Ch18part2

The Common Ion EffectSection 18.2: 

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 ____________. The pH will go _____________. After all, NH4+ is an acid! The Common Ion Effect Section 18.2 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

Slide2: 

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 QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq)

Slide3: 

QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Assuming x is << 0.25, we have [OH-] = x = 0.0021 M This gives pOH = 2.67 and so pH = 11.33 for 0.25 M NH3

Slide4: 

Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Calc pH = 11.33 for 0.25 M NH3 We expect that the pH will decline on adding NH4Cl.

The Common Ion EffectSection 18.2: 

Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) [NH3] [NH4+] [OH-] initial 0.25 0.10 0 change -x +x +x equilib 0.25 - x 0.10 + x x The Common Ion Effect Section 18.2

The Common Ion EffectSection 18.2: 

Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Because equilibrium shifts left, x is MUCH less than 0.0021 M, the value without NH4Cl. The Common Ion Effect Section 18.2

The Common Ion EffectSection 18.2: 

Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) [OH-] = x = 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. The Common Ion Effect Section 18.2

Buffer Solutions: 

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 HOAc + OAc- H2PO4- + HPO42- Weak Base + Conj. Acid NH3 + NH4+ Buffer Solutions

Buffer Solutions: 

ACID USES UP ADDED OH- OAc- + H2O HOAc + OH- 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 snatches up OH- ! Buffer Solutions

Buffer Solutions: 

CONJ. BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ 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 snatches up H+ ! Buffer Solutions

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 Buffer Solutions

Buffer Solutions: 

[HOAc] [OAc-] [H3O+] initial 0.700 0.600 x change -x +x +x equilib 0.700 - x 0.600 + x x 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 Buffer Solutions

Buffer Solutions: 

[HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have [H3O+] = 2.1 x 10-5 and pH = 4.68 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 Buffer Solutions

Buffer Solutions: 

Notice that the H+ or OH- concs. depend on K and the ratio of acid and base concs. Buffer Solutions

Buffer Solutions: 

Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? Ka = 1.8 x 10-5 pH = 4.74 + log [0.600M/0.700M] pH = 4.74 –0.07 = 4.67

Henderson-Hasselbalch Equation: 

Henderson-Hasselbalch Equation 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: 

Adding an Acid to a Buffer Problem: What is the 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) Calc. [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: 

Adding an Acid to a Buffer 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. 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)

Adding an Acid to a Buffer: 

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

Adding an Acid to a Buffer: 

Adding an Acid to a Buffer 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 What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

Adding an Acid to a Buffer: 

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

Adding an Acid to a Buffer: 

Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [H3O+] [OAc-] 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. What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

Adding an Acid to a Buffer: 

Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed on adding HCl to the buffer! What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

Preparing a 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). —then you get the exact [H3O+] by adjusting the ratio of acid to conjugate base.

Preparing a Buffer Solution: 

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

Preparing a Buffer: 

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 Preparing a Buffer

Preparing a Buffer: 

You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Preparing a Buffer

Preparing a Buffer: 

You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Solve for [HOAc]/[OAc-] ratio = 2.78/ 1 Preparing a Buffer

Preparing a Buffer: 

You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M 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

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. Preparing a Buffer