Conjugates do not react!! Acid-Base Reactions Conjugates do not react!! CHAPTER 17 Part 1

Stomach Acidity & Acid-Base Reactions

ACIDS-BASE REACTIONS For any acid-base reaction where only one hydrogen ion is transferred, the equilibrium constant for the reaction can be calculated and is often called Knet. For the general reaction: Knet is always Ka (reactant acid) / Ka (product acid) When Knet >> 1, products are favored. When Knet << 1, reactants are favored. See 16.5 for manipulating K

ACIDS-BASE REACTIONS There are four classifications or types of reactions: strong acid with strong base, strong acid with weak base, weak acid with strong base, and weak acid with weak base. NOTE: For all four reaction types the limiting reactant problem is carried out first. Once this is accomplished, one must determine which reactants and products remain and write an appropriate equilibrium equation for the remaining mixture.

STRONG ACID WITH STRONG BASE The net reaction is: The product, water, is neutral.

STRONG ACID WITH WEAK BASE The net reaction is: The product is HB+ and the solution is acidic.

WEAK ACID WITH STRONG BASE The net reaction is: The product is A- and the solution is basic.

WEAK ACID WITH WEAK BASE The net reaction is: Notice that Knet may even be less than one. This will occur when Ka HB+ > Ka HA.

HBz + NaOH ---> Na+ + Bz- + H2O Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na+ + Bz- + H2O Ka = 6.3 × 10-5 Kb = 1.6 × 10-10 C6H5CO2H = HBz Benzoate ion = Bz-

Acid-Base Reactions The product of the titration of benzoic acid, the benzoate ion, Bz-, is the conjugate base of a weak acid. The final solution is basic. Bz- + H2O HBz + OH- + Kb = 1.6 x 10-10

Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Strategy — find the concentration of the conjugate base Bz- in the solution AFTER the titration, then calculate pH. This is a two-step problem: 1st. stoichiometry of acid-base reaction 2nd. equilibrium calculation

STOICHIOMETRY PORTION Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? STOICHIOMETRY PORTION 1. Calculate moles of NaOH required. (0.100 L HBz)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2. Calculate volume of NaOH required. 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH required

STOICHIOMETRY PORTION Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? STOICHIOMETRY PORTION 25 mL of NaOH required 3. Moles of Bz- produced = moles HBz = 0.0025 mol Bz- 4. Calculate concentration of Bz-. There are 0.0025 mol of Bz- in a TOTAL SOLUTION VOLUME of 125 mL [Bz-] = 0.0025 mol / 0.125 L = 0.020 M

HBz + NaOH ---> Na+ + Bz- + H2O Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na+ + Bz- + H2O Ka = 6.3 × 10-5 Kb = 1.6 × 10-10 C6H5CO2H = HBz Benzoate ion = Bz-

Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? EQUILIBRIUM PORTION Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 [Bz-] [HBz] [OH-] initial change equilib 0.020 0 0 -x +x +x 0.020 - x x x

Solving in the usual way, we find Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? EQUILIBRIUM PORTION Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 [Bz-] [HBz] [OH-] equilib 0.020 - x x x K b = 1.6 x 10 -10 x 2 0.020 - Solving in the usual way, we find x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25

Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point? HBz + H2O H3O+ + Bz- Ka = 6.3 x 10-5 [H3O+] = { [HBz] / [Bz-] } Ka At the half-way point, [HBz] = [Bz-], so [H3O+] = Ka = 6.3 x 10-5 pH = 4.20

The Common Ion Effect LEFT Closer to 0 18 The Common Ion Effect QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Here we are adding an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the ____________. LEFT Closer to 0 The pH will go _______. After all, NH4+ is an acid!

The Common Ion Effect QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Let us first calculate the pH of a 0.25 M NH3 solution. [NH3] [NH4+] [OH-] initial 0.25 0 0 change -x +x +x equilib 0.25 - x x x

The Common Ion Effect QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) K b = 1.8 x 10 -5 [NH 4 + ][OH - ] 3 x 2 0.25 Assuming x is << 0.25, we have [OH-] = x = [Kb(0.25)]1/2 = 0.0021 M This gives pOH = 2.67 and so - - - pH = 14.00 - 2.67 = 11.33 for 0.25 M NH3

The Common Ion Effect QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) We expect that the pH will decline on adding NH4Cl. Let’s test that at 0.10M ! [NH3] [NH4+] [OH-] initial change equilib 0.25 0.10 0 -x +x +x 0.25 - x 0.10 + x x

The Common Ion Effect QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) K b = 1.8 x 10 -5 [NH 4 + ][OH - ] 3 x(0.10 x) 0.25 x Because equilibrium shifts left, x is MUCH less than 0.0021 M, the value without NH4Cl. [OH-] = x = (0.25 / 0.10)Kb = 4.5 x 10-5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion.

