C2H3O2- Calculate the pH of a 0.30 M of acetic acid (HC2H3O2H), whose Ka = 1.8 x 10-5

2. Use stoichiometric ratios to calculate other concentration changes C2H3O2H H+ + C2H3O2-

3. Calculate equilibrium concentrations C2H3O2H H+ + C2H3O2-

4. Substitute equilibrium concentrations into the Ka expression C2H3O2H H+ + C2H3O2-

5. Rewrite equation to form ax2 + bx + c

6. Solve for x using the quadratic equation -b ± b2 – 4ac 2a x =

7. Use value of x to calculate pH

SHORTCUT!! If Ka is much smaller than acid concentration (< 1/100th) …change (x) will be very small Subtracting x from [HA]0 will not substantially change [HA] APPROXIMATION: [HA]0 – x ≈ [HA]0

Calculate the pH of a 0.045 M solution of aniline (C6H5NH3), whose Ka = 2.51 x 10-5. 1. Make ICE table; set [H+] = +x 2. Calculate other changes 3. Calculate eq concs 4. Substitute eq concs into Ka 5. Solve for x 6. Use value of x to calculate pH

Base equilibria B + H2O BH+ + OH- The generic base dissociation reaction is: B + H2O BH+ + OH-

Base ionization constant, Kb B + H2O BH+ + OH- [BH+][OH-] [B] Kb =

pKb pKb = - log Kb Kb = 6.7 x 10-8 pKb = 7.2

Relationship between pKa and pKb pKa + pKb = pKw or pKa + pKb = 14

Calculations for bases are very similar to those for acids Use [OH-] calculated from [OH-] = 10-14/[H+]

A 0. 10 M solution of for ethylamine, CH3CH2NH2, has a pH of 11. 87 A 0.10 M solution of for ethylamine, CH3CH2NH2, has a pH of 11.87. Calculate the Kb and pKb. 1. Make an ICE table 2. Use pH + pOH = 14 3. Use step 2 to calculate change in [OH-] 4. Use step 3 & stoichiometric ratios to calc. other conc. changes 5. Calculate equilibrium concentrations 6. Use final concentrations to calculate Ka

Calculate the pH in a solution of a 0.40 M NH3. (Kb = 1.8 x 10-5) 1. Make an ICE table 2. Use pH + pOH = 14 3. Use step 2 to calculate change in [OH-] 4. Use step 3 & stoichiometric ratios to calc. other conc. changes 5. Calculate equilibrium concentrations 6. Use final concentrations to calculate Ka

Two ways to measure strength of a weak acid The larger the Ka value, the more the acid dissociates and the more H+ it produces Percentage ionization The fraction of the acid that dissociates [H+][A-] [HA] Ka =

Percentage ionization % ionization = x 100% Moles ionized per liter Moles available per liter Moles ionized = [A-] (or [BH+]) Moles available = [HA] + [A-] (or [B] + [BH+])

Enough benzoic acid (C7H6O2H) is dissolved in water to make an 0 Enough benzoic acid (C7H6O2H) is dissolved in water to make an 0.100 M solution. The concentration of benzoate ions (C7H6O2-) was found to be 2.47 x 10-3 M. What is the percent ionization? % ionization = x 100% Moles ionized per liter Moles available per liter

Enough benzoic acid (C7H6O2H) is dissolved in water to make an 0 Enough benzoic acid (C7H6O2H) is dissolved in water to make an 0.100 M solution. The concentration of benzoate ions (C7H6O2-) was found to be 2.47 x 10-3 M. What is the percent ionization? % ionization = x 100% Moles ionized per liter Moles available per liter

Using % ionization to calculate Ka or Kb 1. Make an ICE table and fill in known concentrations 2. Use % ionization & initial concentration to calculate change in [A-] or [BH+] 3. Calculate equilibrium concentrations 4. Plug equilibrium concentrations into expression for Ka or Kb

A 0. 0750 M solution of HF has a % ionization of 29. 6 % A 0.0750 M solution of HF has a % ionization of 29.6 %. What is the Ka for HF? 1. Make an ICE table and fill in known concentrations 2. Use % ionization & initial concentration to calculate change in [A-] or [BH+] 3. Calculate equilibrium concentrations 4. Plug equilibrium concentrations into expression for Ka or Kb

