Chapter 15: Applications of Aqueous Equilibria 12/7/2018 Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.

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

Neutralization Reactions Chapter 15: Applications of Aqueous Equilibria Neutralization Reactions 12/7/2018 Strong Acid-Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Assuming complete dissociation: H2O(l) H1+(aq) + OH1-(aq) 2H2O(l) H3O1+(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. Copyright © 2008 Pearson Prentice Hall, Inc.

Strong acid-Strong base neutralization 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 12/7/2018 Weak Acid-Strong Base CH3CO2H(aq) + NaOH(aq) H2O(l) + NaCH3CO2(aq) Assuming complete dissociation: H2O(l) + CH3CO21-(aq) CH3CO2H(aq) + OH1-(aq) (net ionic equation) 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%. After neutralization: pH > 7 Copyright © 2008 Pearson Prentice Hall, Inc. 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

Neutralization Reactions Chapter 15: Applications of Aqueous Equilibria 12/7/2018 Neutralization Reactions Strong Acid-Weak Base HCl(aq) + NH3(aq) NH4Cl(aq) Assuming complete dissociation: H3O1+(aq) + NH3 (aq) H2O(l) + NH41+(aq) NH41+(aq) H1+(aq) + NH3(aq) or (net ionic equation) 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. 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

Neutralization Reactions Chapter 15: Applications of Aqueous Equilibria Neutralization Reactions 12/7/2018 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. 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

Chapter 15: Applications of Aqueous Equilibria The Common-Ion Effect 12/7/2018 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. Example of Le Chatelier’s principle E.g Adding HCl and NaOH to a solution of acetic acid would shift the equilibrium to which direction? H3O1+(aq) + CH3CO21-(aq) CH3CO2H(aq) + H2O(l) Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 15: Applications of Aqueous Equilibria The Common-Ion Effect 12/7/2018 Le Châtelier’s Principle H3O1+(aq) + CH3CO21-(aq) CH3CO2H(aq) + H2O(l) The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 15: Applications of Aqueous Equilibria 12/7/2018 Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 15: Applications of Aqueous Equilibria 12/7/2018 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. Ka = 1.8 x 10-5 H3O1+(aq) + CH3CO21-(aq) CH3CO2H(aq) + H2O(l) [H3O1+][CH3CO21-] [CH3CO2H] Ka = 25 °C This assumes the complete dissociation of sodium acetate as shown previously. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 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 12/7/2018 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. 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 Buffer Solutions 12/7/2018 H3O1+(aq) + CH3CO21-(aq) CH3CO2H(aq) + H2O(l) Weak acid Conjugate base (NaCH3CO2) Addition of OH1- to a buffer: H2O(l) + CH3CO21-(aq) CH3CO2H(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. Copyright © 2008 Pearson Prentice Hall, Inc. Copyright © 2008 Pearson Prentice Hall, Inc.

Buffer Solutions Addition of H3O1+ to a buffer: 100% H2O(l) + CH3CO2H(aq) CH3CO21-(aq) + H3O1+(aq) 100%

Example pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H2CO3/HCO3-) With addition of H3O+ With addition of -OH

Chapter 15: Applications of Aqueous Equilibria 12/7/2018 Buffer Solutions Copyright © 2008 Pearson Prentice Hall, Inc.

Buffer capacity A measure of the amount of acid or base that a buffer solution can absorb without a significant change in pH Depends on how much weak acid and conjugated base is present For equal volume of solution, the more concentration the solution, the greater the buffer capacity For solution with the same concentration, increasing the volume increases the buffer capacity

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

Example Calculate the pH of 0.100L of a buffer solution that is 0.25M in HF and 0.50 M in NaF, Ka = 6.3 x 10-4 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

Example calculate the pH of a 50.0 ml buffer solution that is 0.50 M in NH3 and 0.20 M NH4Cl. For ammonia, pKb = 4.75 Calculate the pH after addition of 150.0 mg HBr

The Henderson-Hasselbalch Equation Chapter 15: Applications of Aqueous Equilibria 12/7/2018 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. Copyright © 2008 Pearson Prentice Hall, Inc.

The Henderson-Hasselbalch Equation Chapter 15: Applications of Aqueous Equilibria 12/7/2018 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. Copyright © 2008 Pearson Prentice Hall, Inc.

Example 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 = 5.6 x 10-11