CHE 124: General Chemistry II Chapter 15: Acids & Bases CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University

Overview Calculating pH Strong Bases Calculating pH Weak Bases Polyprotic Acids Determining pH of Salts

Strong Bases The stronger the base, the more willing it is to accept H use water as the standard acid For ionic bases, practically all units are dissociated into OH– or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO–] = [strong base] x (# OH) NaOH ® Na+ + OH− Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 11: Calculate the pH at 25 °C of a 0 Example 15.11: Calculate the pH at 25 °C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral Given: Find: [Sr(OH)2] = 1.5 x 10−3 M pH Conceptual Plan: Relationships: [H3O+] [OH] pH [Sr(OH)2] [OH]=2[Sr(OH)2] Solution: [OH] = 2(0.0015) = 0.0030 M Check: pH is unitless; the fact that the pH > 7 means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e 4

Practice – Calculate the pH of the following strong acid or base solutions 0.0020 M HCl 0.0015 M Ca(OH)2 [H3O+] = [HCl] = 2.0 x 10−3 M pH = −log(2.0 x 10−3) = 2.70 [OH−] = 2 x [Ca(OH)2] = 3.0 x 10−3 M pOH = −log(3.0 x 10−3) = 2.52 pH = 14.00 − pOH = 14.00 − 2.52 pH = 11.48 Tro, Chemistry: A Molecular Approach, 2/e

Weak Bases In weak bases, only a small fraction of molecules accept H’s weak electrolyte most of the weak base molecules do not take H from water much less than 1% ionization in water [HO–] << [weak base] Finding the pH of a weak base solution is similar to finding the pH of a weak acid NH3 + H2O Û NH4+ + OH− Tro, Chemistry: A Molecular Approach, 2/e

Base Ionization Constant, Kb Base strength measured by the size of the equilibrium constant when react with H2O :Base + H2O  OH− + H:Base+ The equilibrium constant is called the base ionization constant, Kb larger Kb = stronger base Tro, Chemistry: A Molecular Approach, 2/e 8

Tro, Chemistry: A Molecular Approach, 2/e

Structure of Amines Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12:Find the pH of 0.100 M NH3(aq) NH3 + H2O  NH4+ + OH write the reaction for the base with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH] from water is ≈ 0 [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change equilibrium [NH3] [NH4+] [OH] initial change equilibrium because no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [NH3] [NH4+] [OH] initial 0.100 change equilibrium x +x +x x x 0.100 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) determine the value of Kb from Table 15.8 because Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x Kb for NH3 = 1.76 x 10−5 [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x 0.100 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 check if the approximation is valid by seeing if x < 5% of [NH3]init [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x x = 1.33 x 10−3 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 substitute x into the equilibrium concentration definitions and solve [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.099 1.33E-3 [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.100 x x x = 1.33 x 10−3 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 use the [OH−] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 0.099 1.33E−3 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 0.099 1.33E−3 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0. 0015 M morphine solution, Kb = 1 Practice – Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-8 Solve from first principles; write out all chemical equilibria. Assumptions must be stated and defended mathematically. Answer: pH = 8.69

Acid–Base Properties of Salts Salts are water-soluble ionic compounds Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic NaHCO3 solutions are basic Na+ is the cation of the strong base NaOH HCO3− is the conjugate base of the weak acid H2CO3 Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic NH4Cl solutions are acidic NH4+ is the conjugate acid of the weak base NH3 Cl− is the anion of the strong acid HCl Tro, Chemistry: A Molecular Approach, 2/e

Anions as Weak Bases Every anion can be thought of as the conjugate base of an acid Therefore, every anion can potentially be a base A−(aq) + H2O(l)  HA(aq) + OH−(aq) The stronger the acid is, the weaker the conjugate base is An anion that is the conjugate base of a strong acid is pH neutral Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq) An anion that is the conjugate base of a weak acid is basic F−(aq) + H2O(l)  HF(aq) + OH−(aq) Tro, Chemistry: A Molecular Approach, 2/e

Example 15.13: Use the table to determine if the given anion is basic or neutral NO3− the conjugate base of a strong acid, therefore neutral NO2− the conjugate base of a weak acid, therefore basic Tro, Chemistry: A Molecular Approach, 2/e

Relationship between Ka of an Acid and Kb of Its Conjugate Base Many reference books only give tables of Ka values because Kb values can be found from them when you add equations, you multiply the K’s Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution Na+ is the cation of a strong base – pH neutral. The CHO2− is the anion of a weak acid – pH basic write the reaction for the anion with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH] from water is ≈ 0 Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution CHO2− + H2O  HCHO2 + OH [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change equilibrium

