Chapter 6 Acids and Bases Bronsted-Lowry Model Bronsted-Lowry acid = a H+ ion donor. Bronsted-Lowry base = a H+ ion acceptor Strong Acids Strong acid: completely ionized in water. HNO3(aq) + H2O(l) → NO3-(aq) + H3O+(aq)(H+ donor) (H+ acceptor) Weak Acids Weak acid: partially ionized in water. HNO2(aq) + H2O(l) ⇌ NO2-(aq) + H3O+(aq)(H+ donor) (H+ acceptor) Weak acid: equilibrium between acid (on left) and ion (on right): HA(aq) + H2O(l) ⇌ A-(aq) + H3O+(aq) Strong acids: Ka >> 1 Weak acids: Ka < 1 Conjugate Acid-Base Pairs Conjugate acid-base pairs: Differ from each other only by the presence or absence of a proton. Strong Bases Strong Bases: Oxides/hydroxides of Group 1, 2 metals. Weak Bases Nitrogen-containing compounds: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)(base) (conjugate acid) Relative Strengths of Acids and Bases Leveling Effect: H3O+ is the strongest H+ donor that can exist in water. Strong acids all have the same strength in water; are completely converted into H3O+ ions. Water: An Amphiprotic Substance Amphiprotic: capable of acting as an acid or a base. Water as a base: HNO3(aq) + H2O(l) → NO3-(aq) + H3O+(aq)(H+ donor) (H+ acceptor) Water as an acid: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)(H+ acceptor) (H+ donor) Autoionization of Water The pH Scale pH = -log[H+] pOH = -log[OH-] Kw = 1 x 10-14 = [H3O+][OH-] Log form of Kw:- -log(Kw) = -log([H+][OH-]) 14.00 = pH + pOH Weak Acids HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2-(aq) To calculate pH of: strong acid: [H+] = [Acid]initial weak acid: [H+] = calculated by solving equilibrium problem (Ch. 5) Sample Exercise - pH Is Solution A with a pH of 9.58 more or less acidic than Solution B in which [H+] = 4.3 x 10-10 M? (1) pH of Solution B: pH = -log(4.3 x 10-10) = - (- 9.37) = 9.37 (2) [H+] of Solution A: 9.58 = -log[H+][H+] = 10-9.58 = 2.6 x 10-10 M Solution B is more acidic since it has a higher proton ([H+]) ion concentration Sample Exercise - More p- functions Percent Ionization Degree of ionization:(quanity of substance t h an is ionized)(concentration of substance before ionization) Percent ionization: degree of ionization expressed as a percent % ionization degreases as [Acid]0 increases Sample Exercise - Weak Acid Equilibria The pH of a 1.00 M solution of formic acid (HCOOH), a weak organic acid found in red ants and responsible for the sting of their bite, is 1.88. a. What is the percent ionization of 1.00 M HCOOH? b. What is the Ka value of the acid? c. What is the percent ionization of 0.0100 M HCOOH? A. the pH of the 1.00 M solution is 1.88. The corresponding [H+] is [H+] = 10-1.88 = 1.32 x 10-2 M Inserting this value and the initial concentration of HCOOH into the percent ionization equation gives: B. At equilibrium [HCOO-] = [H+] = 1.32 x 10-2 M, and the equilibrium concentration of HCOOH is (1.00 - 1.32 x 10-2) M = 0.99 M Inserting these values in the expression for Ka: C. To calculate [H+] in 0.0100 M HCOOH, we use an ICE table, as in the equilibrium calculations in Chapter 5, to solve for [H+] at equilibrium. Since it is the unknown in the calculation, we give it the symbolx. Filling in the other cells in the ICE table: Inserting the equilibrium terms into the Ka expression and using Ka from part b, Ka = x¿(0.0100−x)(x)¿¿ = 1.76 x 10-4 and solving for x using the quadratic equation or an equation-solver program: x = 1.24 x 10-3 The x value is equal to [H+] at equilibrium. Therefore, the percent ionization of formic acid in a 0.0100M solution is Weak Base Equilibria NH3(aq) + H2O(l) ⇌ NH4(aq) + OH-(aq) To calculate pH: Obtain [OH-] at equilibrium, use Kw relationship pOH and the Calculation of pH for Bases To calculate pH of base solutions: Obtain [OH-] at equilibrium; Kw = [H3O+][OH-]; [H3O] = Kw/[OH-] Or... pOH = -log[OH-] pKw = 14.00 = pOH + pH Sample Exercise - Weak Base Equilibria The concentration of NH3 in household ammonia ranges between 50 and 100 g/L, or from about 3 M to almost 6 M. What is the pH of a 3.0 M solution of NH3? Taking the negative logarithm of [OH-] to calculate pOH:- pOH = -log[OH-] = -log(7.3 x 10-3 M) = 2.14 Then we subtract this value from 14.00 to obtain the pH:- pH = 14.00 - pOH = 14.00 - 2.14 = 11.86 Polyprotic Acids Monoprotic Acids: One ionizable H atom per molecule Polyprotic Acids: More than one ionizable H atom/molecule Diprotic acids: two ionizable H atoms Triprotic acids: three ionizable H atoms. Diprotic Acids Sulfuric Acid: H2SO4(aq) ⇌ H+(aq) + HSO4-(aq) Ka1 >> 1 HSO4-(aq) ⇌ H+(aq) + SO42-(aq) Ka2 = 1.2 x 10-2 Carbonic Acid: H2CO3(aq) ⇌ H+(aq) + HCO3-(aq) Ka1 = 4.3 x 10-7 HCO3-(aq) ⇌ H+(aq) + CO32-(aq) Ka2 = 4.7 x 10-11 Polyprotic Acids Ka1 > Ka2 ( > Ka3...) More difficult to remove H+ ion (positive charge) from negatively charged anion. pH of Polyprotic acid solutions: Typically only first ionizable H atom affects pH (Ka1) Calculations are identical to those involving the pH of a weak acid. Factors Affecting Acid Strength Number of Oxygen Atoms in Anion: The greater the number of oxygens, the stronger the acid. Structure Electronegativity Acid-Base Properties of Salts Weak Acid: HF(aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)(acid) (conj. base) Salt, NaF: NaF(s) → Na+(aq) + F-(aq)- F-: a base!- F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq)(base) (conj. acid) pH of Salt Solutions: NaF(aq)1. Write balanced equilibrium equation for basic anion or acidic cation.F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq)2. Determine Ka or Kb for equilibrium.3. Solve equilibrium problem for [H+], [OH-] Relationship between Ka and Kb Sample Exercise - pH of a Salt