Applications of Aqueous Equilibria McMurry & Fay ch. 15

Vocabulary: BUFFER SOLUTION an aqueous solution containing both an acid and its conjugate base

Calculate the pH in a solution prepared by dissolving 0 Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in 0.500 L of 0.40 M NH3 (Kb = 1.8 x 10-5). I C E [NH4+][OH-] [NH3] Kb = = 1.8 x 10-5

[H+][A-] [HA] Ka = [H+][A-] [HA] - log Ka = - log [A-] [HA] A shortcut [H+][A-] [HA] Ka = [H+][A-] [HA] - log Ka = - log [A-] [HA] - log Ka = - log [H+] – log = pKa = pH

Henderson-Hasselbalch Equation pKa = pH – log

Uses for Henderson-Hasselbalch Equation Calculate pH of buffer solution Calculate change in pH of buffer solution after additon of acid/base Prepare buffer solution of given pH

Calculating pH of buffer solution Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in 0.500 L of 0.40 M NH3 (Kb = 1.8 x 10-5). pKa = pH – log [A-] [HA]

Calculating pH of buffer solution A buffer solution is created with 0.10 M HF (Ka = 6.8 x 10-4) and 0.15 M NaF. What is the pH? pKa = pH – log [A-] [HA]

Buffer solutions resist changes in pH: adding acid H2CO3 H+ + HCO3-

Buffer solutions resist changes in pH: adding base H2CO3 + OH- H2O + HCO3-

To calculate change in pH: Choose appropriate reaction: HA ⇌ A- + HA if adding acid BH+ + OH- ⇌ B + H2O Make an ICE table that shows number of moles of each species Calculate change using # moles of base or acid added Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

Quick tip: can use either # moles or concentration in Henderson-Hasselbalch eqn 𝑝𝐻=𝑝 𝐾 𝑎 + 𝐴 − 𝐻𝐴 𝑝𝐻=𝑝 𝐾 𝑎 + 𝑚𝑜𝑙𝑒𝑠 𝐻 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑚𝑜𝑙𝑒𝑠 𝐻𝐴 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝐻=𝑝 𝐾 𝑎 + 𝑚𝑜𝑙𝑒𝑠 𝐻 + 𝑚𝑜𝑙𝑒𝑠 𝐻𝐴

How much would the pH change if 100 mL of 1.0 HCl was added to water? 1.0 L of buffer solution has 0.30 M HNO2 (Ka=7.1 x 10-4) and 0.30 M NaNO2. 100 mL of 1.0 M HCl are added to the solution. How much would the pH change if 100 mL of 1.0 HCl was added to water? How much does the pH change when the HCl is added to the buffer solution? Choose appropriate reaction: HA ⇌ A- + HA if adding acid BH+ + OH- ⇌ B + H2O Make an ICE table that shows number of moles of each species Calculate change using # moles of base or acid added Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

Preparing buffer solution of given pH 1. Pick acid with pKa within 1 pH unit of desired pH 2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base 3. Multiply ratio by a factor appropriate for necessary buffer strength

Pick acid with pKa near desired pH HF 3.17 CH3CO2H 4.76 Ethylenediamine 6.85 HOCl 7.53 Tris* 8.07 NH4+ 9.25 *Tris(hydroxymethyl)aminomethane

Make a buffer with pH 7.8 [A-] pKa = pH – log [HA] 1. Pick acid with pKa within 1 pH unit of desired pH HF 3.17 CH3CO2H 4.76 Ethylenediamine 6.85 HOCl 7.53 Tris* 8.07 NH4+ 9.25 2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base 3. Multiply ratio by a factor appropriate for necessary buffer strength [A-] [HA] pKa = pH – log

Acid-Base Titrations

pH changes during Titration: Strong Acid-Strong Base As you add base to an acid solution: Some acid is neutralized; # of moles of H+ decreases Volume of solution increases Molarity = so [H+] will decrease and pH will increase Moles Liters

Calculating pH changes during Strong Acid-Strong Base Titration: before equivalence point Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of H+ remaining: # moles acid at start - # moles base added Calculate volume of sample: Initial volume + volume of base added Calculate [H+] (moles H+/volume) and pH

20. 00 mL of 0. 400 M HCl is titrated with 0 20.00 mL of 0.400 M HCl is titrated with 0.200 M NaOH Calculate pH when you have added: (a) 0 mL of acid (b) 5 mL of acid (c) 7.5 mL of acid (d) 10 mL of acid (e) 15 mL of acid (f) 19 mL of acid (g) 20 mL of acid Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of H+ remaining: # moles acid at start - # moles base added Calculate volume of sample: Initial volume + volume of base added Calculate [H+] (moles H+/volume) and pH

What happens after the equivalence point? All acid has been used up; OH- dominates the pH Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate initial moles of acid : # moles H+ = (Molarity acid) * (volume acid) Calculate moles of OH-: # moles of base – initial moles of acid Calculate volume of sample: Initial volume + volume of NaOH added Calculate [OH-] (moles OH-/volume), pOH, and pH

Equivalence point # moles acid = # moles base

pH changes during Titration: Weak Acid-Strong Base What’s different from strong acid titrations Acid doesn’t completely dissociate so pH depends on Ka Adding base produces the conjugate base of the weak acid Acid + conjugate base both present = buffer solution Use Henderson Hasselbalch to calculate pH

pH changes during Titration: Weak Acid-Strong Base (before equivalence point) Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of conjugate base formed: # moles A- = # moles base added Calculate moles of acid remaining: # moles HA left = initial # moles acid - # moles base added Use Henderson Hasselbalch to calculate pH

Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH What is [H+] at the start? Ka = [H+][A-] [HA]

Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH Where do we expect our equivalence point to be?

Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH What is pH after you have added: 5 mL 10 mL 12.5 mL 15 mL 20 mL 25 mL [A-] [HA] pKa = pH – log

After the equivalence point: No acid left A- + H2O HA + OH-

equivalence point # moles acid = # moles base Half equivalence point [A-] = [HA]

Half-equivalence point Point at which [A-] = [HA] Henderson-Hasselbalch equation: [A-] [HA] pKa = pH – log

At half-equivalence point, pH = pKa

pH changes during Titration: Weak Acid-Strong Base (at equivalence point) All acid has been turned into conjugate base Solution of basic salt: calculate pH as done in ch. 14

pH changes during Titration: Weak Acid-Strong Base (after equivalence point) Most OH- comes from excess NaOH Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of OH- remaining: # moles base added - # moles acid Calculate volume of sample: Initial volume + volume of base added Use [H+] = 10-14/[OH-] to calculate pH

Indicators Dyes that are weak acids or bases Color of acid form different from conjugate base HIn (aq)  H+ (aq) + In– (aq) Acid form base form (one color) (another color)

Choosing an indicator Color change begins when two forms are equal: Choose indicator that has pKa = pH of equiv. point!