Strength of Acids and Bases The terms weak and strong are used to compare the strengths of acids and bases The terms dilute and concentrated are used to compare the concentration of solutions They do not mean the same thing! The combination of strength and concentration determines the behavior of the acid or base.

Strength of Acids and Bases Strength is Determined by how much they ionize (dissociate) in water Strong – ionize 100% in water Weak – only partially ionize in water Complete dissociation - all HCl compounds have separated into H+ and Cl- ions Partial dissociation - some HNO2 compounds have separated into H+ and NO2- ions, while some remain together 2

Strength of Acids and Bases Strong Acids: Only HNO3, HI, HBr, HCl, H2SO4, HClO4 Remember: NO, I Brought Claude SOme ClOthes All other acids are weak, and remain in equilibrium Ex: HNO2 ↔ H+ + NO2- Strong Bases: Group I or II metals Ex: NaOH, Mg(OH)2…etc. All other bases are weak, and remain in equilibrium NH3 + H2O ↔ NH4+ + OH- 3

Conjugate Strength Strong Acid = weak conjugate base Strong Base = weak conjugate acid Weak Acid = strong conjugate base Weak Base = strong conjugate acid Example: HCl + NH3 → Cl- + NH4+ (strong acid) (weak base) (weak c. base) (strong c. acid) 4

Strong Acid/Base Calculations If the acid/base is strong, concentrations of reactants will match the concentration of the products Ex: What is the [H+] in a 0.2M HCl solution? HCl is a strong acid; it ionizes completely. HCl  H+ + Cl- What is the [OH-] in a 0.03 M NaOH solution. NaOH is a strong base and it completely ionizes. NaOH  Na+ + OH- 0.2M 0.0M [H+] = 0.2M [OH-] = 0.03M

Weak Acids and Bases are Equilibriums Since weak acids and bases only partially ionize in water, an equilibrium is established between the original acid (or base) and its ions. Example: HCHOOH  H1+ + CHOOH1- Acetic acid partially ionizes in to H1+ and CHOOH1-.

Equilibrium Expressions for Acids Equilibrium expressions relate concentrations of reactants to those of the products In this case the acid and its ions HA (aq) ↔ H+ (aq) + A- (aq) Ka = [H+][A-] [HA] Ka = equilibrium constant for a acid (capital K) Numerical value of the ratio of product concentrations to reactant concentrations [ ] = concentration in M (mol/L) Order = coefficient becomes exponent behind the brackets

Acid Equilibrium Expression Examples Write the equilibrium expressions for the following reactions: HF (aq) ↔ H+ (aq) + F- (aq) Ka = [H+][F-] / [HF] H2CO3 (aq) ↔ 2H+ (aq) + CO32- (aq) Ka = [H+]2[CO32-] / [H2CO3]

Practice HNO2 (aq) ↔ H+ (aq) + NO2- (aq) [HNO2] = 0.045M [H+]=[NO2-] = 0.045M Write the equilibrium expression. Ka = [H+][NO2-] / [HNO2] Find Ka 0.045 H2SO3 (aq) ↔ 2H+ (aq) + SO32- (aq) Ka = [H+]2[SO32-] / [H2SO3] If Ka = 1.44, the [H+] = 0.2M, and [SO32-] = 0.1M, what is [H2SO3]? 2.8x10-3M

Weak Acid Calculation A weak acid, HA, is found to be 25% ionized in water. Find the Ka for a 0.4M solution of HA. HA  H+ + A- 0.4M 0.0M 0.0M 0.4-0.1M= 0.3M 0.1M 0.1M Ka = [H+][A-] [HA] = [0.1][0.1] [0.3] = 0.033

Acid Equilibrium Constant (Ka) The value of Ka tells you whether the products or reactants are favored. The higher the value of Ka; the stronger the acid. More products favored, so there is more ionization.

