Weak Bases NH3 + H2O NH4+ + OH- 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 NH4+ + OH-

Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5 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 + H2O NH4+ + OH- [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change equilibrium [NH3] [NH4+] [OH] initial change equilibrium since no products initially, Qc = 0, and the reaction is proceeding forward

Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5 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 0.100 -x x x

Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5 determine the value of Kb from Table 15.8 since Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium x 0.100 -x

Find the pH of 0.100 M NH3(aq) solution given 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

Find the pH of 0.100 M NH3(aq) solution given 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

Find the pH of 0.100 M NH3(aq) solution given 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

Find the pH of 0.100 M NH3(aq) solution given 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

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 10-6

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 10-6 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 B + H2O BH+ + OH- [B] [BH+] [OH] initial 0.0015 ≈ 0 change equilibrium since no products initially, Qc = 0, and the reaction is proceeding forward

Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6 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 [B] [BH+] [OH] initial 0.0015 change equilibrium -x +x +x x x 0.0015 -x

Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6 determine the value of Kb since Kb is very small, approximate the [B]eq = [B]init and solve for x [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium x 0.0015 -x

Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6 check if the approximation is valid by seeing if x < 5% of [B]init [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium x x = 4.9 x 10-5 the approximation is valid

Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6 substitute x into the equilibrium concentration definitions and solve [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 4.9E-5 [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 0.0015 x x x = 4.9 x 10-5

Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6 use the [OH-] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 4.9E-5

Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [B] [BH+] [OH] initial 0.0015 ≈ 0 change -x +x equilibrium 4.9E-5 the answer matches the given Kb

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 Example: NaHCO3 solutions are basic Na+ is the cation of the strong base NaOH HCO3− is the conjugate base of the weak acid H2CO3 Conversely salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic Example: NH4Cl solutions are acidic NH4+ is the conjugate acid of the weak base NH3 Cl− is the anion of the strong acid HCl

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 HA is, the weaker the conjugate base A- 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) since HF is a weak acid, the position of this equilibrium favors the right

Use the Table to Determine if the Given Anion Is Basic or Neutral NO3− the conjugate base of a strong acid, therefore neutral HCO3− the conjugate base of a weak acid, therefore basic PO43−

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

Find the pH of 0. 100 M NaCHO2(aq) solution assuming Ka =1 Find the pH of 0.100 M NaCHO2(aq) solution assuming Ka =1.8x10-4 for HCOOH 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 HCO2− + H2O HCHO2 + OH- [CHO2−] [HCHO2] [OH-] initial 0.100 ≈ 0 change equilibrium

Find the pH of 0. 100 M NaCHO2(aq) solution assuming Ka =1 Find the pH of 0.100 M NaCHO2(aq) solution assuming Ka =1.8x10-4 for HCOOH 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 =1.8x10-4 substitute into the equilibrium constant expression [CHO2−] [HCHO2] [OH-] initial 0.100 ≈ 0 change equilibrium -x +x +x x x 0.100 -x

Find the pH of 0. 100 M NaCHO2(aq) solution assuming Ka =1 Find the pH of 0.100 M NaCHO2(aq) solution assuming Ka =1.8x10-4 for HCOOH since 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 Kb for CHO2− = 5.6 x 10-11 0.100 -x

Find the pH of 0.100 M NaCHO2(aq) solution 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 Kb for CHO2− = 5.6 x 10-11 x = 2.4 x 10-6 the approximation is valid

Find the pH of 0.100 M NaCHO2(aq) solution 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 Kb for CHO2− = 5.6 x 10-11 x = 2.4 x 10-6

Find the pH of 0.100 M NaCHO2(aq) solution 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 Kb for CHO2− = 5.6 x 10-11

Find the pH of 0.100 M NaCHO2(aq) solution 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 Kb for CHO2− = 5.6 x 10-11

Polyatomic Cations as Weak Acids some polyatomic cations can be thought of as the conjugate acid of a base therefore, some cations can potentially be an acid MH+(aq) + H2O(l) MOH(aq) + H3O+(aq) the stronger the base MOH is, the weaker the conjugate acid MH+ is a cation that is the counterion of a strong base (Na+, K+ etc) is pH neutral a cation that is the conjugate acid of a weak base is acidic NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) since NH3 is a weak base, the position of this equilibrium favors the right

Metal Cations as Weak Acids cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations pH neutral cations are hydrated Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+ (aq) + H3O+(aq)