Buffer Solutions HCl is added to pure water. HCl is added to a solution of a weak acid H2PO4- and its conjugate base HPO42-.

Buffer Solutions The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the common ion effect. Buffer Composition Weak Acid + Conj. Base HC2H3O2 + C2H3O2- H2PO4- + HPO42- Weak Base + Conj. Acid NH3 + NH4+

Buffer Solutions Consider HOAc/OAc- to see how buffers work. ACID USES UP ADDED OH-. We know that OAc- + H2O HOAc + OH- has Kb = 5.6 x 10-10 Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb = 1.8 x 109 Kreverse is VERY LARGE, so HOAc completely uses up the OH- !!!!

Buffer Solutions Consider HOAc/OAc- to see how buffers work. CONJUGATE BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ has Ka = 1.8 x 10-5. Therefore, the reverse reaction of the WEAK BASE with added H+has Kreverse = 1/ Ka = 5.6 x 104 Kreverse is VERY LARGE, so OAc- completely uses up the H+ !

Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] initial change equilib 0.700 0.600 0 -x +x +x 0.700 - x 0.600 + x x

Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have K a = 1.8 x 10 -5 [H 3 O + ](0.600) 0.700 [H3O+] = 2.1 x 10-5 and pH = 4.68

Buffer Solutions Notice that the expression for calculating the H+ concentration of the buffer is [ H 3 O + ] = Orig. conc. of HOAc Orig. conc. of OAc - • K a This leads to a general equation for finding the H+ or OH- concentration of a buffer. [ H 3 O + ] = [Acid] [Conj. base] • K a [ OH - ] = [Base] [Conj. acid] • K b Notice that the H+ or OH- concentrations depend on K and the ratio of acid and base concentrations.

Henderson-Hasselbalch Equation [ H 3 O + ] = [Acid] [Conj. base] • K a Take the negative log of both sides of this equation pH = pK a - log [Acid] [Conj. base] pH = pK a + log [Conj. base] [Acid] OR This is called the Henderson-Hasselbalch equation.

Henderson-Hasselbalch Equation pH = pK a + log [Conj. base] [Acid] This shows that the pH is determined largely by the pKa of the acid and then adjusted by the ratio of acid and conjugate base.

Adding an Acid to a Buffer Problem: What is the new pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (a) Calculate [HCl] after adding 1.00 mL of HCl to 1.00 L of water M1 • V1 = M2 • V2 M2 = 1.00 x 10-3 M pH = 3.00

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer) The reaction occurs completely because K is very large.

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 1—Stoichiometry [H3O+] [OAc-] [HOAc] Before rxn 0.00100 0.600 0.700 Change -0.00100 -0.00100 +0.00100 After rxn 0 0.599 0.701

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to; a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [OAc-] [H3O+] Before rxn 0.701 0.599 0 Change -x +x +x After rxn 0.710-x 0.599+x x

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [OAc-] [H3O+] After rxn 0.710-x 0.599+x x Because [H3O+] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599.

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O OAc- + H3O+ [ H 3 O + ] = [HOAc] [OAc - • K a 0.701 0.599 ( 1 . 8 x 10 -5 ) [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed significantly upon adding HCl to the buffer!

Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10-5 M It is best to choose an acid such that [H3O+] is about equal to Ka (or pH ­ pKa). You get the exact [H3O+] by adjusting the ratio of acid to conjugate base.

Preparing a Buffer Solution Buffer prepared from HCO3- weak acid CO32- conjugate base HCO3- + H2O H3O+ + CO32-

Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M POSSIBLE ACIDS Ka HSO4- / SO42- 1.2 x 10-2 HOAc / OAc- 1.8 x 10-5 HCN / CN- 4.0 x 10-10 Best choice is acetic acid / acetate.

Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M [ H 3 O + ] = 5.0 x 10 -5 [HOAc] [OAc - (1.8 x 10 ) Solve for [HOAc]/[OAc-] ratio = 2.78/ 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30.

Preparing a Buffer A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. This simplifying approximation will be correct for all buffers with 3<pH<11, since the [H]+ will be small compared to the acid and conjugate base.

REVIEW PROBLEMS Calculate the pH of a 0.10 M HNO2 solution before and after making the solution 0.25 M in NaNO2. Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

REVIEW PROBLEMS SOLUTIONS Calculate the pH of a solution that is 0.18 M in Na2HPO4 and 0.12 M in NaH2PO4. 6.2 x 10-8 Suggest an appropriate buffer system for pH 5.0. SOLUTIONS

Titrations 45 pH Titrant volume, mL

Acid-Base Titrations Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly.

Acid-Base Titrations Additional NaOH is added. pH rises as equivalence point is approached.

Acid-Base Titrations Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point.

pH of solution of benzoic acid, a weak acid QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Equivalence point pH of solution of benzoic acid, a weak acid

Acid-Base Titrations EQUILIBRIUM PORTION QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? EQUILIBRIUM PORTION Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 [Bz-] [HBz] [OH-] equilib 0.020 - x x x K b = 1.6 x 10 -10 x 2 0.020 - Solving in the usual way, we find: x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25

QUESTION: You titrate 100. mL of a 0 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Half-way point

Acid-Base Reactions pH = 4.20 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point? HBz + H2O H3O+ + Bz- Ka = 6.3 x 10-5 [H3O+] = { [HBz] / [Bz-] } Ka At the half-way point, [HBz] = [Bz-], so [H3O+] = Ka = 6.3 x 10-5 pH = 4.20

Sample Problem Titration Curve 20.00 mL 0.30 M HC2H3O2 is titrated with 0.30 M NaOH. Calculate the pH at 0 mL NaOH added HC2H3O2 <---> H+ + C2H3O2- 0.30 0 0 - x + x + x 0.30 x x [C2H3O2-][H+] [HC2H3O2] x2 (0.30) Ka = = = 1.8 x 10-5 X = 2.3 x 10-3 M = [H+] pH = 2.63

Sample Problem 5.0 mL NaOH added HC2H3O2- + OH- ---> C2H3O2- + HOH 6.0 1.5 0 -1.5 -1.5 + 1.5 4.5 0 1.5 1.5 25.00 [C2H3O2- ] = = 0.060 M 4.5 25.00 [HC2H3O2 ] = = 0.18 M

Sample Problem HC2H3O2 <---> H+ + C2H3O2- 0.18 0 0.060 0.18 0 0.060 - x + x + x 0.18 x 0.060 [C2H3O2-][H+] [HC2H3O2] x(0.060) (0.18) Ka = = = 1.8 x 10-5 X = 5.4 x 10-5 M = [H+] pH = 4.27

Sample Problem 20.0 mL NaOH added HC2H3O2 + OH- ---> C2H3O2- + HOH 6.0 6.0 0 -6.0 -6.0 + 6.0 0 0 6.0 6.0 40.00 [C2H3O2- ] = = 0.15 M

Sample Problem C2H3O2- + HOH <---> OH- + HC2H3O2 0.15 0 0 0.15 0 0 - x + x + x 0.15 x x [C2H3O2-][H+] [HC2H3O2] x2 (0.15) Kb = = = 5.6 x 10-10 X = 9.2 x 10-6 M = [OH-] pH = 8.96

Sample Problem 30.0 mL NaOH added HC2H3O2 + OH- ---> C2H3O2- + HOH 6.0 9.0 0 -6.0 -6.0 + 6.0 0 3.0 6.0 3.0 50.0 [OH-] = = 0.060 M pOH = 1.22 pH = 12.78

Sample Problem Titration Curve 25.00 mL 0.300 M HCl is titrated with 0.7500 M NaOH. Calculate the pH at: 0 mL NaOH added [H+] = 0.300 M pH = 0.523

Sample Problem 5.00 mL NaOH added HCl + NaOH ---> NaCl + HOH 12.50 3.75 -3.75 -3.75 8.75 0 8.75 30.00 [H+] = = 0.292 M pH = 0.535