A 0. 00392 M solution of imidazole (C3H3N2H2) has a % ionization of 0 A 0.00392 M solution of imidazole (C3H3N2H2) has a % ionization of 0.51%. What is the Ka for imidazole? 1. Make an ICE table and fill in known concentrations 2. Use % ionization & initial concentration to calculate change in [A-] or [BH+] 3. Calculate equilibrium concentrations 4. Plug equilibrium concentrations into expression for Ka or Kb

Using Ka to calculate pH and % ionization 1. Make an ICE table; set change in [H+] to +x 2. Use stoichiometric ratios to calculate other concentration changes 3. Calculate equilibrium concentrations 4. Substitute equilibrium concentrations in to the Ka expression 5. Solve for x 6. Use value of x to calculate pH 7. Use values in ICE table to calculate % ionization

Calculate the pH and % ionization of a 1 Calculate the pH and % ionization of a 1.0 M solution of chloroacetic acid (HC2H2ClO2), whose Ka = 1.36 x 10-3 1. Make ICE table; change in [H+] = +x 2. Calculate other conc changes 3. Calculate eq concs 4. Substitute eq concs into Ka 5. Solve for x 6. Use x to calculate pH 7. Use values in ICE table to calculate % ionization

Calculate the pH and % ionization of a 0 Calculate the pH and % ionization of a 0.040 M solution of pyridine (C5H5NH), whose Ka = 6.3 x 10-6 1. Make ICE table; change in [H+] = +x 2. Calculate other conc changes 3. Calculate eq concs 4. Substitute eq concs into Ka 5. Solve for x 6. Use x to calculate pH 7. Use values in ICE table to calculate % ionization

Acid-Base Properties of Salt Solutions

Basic principle HA H+ + A- A- + H+ HA A strong acid has a weak conjugate base A weak acid has a strong conjugate base HA H+ + A- A- + H+ HA [H+][A-] [HA] [HA] [H+][A-] Ka = Kb =

If salts… …have an anion that is the conjugate base of a weak acid… If salts… …have an anion that is the conjugate base of a weak acid… OR …have a cation that is the conjugate acid of a weak base… They will affect the pH of a solution

0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution? Acetate (CH3CO2-) is the conjugate base of a fairly weak acid (acetic acid) and therefore is a fairly strong base Sodium (Na+) is the conjugate acid of a very strong base (NaOH) and therefore is a very weak acid

pH calculated similarly to previous problems 1. Make an ICE table; set change in [H+] or [OH-] to +x 2. Fill out remainder of ICE table 3. Calculate Kb for conjugate base or Ka for conjugate acid 4. Substitute equilibrium concentrations in to the Ka or Kb expression 5. Solve for x 6. Use value of x to calculate pH (or convert pOH to pH)

1. Make an ICE table; set change in [H+] or [OH-] to +x H2O + C2H3O2- OH- + C2H3O2H [C2H3O2-] (M) [OH-] (M) [C2H3O2H] (M) I C E 0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution? Ka of CH3CO2H is 1.8 x 10-5.

2. Fill out remainder of ICE table H2O + C2H3O2- OH- + C2H3O2H [C2H3O2-] (M) [OH-] (M) [C2H3O2H] (M) I C E 0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution? Ka of CH3CO2H is 1.8 x 10-5.

3. Calculate Kb for conjugate base or Ka for conjugate acid H2O + C2H3O2- OH- + C2H3O2H [C2H3O2-] (M) [OH-] (M) [C2H3O2H] (M) I C E 0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution? Ka of CH3CO2H is 1.8 x 10-5.

4. Substitute equilibrium concentrations in to the Ka or Kb expression H2O + C2H3O2- OH- + C2H3O2H [C2H3O2-] (M) [OH-] (M) [C2H3O2H] (M) I C E 0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution? Ka of CH3CO2H is 1.8 x 10-5.

H2O + C2H3O2- OH- + C2H3O2H 5. Solve for x. [C2H3O2-] (M) [OH-] (M) [C2H3O2H] (M) I C E

6. Use value of x to calculate pH (or convert pOH to pH) H2O + C2H3O2- OH- + C2H3O2H [C2H3O2-] (M) [OH-] (M) [C2H3O2H] (M) I C E