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Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro. Chapter 16 Aqueous Ionic Equilibrium. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2008, Prentice Hall. ethylene glycol (aka 1,2–ethandiol). The Danger of Antifreeze.
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Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro Chapter 16Aqueous IonicEquilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall
ethylene glycol (aka 1,2–ethandiol) The Danger of Antifreeze • each year, thousands of pets and wildlife die from consuming antifreeze • most brands of antifreeze contain ethylene glycol • sweet taste • initial effect drunkenness • metabolized in the liver to glycolic acid • HOCH2COOH • if present in high enough concentration in the bloodstream, it overwhelms the buffering ability of HCO3−, causing the blood pH to drop • when the blood pH is low, it ability to carry O2 is compromised • acidosis • the treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol
Buffers • buffers are solutions that resist changes in pH when an acid or base is added • they act by neutralizing the added acid or base • but just like everything else, there is a limit to what they can do, eventually the pH changes • many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion Tro, Chemistry: A Molecular Approach
Making an Acid Buffer Tro, Chemistry: A Molecular Approach
How Acid Buffers WorkHA(aq) + H2O(l) A−(aq) + H3O+(aq) • buffers work by applying Le Châtelier’s Principle to weak acid equilibrium • buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it • you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium • the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant Tro, Chemistry: A Molecular Approach
H2O How Buffers Work new HA HA HA A− A− H3O+ + Added H3O+ Tro, Chemistry: A Molecular Approach
H2O How Buffers Work new A− A− HA A− HA H3O+ + Added HO− Tro, Chemistry: A Molecular Approach
Common Ion Effect HA(aq) + H2O(l) A−(aq) + H3O+(aq) • adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left • this causes the pH to be higher than the pH of the acid solution • lowering the H3O+ ion concentration Tro, Chemistry: A Molecular Approach
Common Ion Effect Tro, Chemistry: A Molecular Approach
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? HC2H3O2 + H2O C2H3O2 + H3O+ Tro, Chemistry: A Molecular Approach
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? HC2H3O2 + H2O C2H3O2 + H3O+ x +x +x x 0.100 x 0.100 + x Tro, Chemistry: A Molecular Approach
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 0.100 x 0.100 +x Tro, Chemistry: A Molecular Approach
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 x = 1.8 x 10-5 the approximation is valid Tro, Chemistry: A Molecular Approach
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? 0.100 x 0.100 + x x x = 1.8 x 10-5 Tro, Chemistry: A Molecular Approach
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Tro, Chemistry: A Molecular Approach
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 the values match Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O F + H3O+ Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O F + H3O+ x +x +x x 0.14 x 0.071 + x Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 0.14 x 0.071 +x Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 x = 1.4 x 10-3 the approximation is valid Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? 0.14 x 0.071 + x x x = 1.4 x 10-3 Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 the values are close enough Tro, Chemistry: A Molecular Approach
Henderson-Hasselbalch Equation • calculating the pH of a buffer solution can be simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation • the equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate base • as long as the “x is small” approximation is valid Tro, Chemistry: A Molecular Approach
Deriving the Henderson-Hasselbalch Equation Tro, Chemistry: A Molecular Approach
Ex 16.2 - What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2? HC7H5O2 + H2O C7H5O2 + H3O+ Ka for HC7H5O2 = 6.5 x 10-5 Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Tro, Chemistry: A Molecular Approach
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O F + H3O+ Tro, Chemistry: A Molecular Approach
Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? • the Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable • generally, the “x is small” approximation will work when both of the following are true: • the initial concentrations of acid and salt are not very dilute • the Ka is fairly small • for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of Ka Tro, Chemistry: A Molecular Approach
How Much Does the pH of a Buffer Change When an Acid or Base Is Added? • though buffers do resist change in pH when acid or base are added to them, their pH does change • calculating the new pH after adding acid or base requires breaking the problem into 2 parts • a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other • added acid reacts with the A− to make more HA • added base reacts with the HA to make more A− • an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−] Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2 + H2O Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2 + H2O Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O C2H3O2 + H3O+ Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O C2H3O2 + H3O+ x +x +x x 0.090 x 0.110 + x Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10-5 0.110 +x 0.090 x Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10-5 x = 1.47 x 10-5 the approximation is valid Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? 0.090 x x 0.110 + x x = 1.47 x 10-5 Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10-5 the values match Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O C2H3O2 + H3O+ Ka for HC2H3O2 = 1.8 x 10-5 Tro, Chemistry: A Molecular Approach
Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O C2H3O2 + H3O+ pKa for HC2H3O2 = 4.745 Tro, Chemistry: A Molecular Approach
Ex 16.3 – Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water? HC2H3O2 + H2O C2H3O2 + H3O+ pKa for HC2H3O2 = 4.745 Tro, Chemistry: A Molecular Approach
H2O(l) + NH3(aq) NH4+(aq) + OH−(aq) Basic BuffersB:(aq) + H2O(l) H:B+(aq) + OH−(aq) • buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl− Tro, Chemistry: A Molecular Approach
Ex 16.4 - What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl? NH3 + H2O NH4+ + OH− Tro, Chemistry: A Molecular Approach
Buffering Effectiveness • a good buffer should be able to neutralize moderate amounts of added acid or base • however, there is a limit to how much can be added before the pH changes significantly • the buffering capacity is the amount of acid or base a buffer can neutralize • the buffering range is the pH range the buffer can be effective • the effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base Tro, Chemistry: A Molecular Approach
Effect of Relative Amounts of Acid and Conjugate Base a buffer is most effective with equal concentrations of acid and base Buffer 1 0.100 mol HA & 0.100 mol A- Initial pH = 5.00 Buffer 12 0.18 mol HA & 0.020 mol A- Initial pH = 4.05 pKa (HA) = 5.00 HA + OH− A + H2O after adding 0.010 mol NaOH pH = 5.09 after adding 0.010 mol NaOH pH = 4.25
Effect of Absolute Concentrations of Acid and Conjugate Base a buffer is most effective when the concentrations of acid and base are largest Buffer 1 0.50 mol HA & 0.50 mol A- Initial pH = 5.00 Buffer 12 0.050 mol HA & 0.050 mol A- Initial pH = 5.00 pKa (HA) = 5.00 HA + OH− A + H2O after adding 0.010 mol NaOH pH = 5.02 after adding 0.010 mol NaOH pH = 5.18
Effectiveness of Buffers • a buffer will be most effective when the [base]:[acid] = 1 • equal concentrations of acid and base • effective when 0.1 < [base]:[acid] < 10 • a buffer will be most effective when the [acid] and the [base] are large Tro, Chemistry: A Molecular Approach
Buffering Range • we have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 • substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pH Highest pH therefore, the effective pH range of a buffer is pKa± 1 when choosing an acid to make a buffer, choose one whose is pKa is closest to the pH of the buffer