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5 substitute into the equilibrium constant expression [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change equilibrium x +x +x 0.100 x x x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 because Kb is very small, approximate the [CHO2−]eq = [CHO2−]init and solve for x [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x 0.100 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 check if the approximation is valid by seeing if x < 5% of [CHO2−]init [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x x = 2.4 x 10−6 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 substitute x into the equilibrium concentration definitions and solve [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 0.100 −x x [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 2.4E-6 x = 2.4 x 10−6 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 use the [OH−] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 2.4E-6 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 2.4E−6 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

Polyatomic Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base others are the counter-ions of a strong base Therefore, some cations can potentially be acidic MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq) The stronger the base is, the weaker the conjugate acid is a cation that is the counter-ion of a strong base is pH neutral a cation that is the conjugate acid of a weak base is acidic NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) because NH3 is a weak base, the position of this equilibrium favors the right Tro, Chemistry: A Molecular Approach, 2/e

Metal Cations as Weak Acids Cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations are pH neutral cations are hydrated Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+ (aq) + H3O+(aq) Tro, Chemistry: A Molecular Approach, 2/e

Example 15.15: Determine if the given cation Is acidic or neutral C5N5NH2+ the conjugate acid of the weak base pyridine, therefore acidic Ca2+ the counter-ion of the strong base Ca(OH)2, therefore neutral Cr3+ a highly charged metal ion, therefore acidic Tro, Chemistry: A Molecular Approach, 2/e

Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution NaCl Ca(NO3)2 KBr If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution NaF Ca(C2H3O2)2 KNO2 Tro, Chemistry: A Molecular Approach, 2/e

Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution NH4Cl If the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution Al(NO3)3 Tro, Chemistry: A Molecular Approach, 2/e

Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base NH4F because HF is a stronger acid than NH4+, Ka of NH4+ is larger than Kb of the F−; therefore the solution will be acidic Tro, Chemistry: A Molecular Approach, 2/e

Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral SrCl2 Sr2+ is the counter-ion of a strong base, pH neutral Cl− is the conjugate base of a strong acid, pH neutral solution will be pH neutral AlBr3 Al3+ is a small, highly charged metal ion, weak acid solution will be acidic CH3NH3NO3 CH3NH3+ is the conjugate acid of a weak base, acidic NO3− is the conjugate base of a strong acid, pH neutral Tro, Chemistry: A Molecular Approach, 2/e

Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral NaCHO2 Na+ is the counter-ion of a strong base, pH neutral CHO2− is the conjugate base of a weak acid, basic solution will be basic NH4F NH4+ is the conjugate acid of a weak base, acidic F− is the conjugate base of a weak acid, basic Ka(NH4+) > Kb(F−); solution will be acidic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine whether a solution of the following salts is acidic, basic, or neutral K+ is the counter-ion of a strong base, pH neutral KNO3 CoCl3 Ba(HCO3)2 CH3NH3NO3 NO3− is the counter-ion of a strong acid, pH neutral the solution is pH neutral Co3+ is a highly charged cation, pH acidic Cl− is the counter-ion of a strong acid, pH neutral the solution is pH acidic Ba2+ is the counter-ion of a strong base, pH neutral the solution is pH basic HCO3− is the conjugate of a weak acid, pH basic CH3NH3+ is the conjugate of a weak base, pH acidic the solution is pH acidic NO3− is the counter-ion of a strong acid, pH neutral Tro, Chemistry: A Molecular Approach, 2/e

Ionization in Polyprotic Acids Because polyprotic acids ionize in steps, each H has a separate Ka Ka1 > Ka2 > Ka3 Generally, the difference in Ka values is great enough so that the second ionization does not happen to a large enough extent to affect the pH most pH problems just do first ionization except H2SO4  use [H2SO4] as the [H3O+] for the second ionization [A2−] = Ka2 as long as the second ionization is negligible Tro, Chemistry: A Molecular Approach, 2/e

Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? (Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11) Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? H2CO3 + H2O  HCO3 + H3O+ write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations – assuming the second ionization is negligible HCO3− + H2O  CO32− + H3O+ [HA] [A−] [H3O+] initial 0.12 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? H2CO3 + H2O  HCO3 + H3O+ represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HA] [A−] [H3O+] initial 0.12 change equilibrium x +x +x x x 0.12 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? determine the value of Ka1 because Ka1 is very small, approximate the [HA]eq = [HA]init and solve for x Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11 [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.012 x 0.12 x Tro, Chemistry: A Molecular Approach, 2/e

the approximation is valid Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? Ka1 for H2CO3 = 4.3 x 10−7 check if the approximation is valid by seeing if x < 5% of [H2CO3]init [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium x x = 2.27 x 10−4 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? substitute x into the equilibrium concentration definitions and solve [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.12−x x x = 2.3 x 10−4 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? substitute [H3O+] into the formula for pH and solve [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.00023 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? Ka1 for H2CO3 = 4.3 x 10−7 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.00023 the values match within sig figs Tro, Chemistry: A Molecular Approach, 2/e