Sample Problem 16.67 mL NaOH added HCl + NaOH ---> NaCl + HOH 12.50 12.50 -12.50 -12.50 0 0 pH = 7.00

Sample Problem 20.00 mL NaOH added HCl + NaOH ---> NaCl + HOH 12.50 15.00 -12.50 -12.50 0 2.50 2.50 45.00 [OH-] = = 0.0556 M pH = 12.745

Sample Problem Titration Curve 20.00 mL 0.150 M NH3 is titrated with 0.100 M HCl. Calculate the pH at: 0 mL HCl added NH3 + HOH <---> NH4+ + OH- 0.150 0 0 - x + x + x 0.150 x x [NH4+][OH-] [NH3] x2 (0.150) Kb = = = 1.8 x 10-5 X = 1.6 x 10-3 M = [OH-] pH = 11.20

Sample Problem 10.0 mL HCl added NH3 + H+ ---> NH4+ 3.00 1.00 0 -1.00 -1.00 + 1.00 2.00 0 1.00 1.00 30.0 [NH4+] = = 0.0333 M 2.00 30.0 [NH3] = = 0.0667 M

Sample Problem NH4+ <---> NH3 + H+ 0.0333 0.0667 - x + x + x 0.0333 0.0667 - x + x + x 0.0333 0.0667 x [NH3] [H+] [NH4+] (0.0667)x (0.0333) Ka = = = 5.6 x 10-10 X = 2.8 x 10-10 M = [H+] pH = 9.55

Sample Problem 30.0 mL HCl added NH3 + H+ ---> NH4+ 3.00 3.00 0 -3.00 -3.00 + 3.00 0 0 3.00 3.00 50.00 [NH4+] = = 0.0600 M

Sample Problem NH4+ <---> NH3 + H+ 0.0600 0 0 - x + x + x 0.0600 0 0 - x + x + x 0.0600 x x [NH3] [H+] [NH4+] x2 (0.0600) Ka = = = 5.6 x 10-10 X = 5.8 x 10-6 M = [H+] pH = 5.24

Sample Problem 40.0 mL HCl added NH3 + H+ ---> NH4+ 3.00 4.00 0 -3.00 -3.00 + 3.00 0 1.00 3.00 1.00 60.00 [H+] = = 0.0167 pH = 1.78

Practice Problems 1. Determine the pH of a solution made by mixing 25.0 mL of 0.20 M nitric acid with 25.0 mL of 0.10 M potassium hydroxide. 2. Determine the pH at the equivalence point if 20.0 mL of 0.30 M HCN is titrated with 0.20 M sodium hydroxide. 3. Determine the pH at the equivalence point if 15.0 mL of 0.20 M nitric acid is titrated with 0.20 M ammonia.

Practice Problems 4. a) Calculate the pH of a solution made by mixing 50.0 mL of 0.15 M formic acid and 0.41 g of sodium formate. b) Calculate the pH if 10.0 mL of 0.10 M NaOH is added. 5. What is the pH of a solution made by mixing 25.0 mL of 0.20 M benzoic acid and 45.0 mL of 0.10 M sodium benzoate? 6. How many moles of sodium carbonate must be added to 0.20 mole sodium hydrogen carbonate in 250. mL to obtain a pH of 10.00?

Practice Problems 7. How many mL’s of 0.30 HCl must be added to 25.0 mL of 0.500 M sodium phosphate to produce a solution with a pH of 13.00? 8. A 20.00 mL sample of 0.500 M HNO3 is titrated with 0.500 M KOH. Calculate the pH of the solution: a) before the titration begins. b) when 10.00 mL of base have been added. c) when 19.00 mL of base have been added. d) at the equivalence point. e) when 21.00 mL of base have been added.

Practice Problems 9. A 20.00 mL sample of 0.5000 M formic acid is titrated with 0.500 M sodium hydroxide. Calculate the pH of the solution: a) before the titration begins. b) when 10.00 mL of base have been added. c) when 19.00 mL of base have been added. d) at the equivalence point. e) when 21.00 mL of base have been added.

Practice Problems 10. A 20.00 mL sample of 0.400 M ammonia is titrated with 0.200 M HCl. Calculate the pH of the solution: a) before the titration begins. b) when 20.00 mL of acid have been added. c) when 39.00 mL of acid have been added. d) at the equivalence point. e) when 41.00 mL of acid have been added.

