Chapter 15 Applications of Aqueous Equilibria Chemistry: McMurry and Fay, 6th Edition Chapter 15: Applications of Aqueous Equilibria 4/26/2017 12:15:57 PM Chapter 15 Applications of Aqueous Equilibria Copyright © 2011 Pearson Prentice Hall, Inc.

Neutralization Reaction General Formula Acid + Base - Water + Salt

Neutralization Reactions Chapter 15: Applications of Aqueous Equilibria 4/26/2017 Neutralization Reactions Strong Acid-Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Assuming complete dissociation: 2H2O(l) H3O1+(aq) + OH1-(aq) H2O(l) H1+(aq) + OH1-(aq) or Recall that we’re considering H1+ and H3O1+ equivalent. HCl is a strong acid. NaOH and NaCl are soluble ionic compounds. Water and NaCl are the products after neutralization. Since neither sodium nor chloride ions has any substantial acidic/basic properties, the solution is neutral (pH=7) upon neutralization. The titration generally goes to 100%. (net ionic equation) After neutralization: pH = 7 Copyright © 2008 Pearson Prentice Hall, Inc.

Strong acid-Strong base neutralization 2H2O(l) H3O1+(aq) + OH1-(aq) When the number moles of acid and base are mixed together [H3O+] = [-OH] = 1.0 x 10-7M Reaction proceeds far to the right

Neutralization Reactions Chapter 15: Applications of Aqueous Equilibria Neutralization Reactions 4/26/2017 Weak Acid-Strong Base CH3CO2H(aq) + NaOH(aq) H2O(l) + NaCH3CO2(aq) Assuming complete dissociation: H2O(l) + CH3CO21-(aq) CH3CO2H(aq) + OH1-(aq) CH3CO2H is a weak acid. NaOH and NaCH3CO2 are soluble ionic compounds. Water and NaCH3CO2 are the products after neutralization. Sodium ion does not have substantial acidic/basic properties; but, the solution is basic (pH>7) upon neutralization because of the presence of the conjugate base, CH3CO21-, upon neutralization. The titration generally goes to 100%. (net ionic equation) After neutralization: pH > 7 Copyright © 2008 Pearson Prentice Hall, Inc.

Weak acid-strong base neutralization Neutralization of any weak acid by a strong base goes 100% to completion -OH has a great infinity for protons H2O(l) + CH3CO21-(aq) CH3CO2H(aq) + OH1-(aq)

Neutralization Reactions Chapter 15: Applications of Aqueous Equilibria 4/26/2017 Neutralization Reactions Strong Acid-Weak Base HCl(aq) + NH3(aq) NH4Cl(aq) Assuming complete dissociation: NH41+(aq) H1+(aq) + NH3(aq) or (net ionic equation) H3O1+(aq) + NH3 (aq) H2O(l) + NH41+(aq) HCl is a strong acid. NH4Cl is a soluble ionic compound. Water and NH4Cl are the products after neutralization. Chloride ion does not have substantial acidic/basic properties; but, the solution is acidic (pH<7) upon neutralization because of the presence of the conjugate acid, NH41+, upon neutralization. The titration generally goes to 100%. After neutralization: pH < 7 Copyright © 2008 Pearson Prentice Hall, Inc.

Strong acid-weak base neutralization Neutralization of any weak base by a strong acid goes 100% to completion H3O+ has a great infinity for protons H3O1+(aq) + NH3 (aq) H2O(l) + NH41+(aq)

Neutralization Reactions Chapter 15: Applications of Aqueous Equilibria Neutralization Reactions 4/26/2017 Weak Acid-Weak Base CH3CO2H(aq) + NH3(aq) NH4CH3CO2(aq) NH41+(aq) + CH3CO21-(aq) CH3CO2H(aq) + NH3(aq) (net ionic equation) After neutralization: pH = ? Since both the acid and the base are weak, it complicates the titration. Typical general chemistry titrations are not done with both of them weak. The titration generally does not go to 100%. The pH of the solution can be acidic or basic and depends on the weak acid and weak base. Copyright © 2008 Pearson Prentice Hall, Inc.

Weak acid-weak base neutralization Less tendency to proceed to completion than neutralization involving strong acids and strong bases

Examples Write balanced net ionic equations for the neutralization of equal molar amounts of the following acids and bases. Indicate whether the pH after the neutralization is greater than, equal to or less than 7 HNO2 and KOH Ka HNO2 = 4.5 x 10-4

Chemistry: McMurry and Fay, 6th Edition Chapter 15: Applications of Aqueous Equilibria 4/26/2017 12:15:57 PM The Common-Ion Effect Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2–(aq) Copyright © 2011 Pearson Prentice Hall, Inc.

Chemistry: McMurry and Fay, 6th Edition Chapter 15: Applications of Aqueous Equilibria 4/26/2017 12:15:57 PM The Common-Ion Effect The pH of 0.10 M acetic acid is 2.89. Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. 25 °C p590-591 This assumes the complete dissociation of sodium acetate as shown previously. Copyright © 2011 Pearson Prentice Hall, Inc.