Comparing Ka Which acid is stronger? HCOOH (formic acid) = 1.8 x 10-4 H2CO3 = 4.4 X 10-7 HC2H3O2 = 1.84 X 10-5 Formic Acid Which acid has the stronger conjugate base? H2CO3

Equilibrium Expressions for Bases Equilibrium expressions relate concentrations of reactants to those of the products In this case the base and its ions BOH (aq) ↔ B+ (aq) + OH- (aq) Kb = [B+][OH-] [BOH] Kb = equilibrium constant for a acid (capital K) Numerical value of the ratio of product concentrations to reactant concentrations [ ] = concentration in M (mol/L) Order = coefficient becomes exponent behind the brackets

Practice AgOH (aq) ↔ Ag+ (aq) + OH- (aq) [AgOH] = 0.52M [Ag+]=[OH-] = 0.52M Write the equilibrium expression. Kb = [Ag+][OH-] / [AgOH] Find Kb 0.52 Cu(OH)2 (aq) ↔ Cu2+ (aq) + 2OH- (aq) Kb = [Cu2+][OH-]2 / [Cu(OH)2] If Kb = 0.5, the [Cu2+] = 0.12M, and [OH-] = 0.24M, what is [Cu(OH)2]? 1.4x10-2M

Kb Calculations A weak base, BOH, is prepared by dissolving 1.5 moles in 3 liters of solution. At equilibrium, the [OH] is 0.01 M. Find the Kb. BOH  B+ + OH- 2x10-4

Additional KbCalculations The Kb for NH3 is 1.8x10-5. Find the pH for a 0.5M NH3 solution. The ionization constant (Kb) is less than 5% (5x10-2) of the original [base] so the amount of base that ionizes is negligible and can be ignored. NH3 + H2O  NH4+ + OH- 0.0M 0.5M 0.0M x x 0.5-x Kb = [NH4+][OH-] [NH3] = [x][x] [0.5-x] = x2 [0.5] = 1.8x10-5 X2 = 9.0x10-6 [OH-] = X = 3.0x10-3M

Additional Kb Calculations Now that you have the [OH-], you can find pOH which gives you pH. pOH = - log 3.0x10-3M pOH = 2.52 pH = 14 – pOH = 11.5

Base Equilibrium Constant (Kb) The value of Kb tells you whether the products or reactants are favored. The higher the value of Kb; the stronger the base. More products favored, so there is more ionization.

Acid-Base Reactions The reaction of an acid and a base to produce a salt and water is called a neutralization reaction.

Neutralization Reactions Sodium hydroxide (base) and hydrochloric acid (acid) react to form sodium chloride (salt) and water.

Neutralization Practice Neutralizations are double displacement reactions! Practice predicting products (always a salt and water): Example #1 __Ca(OH)2 + __H3PO4 → ? __Ca(OH)2 + __H3PO4 → __Ca3(PO4)2 + __H2O 3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O Example #2 __Fe(OH)2 + __HBr → ? __Fe(OH)2 + __HBr → __FeBr2 + __H2O Fe(OH)2 + 2HBr → FeBr2 + 2H2O 21

Acid-Base Titrations Titration is used in determining the concentration of an unknown acid or base. An indicator is added to the standardized acid (acid of known concentration). The unknown base is added slowly with a burette until the solution is neutralized and reaches the equivalence point (where [H+] = [OH-]). The equivalence point is not necessarily where pH = 7 Adding one more drop of base changes the color of the solution to pink. This is called the endpoint.

Titration Calculations MaVa(H+)= MbVb(OH-) Note: pH at the equivalence point is not always 7. Strong Acid + Strong Base, pH=7 Strong Acid + Weak Base, pH<7 Weak Acid + Strong Base, pH>7 Example: It took 75mL of NaOH to neutralize 50mL of 2M HCl. What is the concentration of the NaOH? MaVa(#H+) = MbVb(#OH-) (2M)(50mL)(1) = (x)(75mL)(1) x = 1.33M NaOH

Titration Practice It took 20mL of Ca(OH)2 to neutralize 25mL of 0.05M HCl. What is the concentration of the base? MaVa(#H+) = MbVb(#OH-) (0.05M)(25mL)(1) = (x)(20mL)(2) x = 0.031 M