Practice Problems Answers 1. 1.30 2. 11.23 3. 5.12 4. 3.66, 3.77 5. 4.15 6. .096 7. 9.1 8. .301, .77, 2.0, 7.00, 12.00 9. 2.03, 3.74, 5.03, 8.57, 12.0 10. 11.43, 9.25, 7.66, 5.07, 2.48 The End!!

HCl + NaOH ---> NaCl + HOH Acid-Base Reactions 1. Calculate the pH if 25.0 mL 0.20 M of HCl is added to 40.0 mL of 0.20 M NaOH. HCl + NaOH ---> NaCl + HOH 5.0 8.0 0 - 5.0 - 5.0 + 5.0 0 3.0 5.0 3.0 65.0 [OH-] = = 0.046 M pOH = 1.34 pH = 12.66

Acid-Base Reactions HCl + NH3 ---> NH4+ + Cl- 2. Calculate the pH at the equivalence point if 25.00 mL 0.20 M of HCl is titrated with 0.20 M NH3. HCl + NH3 ---> NH4+ + Cl- 5.0 5.0 0 0 - 5.0 - 5.0 + 5.0 0 0 5.0 5.0 50. [NH4+] = = 0.10 M

Acid-Base Reactions NH4+ <---> NH3 + H+ 2. [NH4+] = 0.10 M - x + x + x 0.10 x x [H+][NH3] [NH4+] x2 (0.10) Ka = = = 5.6 x 10-10 X = 7.5 x 10-6 M = [H+] pH = 5.12

Acid-Base Reactions HC2H3O2 + OH- ---> HOH + C2H3O2- 3. Calculate the pH at the equivalence point if 30.00 mL 0.20 M of HC2H3O2 is titrated with 0.30 M NaOH. HC2H3O2 + OH- ---> HOH + C2H3O2- 6.0 6.0 0 - 6.0 - 6.0 + 6.0 0 0 6.0 6.0 50. [C2H3O2- ]= = 0.12M

Acid-Base Reactions C2H3O2- + HOH <---> HC2H3O2+ OH- 3. [C2H3O2- ]= 0.12 M C2H3O2- + HOH <---> HC2H3O2+ OH- 0.12 0 0 - x + x + x 0.12 x x [HC2H3O2][OH-] [C2H3O2-] x2 (0.12) Kb = = = 5.6 x 10-10 X = 8.2 x 10-6 = [OH-] pH = 8.91

Sample Problem Calculate the pH of a 0.100 M HC2H3O2 solution. HC2H3O2 <---> H+ + C2H3O2- 0.100 0 0 - x + x + x 0.100 x x [C2H3O2-][H+] [HC2H3O2] x2 (0.100) Ka = = = 1.8 x 10-5 X = 1.3 x 10-3 M = [H+] pH = 2.89

Sample Problem Calculate the pH of a solution that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 HC2H3O2 <---> H+ + C2H3O2- 0.100 0 0.100 - x + x + x 0.100 x 0.100 [C2H3O2-][H+] [HC2H3O2] x(0.100) (0.100) Ka = = = 1.8 x 10-5 X = 1.8 x 10-5 M = [H+] pH = 4.74

Sample Problem 1.00 L H2O has a pH = 7.00 Calculate the pH if 0.010 mole HCL is added. .010 1.00 [H+ ] = = 0.010 pH = 2.00 Adding 0.010 mole HCl changes the pH from 7.00 to 2.00, 5.00 pH units.

Sample Problem 1.00 L 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 has a pH = 4.74 Calculate the pH if 0.010 mole HCL is added. C2H3O2- + H+ ---> HC2H3O2 0.100 0.010 0.100 - 0.010 - 0.010 + 0.010 0.090 0 0.110 .090 1.00 [C2H3O2- ] = = 0.090 M .110 1.00 [HC2H3O2 ] = = 0.110 M

Sample Problem HC2H3O2 <---> H+ + C2H3O2- 0.110 0 0.090 0.110 0 0.090 - x + x + x 0.110 x 0.090 [C2H3O2-][H+] [HC2H3O2] x(0.090) (0.110) Ka = = = 1.8 x 10-5 X = 2.2 x 10-5 M = [H+] pH = 4.66 Adding 0.010 mole HCl changes the pH from 4.74 to 4.66, only 0.08 pH units.

