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Chapter 17: Acid-base equilibria

Chapter 17: Acid-base equilibria. Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor. Acid-ionization equilibria. Weak acids such as acetic acid involve the following equilibrium when dissolved in water: HC 2 H 3 O 2 ( aq ) + H 2 O( aq ) = H 3 O + ( aq ) + C 2 H 3 O 2 - ( aq )

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Chapter 17: Acid-base equilibria

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  1. Chapter 17: Acid-base equilibria Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor

  2. Acid-ionization equilibria • Weak acids such as acetic acid involve the following equilibrium when dissolved in water: HC2H3O2(aq) + H2O(aq) = H3O+(aq) + C2H3O2-(aq) • Since acetic acid is a weak acid, only a small amount of its molecules are ionized, and this equilibrium lies far to the left • General weak acid equilibrium: HA + H2O = H3O+ + A- • Ka: acid-dissociationconstant • Related to equilibrium constant (Kc)

  3. Calculating acid-dissociation constant • Most Kavalues were obtained by recording the pH of a weak acid solution at equilibrium • To solve the problem, the molarity of the weak acid solution must be known (this refers to how the solution was prepared, and can be considered the initial value on an ICE table) • x = [H3O+] = [A-] = antilog(-pH) • If x is sufficiently small, an approximation can be used in the Kaequation, (Ca – x)  ca • Degree of ionization: percentage of solute that’s in ionized form: ([A-]/Ca) x 100%

  4. Calculating Ka • A 0.025 M solution of lactic acid has a pH of 2.75. • Calculate Ka • Calculate the degree of ionization. • Construct ICE table • Construct Ka equation • Calculate [H3O+] from pH • Calculate Ka • Calculate degree of ionization

  5. Using Ka in calculations • What is the pH of a 0.10 M acetic acid solution? Ka = 1.7 x 10-5 • Follow same procedure as last slide, in constructing ICE table and Kaexpression • If Ca / Ka > 100, you can simplify Kaexpression to just Ka = (x2 / Ca)

  6. Solving for pH when the approximation is not valid • What is the pH of an aqueous 0.0030 M solution of pyruvic acid, Ka = 1.4 x 10-4 • If Ca / Ka < 100, the approximation is not valid, and Ka =x2 / (Ca – x) must be solved with the quadratic equation

  7. Polyprotic acids • Polyprotic acids have more than one acidic proton • Separate Kavalues for each acidic proton • Kafor the most acidic proton, Ka1 is generally much larger than Ka2 • Proton concentration as a result of Ka2 equilibrium is negligible, can be ignored in pH calculations • For triprotic acids like H3PO4, Ka3is even smaller than Ka2

  8. Polyprotic acid calculations • What is the pH of a 0.25 M solution of sulfurous acid (H2SO3)? What is the concentration of sulfite ion, SO32- at equilibrium? Ka1 = 1.3 x 10-2, Ka2 = 6.3 x 10-8 • Solve for pH by just using Ka1reaction • Solve for [SO32-] by using equilibrium values from Ka1reaction as initial values in Ka2 equilibrium

  9. Base-ionization equilibria • Kb (the base-ionization constant) can be found by a method similar to that of Ka • B + H2O = HB+ + OH- • Weak bases are nearlyall amines • All contain nitrogenwith lone pair that acceptsproton from water

  10. Weak base pH calculation • What is the pH of a 0.20 M solution of ammonia in water? Kb = 1.8 x 10-5 • Use an ICE table and construct a Kbequation to solve for [OH-] • Convert to [H3O+] using Kw, and then to pH • Or, convert to pOH and subtract from 14 to get pH

  11. Acid-base properties of salt solutions • A salt is an ionic compound resulting from an acid-base reaction • NaCl (aq) is obtained by neutralizing NaOH and HCl • Consider the individual ions produced in a salt to determine whether a salt solution is acidic or basic • Hydrolysis of an ion: acid-base reaction of an ion with water

  12. Predicting whether a salt is an acid or base • Is an aqueous solution of NH4Cl acidic or basic? • Consider reaction of individual ions with water NH4+ + H2O = NH3 + H3O+ Cl- + H2O = no reaction • Ammonium can donate a proton to water, so it will make the solution acidic • Chloride does not accept a proton from water, since HCl is a strong acid • Therefore, the solution will be acidic

  13. Generalizations about salt pH • A salt of a strong base and strong acid is neutral since no ions will react with water • A salt of a strong base and a weak acid is basic (the anion is strongly basic and will produce hydroxide) • A salt of a weak base and a strong acid is acidic (the cation is strongly acidic and will produce hydronium) • A salt of a weak base and a weak acid can be either acidic or basic. If Kaof the cation is larger than Kbfor the anion, the solution is acidic, and vice-versa.