Determine if the Given Cation is Acidic or Neutral C2H5NH3+ (Kb of C2H5NH2 = 5.6 x 10-4) the conjugate acid of a weak base, therefore acidic Ca2+ (Kb of Ca(OH)2 = 3.74 x 10-3) the counterion of a (not so) strong base, therefore neutral Cr3+ a highly charged metal ion, therefore acidic

Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the counterion 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 counterion 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 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

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 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

Determine whether a solution of the following salts is acidic, basic, or neutral SrCl2 Sr2+ is the counterion 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 Br− is the conjugate base of a strong acid, pH neutral 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

Determine whether a solution of the following salts is acidic, basic, or neutral NaCHO2 Na+ is the counterion 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

Polyprotic Acids since 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]o = [H3O+] for the second ionization [A2-] = Ka2 as long as the second ionization is negligible

Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C H2SO4 + H2O HSO4- + H3O+ Ka1= big! Write the reactions for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [HSO4−] and [H3O+] is ≈ [H2SO4] HSO4- + H2O SO42- + H3O+ Ka2= 0.012 [HSO4 -] [SO42 -] [H3O+] initial 0.0100 change equilibrium

Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C 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 [HSO4 -] [SO42 -] [H3O+] initial 0.0100 change −x +x equilibrium 0.0100 −x x

Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C expand and solve for x using the quadratic formula Ka for HSO4− = 0.012

Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C substitute x into the equilibrium concentration definitions and solve [HSO4 -] [SO42 -] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 [HSO4 -] [SO42 -] [H3O+] initial 0.0100 change −x +x equilibrium 0.0100 −x x x = 0.0045

Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C substitute [H3O+] into the formula for pH and solve [HSO4 -] [SO42 -] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145

Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HSO4 -] [SO42 -] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 the answer matches Ka for HSO4− = 0.012

Strengths of Binary Acids the more δ+ H-X δ- polarized the bond, the more acidic the bond the stronger the H-X bond, the weaker the acid binary acid strength increases to the right across a period H-C < H-N < H-O < H-F binary acid strength increases down the column H-F < H-Cl < H-Br < H-I

Strengths of Oxyacids, H-O-Y the more electronegative the Y atom, the stronger the acid helps weakens the H-O bond the more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond

Lewis Acid - Base Theory electron sharing electron donor = Lewis Base = nucleophile must have a lone pair of electrons electron acceptor = Lewis Acid = electrophile electron deficient when Lewis Base gives electrons from lone pair to Lewis Acid, a covalent bond forms between the molecules Nucleophile: + Electrophile  Nucleophile:Electrophile product called an adduct other acid-base reactions also Lewis Tro, Chemistry: A Molecular Approach

Example - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile OH H C H  + OH-1  OH H C H  + OH-1  Electrophile Nucleophile •• • Tro, Chemistry: A Molecular Approach

Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF3 + HF  CaO + SO3  KI + I2  Tro, Chemistry: A Molecular Approach

Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF3 + HF H+1BF4-1 CaO + SO3 Ca+2SO4-2 KI + I2 KI3 F B F H F •• + -1 H+1 Nuc Elec O S O •• Ca+2 O -2 + -2 Ca+2 Elec Nuc I I K+1 I -1 •• + Elec Nuc K+1 I I I -1

What is Acid Rain? natural rain water has a pH of 5.6 naturally slightly acidic due mainly to CO2 rain water with a pH lower than 5.6 is called acid rain acid rain is linked to damage in ecosystems and structures

What Causes Acid Rain? many natural and pollutant gases dissolved in the air are nonmetal oxides CO2, SO2, NO2 nonmetal oxides are acidic CO2 + H2O H2CO3 2 SO2 + O2 + 2 H2O 2 H2SO4 processes that produce nonmetal oxide gases as waste increase the acidity of the rain natural – volcanoes and some bacterial action man-made – combustion of fuel weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced

Damage from Acid Rain

pH of Rain in Different Regions

Sources of SO2 from Utilities

Damage from Acid Rain acids react with metals, and materials that contain carbonates acid rain damages bridges, cars, and other metallic structures acid rain damages buildings and other structures made of limestone or cement acidifying lakes affecting aquatic life dissolving and leaching more minerals from soil making it difficult for trees

Acid Rain Legislation 1990 Clean Air Act attacks acid rain force utilities to reduce SO2 result is acid rain in northeast stabilized and beginning to be reduced