Preparing a Buffer Preparing Buffers 1. Solid/Solid: mix two solids. (Example 1) 2. Solid/Solution: mix one solid and one solution.(Example 2) 3. Solution/Solution: mix two solutions. (Example 3) 4. Neutralization: Mix weak acid with strong base (Examples 4 and 5) or weak base with strong acid. (Example 6)

Conjugates do not react!! Sample Problem 1. Calculate the pH of a solution made by mixing 1.5 moles of phthalic acid and 1.2 moles of sodium hydrogen phthalate in 500. mL of sol’n. Ka= 3.0 x 10-4 Conjugates do not react!! HA <---> H+ + A- 3.0 0 2.4 - x + x + x 3.0 - x x 2.4 + x [A-] [H+] [HA] (2.4 + x)(x) (3.0 + 0) Ka = = = 3.0 x 10-4 x = 3.7 x 10-4 M pH = 3.43

Conjugates do not react!! Sample Problem 2. How many moles of sodium acetate must be added to 500. mL of 0.25 M acetic acid to produce a solution with a pH of 5.50? X = moles NaC2H3O2 Conjugates do not react!! HC2H3O2 <---> H+ + C2H3O2- 0.25 0 x/0.500 - 3.2x10-6 + 3.2x10-6 + 3.2x10-6 0.25 3.2x10-6 x/0.500 [C2H3O2-][H+] [HC2H3O2] (x/0.500)(3.2x10-6 ) (0.25) Ka = = = 1.8 x 10-5 x = 0.70 mole NaC2H3O2

Conjugates do not react!! Sample Problem 3. How many mLs of 0.10 M sodium acetate must be added to 20.0 mL of 0.20 M acetic acid to produce a solution with a pH of 3.50? X = mLs NaC2H3O2 HC2H3O2 <---> H+ + C2H3O2- 0.20(20.0/20.0+x) 0 0.10(x/20.0+x) - 3.2x10-4 + 3.2x10-4 + 3.2x10-4 0.20(20.0/20.0+x) 3.2x10-4 0.10(x/20.0+x) Conjugates do not react!!

Sample Problem [C2H3O2-][H+] [HC2H3O2] Ka = {0.10(x/20.0+x)}(3.2x10-4 ) 0.20(20.0/20.0+x) = 1.8 x 10-5 x = 2.2 mL NaC2H3O2

Sample Problem 4. How many moles of potassium hydroxide must be added to 500. mL of .250 M HCN to produce a solution with a pH of 9.00? X = moles KOH OH- + HCN ---> CN- + HOH x 0.125 0 - x - x + x 0 0.125 - x x x .500 0.125 - x 0.500 [CN-] = [HCN] =

Sample Problem HCN <---> H+ + CN- (0.125 - x)/0.500 x/0.500 - 1.0 x 10-9 + 1.0 x 10-9 + 1.0 x 10-9 (0.125 - x)/0.500 1.0 x 10-9 x/0.500 [CN-][H+] [HCN] x/.500 (1.0 x 10-9) (.125 - x)/.500) Ka = = = 4.0x10-10 X = 0.036 mole KOH

Sample Problem HC2H3O2 + OH- ---> HOH + C2H3O2- 5. How many mLs of 0.30 M sodium hydroxide must be added to 25.0 mL of 0.500 M acetic acid to produce a solution with a pH of 4.10? X =mL NaOH HC2H3O2 + OH- ---> HOH + C2H3O2- 12.5 0.30x 0 - 0.30x - 0.30x + 0.30x 12.5 - 0.30x 0 0.30x 12.5 - 0.30x 25.0 + x 0.30x 25.0 + x [C2H3O2- ] = [HC2H3O2] =

Sample Problem HC2H3O2 <---> C2H3O2- + H+ (12.5 - 0.30x)/(25.0+x) 0.30x/(25.0+x) 0 - 7.9 x 10-5 + 7.9 x 10-5 + 7.9 x 10-5 (12.5 - 0.30x)/(25.0+x) 0.30x/(25.0+x) 7.9 x 10-5 [C2H3O2-][H+] [HC2H3O2] {0.30x/(25.0+x)} 7.9 x 10-5 (12.5 - 0.30x)/(25.0+x)) Ka = = = 1.8 x 10-5 X = 7.6 mL NaOH