The Common-Ion Effect Chemistry: McMurry and Fay, 6th Edition Chapter 15: Applications of Aqueous Equilibria 4/26/2017 12:15:57 PM The Common-Ion Effect Copyright © 2011 Pearson Prentice Hall, Inc.

The Common-Ion Effect CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2–(aq) Chemistry: McMurry and Fay, 6th Edition Chapter 15: Applications of Aqueous Equilibria 4/26/2017 12:15:57 PM The Common-Ion Effect Le Châtelier’s Principle CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2–(aq) The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left. Copyright © 2011 Pearson Prentice Hall, Inc.

The Common-Ion Effect Chemistry: McMurry and Fay, 6th Edition Chapter 15: Applications of Aqueous Equilibria 4/26/2017 12:15:57 PM The Common-Ion Effect Copyright © 2011 Pearson Prentice Hall, Inc.

Example In 0.15 M NH3, the pH is 11.21 and the percent dissociation is 1.1%. Calculate the concentrations of NH3, pH and percent dissociation of ammonia in a solution that is 0.15M and 0.45 MNH4Cl

Chapter 15: Applications of Aqueous Equilibria Buffer Solutions 4/26/2017 Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH. Weak acid + Conjugate base CH3CO2H + CH3CO21- HF + F1- NH41+ + NH3 H2PO42- + HPO42- For Example: Acidic/basic salts with the weak base/acid. Copyright © 2008 Pearson Prentice Hall, Inc.

Buffer Solutions Add a small amount of base (-OH) to a buffer solution Acid component of solution neutralizes the added base Add a small amount of acid (H3O+) to a buffer solution Base component of solution neutralizes the added acid The addition of –OH or H3O+ to a buffer solution will change the pH of the solution, but not as drastically as the addition of –OH or H3O+ to a non-buffered solution

Chapter 15: Applications of Aqueous Equilibria 4/26/2017 Buffer Solutions H3O1+(aq) + A1-(aq) HA(aq) + H2O(l) Weak acid Conjugate base (M+A-) Addition of OH1- to a buffer: H2O(l) + A1-(aq) HA(aq) + OH1-(aq) 100% Both the acid and the conjugate base are present. For a buffer made from a weak acid and conjugate base, the conjugate base is in the form of a salt. For a buffer made from a weak base and conjugate acid, the conjugate acid is in the form of a salt. Addition of H3O1+ to a buffer: H2O(l) + HA(aq) A1-(aq) + H3O1+(aq) 100% Copyright © 2008 Pearson Prentice Hall, Inc.

Buffer Solutions Chemistry: McMurry and Fay, 6th Edition Chapter 15: Applications of Aqueous Equilibria 4/26/2017 12:15:57 PM Buffer Solutions Copyright © 2011 Pearson Prentice Hall, Inc.

Example pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H2CO3/HCO3-). Write the neutralization equation fro the following effects With addition of HCl With addition of NaOH

Example Calculate the pH of 0.100L of a buffer solution that is 0.25M in HF and 0.50 M in NaF. What is the change in pH on addition of 0.002 mol HCl What is the change in pH on addition of 0.010 moles KOH Calculate the pH after addition of 0.080 moles HBr * Assume the volume remains constant

Example Calculate the pH of the buffer that results from mixing 60.0mL of 0.250M HCHO2 and 15.0 mL of 0.500M NaCHO2 Ka = 1.8 x 10-4 Calculate the pH after addition of 10.0 mL of 0.150 MHBr. Assume volume is additive

The Henderson-Hasselbalch Equation Chapter 15: Applications of Aqueous Equilibria 4/26/2017 The Henderson-Hasselbalch Equation H3O1+(aq) + CH3CO21-(aq) CH3CO2H(aq) + H2O(l) Weak acid Conjugate base H3O1+(aq) + Base(aq) Acid(aq) + H2O(l) [H3O1+][Base] [Acid] [Acid] [Base] Ka = [H3O1+] = Ka Copyright © 2008 Pearson Prentice Hall, Inc.

The Henderson-Hasselbalch Equation Chapter 15: Applications of Aqueous Equilibria 4/26/2017 The Henderson-Hasselbalch Equation [Acid] [Base] [H3O1+] = Ka [Acid] [Base] -log([H3O1+]) = -log(Ka) - log [Base] [Acid] -log(x/y) = log(y/x) pH = pKa + log Copyright © 2008 Pearson Prentice Hall, Inc.

Examples Calculate the pH of a buffer solution that is 0.50 M in benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate (NaC7H5O2). Ka = 6.5 x 10-5 How would you prepare a NaHCO3-Na2CO3 buffer solution that has pH = 10.40 Ka2 = 4.7 x 10-11