  14. pH of salt solutions • The Kaand Kb of a conjugate acid-base pair are related • Ex. for the HCN / CN- conjugate acid-base pair, both hydrolysis equilibria can be drawn HCN + H2O = H3O+ + CN- Ka CN- + H2O = HCN + OH- Kb • The sum of these two equilibria is water’s autoionization equilibrium, 2H2O = H3O+ + OH- Kw • When the two reactions are added, the equilibrium constants are multiplied, so KaKb = Kw

  15. pH of salt solutions • What is the pH of a 0.015 M solution of sodium benzoate? Kaof benzoic acid is 6.3 x 10-5 • Consider the hydrolysis reactions of each of the ions • Calculate Kbof sodium benzoate from the Kaof its conjugate acid • Solve for [OH-], convert to pOH and then pH

  16. Common-ion effect • Addition of another solute to a weak acid or base equilibrium can disrupt the equilibrium • Ex. if HCl is added to a solution of acetic acid: HC2H3O2 + H2O = C2H3O2- + H3O+ • Since HCl is a strong acid, it will add H3O+ to the system, and the equilibrium above will shift to the left (Le Chatelier’s principle) • The ionization of weak acids is repressed by the addition of strong acid • The same applies to the addition of extra conjugate base to the equilibrium: if acetate were added, the ionization of acetic acid would be repressed

  17. Common-ion calculation • What is the concentration of formate ion, CHO2-, in a solution of 0.10 M HCHO2 and 0.20 M HCl? Ka = 1.7 x 10-4 • Set up the equilibrium with the weak acid on the left HCHO2 + H2O = CHO2- + H3O+ • Initial concentrations of HCHO2 and H3O+ will be as given in the problem • An approximation can be used if x is sufficiently smaller than either of the concentrations

  18. Buffers • Buffer: a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid • Ability to resist changes in pH • In a solution of equimolar amounts of a weak acid and its conjugate base… • If acid is added, the conjugate base will accept the protons to neutralize it • If base is added, the weak acid will supply protons to neutralize it

  19. pH of a buffer • What is the pH of a buffer prepared by adding 30.0 mL of 0.15 M acetic acid to 70.0 mL of 0.20 M sodium acetate? Ka = 1.7 x 10-5 • Calculate initial molarities • Use those as initial values in ICE and solve like a common-ion problem • Use the approximation if x is sufficiently small

  20. Adding a strong acid or base to a buffer • First use stoichiometry to determine immediate effects of adding strong acid/base • Ex. adding HCl to an acetic acid/sodium acetate buffer solution HC2H3O2 + H2O = C2H3O2- + H3O+ • Assume that acetate will be stoichiometrically reduced in moles by the moles of H3O+ added • Then allow the solution to equilibrate using new calculated concentrations

  21. Adding a strong acid or base to a buffer • Using the buffer in the last example, calculate the effect of adding 9.5 mL of 0.10 M HCl • Calculate moles of H3O+ added to solution, subtract this from the base, and add this to the acid • Calculate new molarities and solve the equilibrium equation as before

  22. Henderson-Hasselbalch equation • In buffer solutions made with weak acids and their salt, the equilibrium concentrations of HA and A- differ very little from their initial values • Using the same assumption that x is very small, the equation can be rearranged • pH = pKa + log ( [A-] / [HA] ) • for weak acid / salt buffers only! (the most common kind)

  23. Acid-base titration curves • Titration curve: plot of pH vs volume of acid or base added to a solution • Equivalence point: point in a titration where a stoichiometric amount of reactant has been added

  24. Strong acid, strong base • Strong acid / strong base titration: equivalence point is at pH 7.0 because the salt of a strong acid and strong base is neutral • Calculating points on a titration curve for a strong acid strong base is a matter of stoichiometry: just subtract molar amounts of acid and base to find the concentration of unreacted H3O+ • What is the pH of a solution in which 15 mL of 0.10 M NaOH has been added to 25 mL of 0.10 M HCl?

  25. Titration of a weak acid by a strong base • What is the pH at the equivalence point when 25 mL of 0.10 M HF is titrated by 0.15 M NaOH? Ka = 6.8 x 10-4 • In this case the pH is basic at the equivalence point • To find the pH of the equivalence point, first find the number of moles of the salt formed if the acid were completely deprotonated, and use total volume to get its concentration • Then, find the pH just as before. • Similar calculations can be used for titration of a weak base by a strong acid.

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