Sample Problem HCl + NH3 ---> NH4+ + Cl- 6. Calculate the pH of a solution made by mixing 50.0 mL of 0.15 M NH3 and 20.0 mL of 0.10 M HCl. HCl + NH3 ---> NH4+ + Cl- 2.0 7.5 0 0 - 2.0 - 2.0 + 2.0 NA 0 5.5 2.0 NA 2.0 70.0 [NH4+] = = 0.029 M 5.5 70.0 [NH3] = = 0.079 M

Sample Problem NH4+ <---> NH3 + H+ 0.029 0.079 0 - x + x + x 0.029 0.079 0 - x + x + x 0.029 0.079 x [H+][NH3] [NH4+] x(0.079) (0.029) Ka = = = 5.6 x 10-10 X = 2.1 x 10-10 M = [H+] pH = 9.68 For more practice do all General problems 84-101

Sample Problems 1. Calculate the pH of a 0.10 M HNO2 solution before and after making the solution 0.25 M in NaNO2.

Sample Problem Calculate the pH of a 0.10 M HNO2 solution. HNO2 <---> H+ + NO2- 0.10 0 0 - x + x + x 0.10 - x x x [NO2-][H+] [HNO2] x2 (.10 - x) Ka = = = 4.5 x 10-4 X = 6.5 x 10-3 M = [H+] pH = 2.19

Sample Problem Calculate the pH of a 0.10 M HNO2 and 0.25 M NaNO2 solution. HNO2 <---> H+ + NO2- 0.10 0 0.25 - x + x + x 0.10 - x x 0.25 + x [NO2-] [H+] [HNO2] x(0.25 + x) (0.10 - x) Ka = = = 4.5 x 10-4 X = 1.8 x 10-4 M = [H+] pH = 3.74

Sample Problems 2. Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

Sample Problem Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid. HC2H3O2 <---> H+ + C2H3O2- 0.15 0 0.10 - x + x + x 0.15 x 0.10 [C2H3O2-] [H+] [HC2H3O2] x(0.10) (0.15) Ka = = = 1.8 x 10-5 X = 2.7 x 10-5 M = [H+] pH = 4.57

Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid. HC2H3O (.500L)(.15M) = .075 mole NaC2H3O (.500L)(.10M) = .050 mole NaOH (1.0g)(40.0g/mole) = .025 mole HC2H3O2 + OH- ---> HOH + C2H3O2- 0.075 0.025 0.050 - 0.025 - 0.025 + 0.025 0.050 0 0 .075 .050 .500 .075 .500 [HC2H3O2] = = 0.10 [C2H3O2-] = = 0.15 M

Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid. HC2H3O2 <---> H+ + C2H3O2- 0.10 0.15 - x + x + x 0.15 x 0.10 [C2H3O2-][H+] [HC2H3O2] x(0.15) (0.10) Ka = = = 1.8 x 10-5 X = 1.2 x 10-5 M = [H+] pH = 4.92

Sample Problems 3. Calculate the pH of a solution that is 0.18 M in Na2HPO4 and 0.12 M in NaH2PO4.

Sample Problem H2PO4- <---> H+ + HPO42- 0.12 0 0.18 - x + x + x 0.12 0 0.18 - x + x + x 0.12 x 0.18 [HPO42-][H+] [H2PO4-] x(0.18) (0.12) Ka = = = 6.2 x 10-8 X = 4.1 x 10-8 M = [H+] pH = 7.38

Sample Problems 4. Suggest an appropriate buffer system for pH 5.0. Name of Acid Ka pKa Oxalic Acid 3.8 x 10-2 1.42 Hydrogen Sulfate Ion 1.2 x 10-2 1.92 Phosphoric Acid 7.1 x 10-3 2.15 Formic Acid 1.8 x 10-4 3.74 Hydrogen Oxalate Ion 5.0 x 10-4 4.30 Acetic Acid 1.8 x 10-5 4.74 Dihydrogen Phosphate Ion 6.3 x 10-8 7.20 Boric Acid 6.0 x 10-10 9.22 Ammonium Ion 9.6 x 10-10 9.25 Hydrogen Carbonate Ion 4.7 x 10-11 10.33 Hydrogen Phosphate Ion 4.4 x 10-13 12.36