Analytical Chemistry I Lecture 4 PDF

Analytical Chemistry I PMC 101 Lecture 4 Dr. Dina A. Gawad Assistant Professor of Pharm. Analytical Chemistry  The pH of a solution pH, pOH, and pKw The ion-product constant for water: Kw = [H3O+ ][OH- ] -log Kw = -log H+ −log OH- = -l...

Analytical Chemistry I PMC 101 Lecture 4 Dr. Dina A. Gawad Assistant Professor of Pharm. Analytical Chemistry  The pH of a solution pH, pOH, and pKw The ion-product constant for water: Kw = [H3O+ ][OH- ] -log Kw = -log H+ −log OH- = -log 10 -14 pKw = pH + pOH = 14 ❖ Calculate pH of an ammonia solution whose hydroxide ion concentration is 1.9 x 10 -3 M. pOH = – log (1.9 x 10 -3 ) = 2.72 pH = 14.00 – pOH = 14.00 – 2.72 = 11.28 pH and pOH important relations 1- [H3O+ ] [OH- ]= 1 x 10 -14 2. pH= -log[H3O+ ] 3. pOH = -log[OH- ] 4. [H3O+ ] = 10 -pH 5. [OH-]= 10 -pOH 6. pH + pOH = 14  Complete the table 0.1 M HCl Pure water 0.1 M NaOH [H+ ] or [H3O+ ] [OH-] pH 1 7 13 pOH 13 7 1 Acidic or basic  Acid and base dissociation constants (Ka& Kb) A weak acid HA ionizes only partially as follows: HA ↔ H+ + A- The equilibrium constant for this reaction will be: + - Ka = [H ][A ] [HA] Stronger acid higher larger Ka A number of weak acids are capable of ionization in a stepwise manner (polyprotic acid): H2A ↔ H+ + HA- Ka1 = [H+] [HA-] [H2A] HA- ↔ H+ + A2- Ka2 = [H+] [A2-] [HA-] Ka1 is always numerically greater than Ka2 The overall ionization constant will be: H2A ↔ 2H+ + A- Ka = [H+]2 [A2-] [H2A]  As the acid gets stronger, its Ka value increases For weak polyprotic acids the Ka1 is always > Ka2 , etc Acid strength decreases in the order: H3PO4 >> H2PO4- >> HPO42- The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion; the fully deprotonated species is the strongest base.  In case of a weak base as NH3 NH3 + H2O ↔ NH4+ + OH- Kb = [NH4+] [OH-] [NH3] As Kb increases, strength of base increases  The Ka or Kb of an acid or base may also be written in terms of “pKa ” or “pKb ” pKa= - log Ka pKb= - log Kb As Ka or Kb increase → pKa or pKb decrease A strong acid/base has a high Ka or Kb and a low pKa or pKb Dissociation constants for conjugate acid-base pair Consider the dissociation constant expressions for an acid and its conjugate base: HA + H2O ↔ H3O+ + A- Ka = [H3O+][A-] [HA] A- + H2O ↔ HA + OH- Kb = [HA ][OH-] [A-] Multiplication of one equilibrium constant expression by the other gives: Ka x Kb = [ H3O+][OH-] = Kw This relationship indicates that if an acid is weak, its conjugate base must be strong and vice versa, and pKa + pKb = p Kw = 14  pH calculations for different solutions 1-Solutions of strong acids and strong bases Strong acids and strong bases are completely ionized in solvents, so the [H+] or [OH-] is directly related to the concentration of the substance. pH = -log [H+] , pOH=-log [OH-] Examples 1-Calculate the pH of a 1.0 M solution of HCl. [H+] = 1.0 M pH = - log 1 = 0 2-Calculate the pH of a 1.0 M solution of NaOH. [OH-] = 1.0 M pOH = - log 1 = 0 pH = 14 - 0 = 14 3-Calculate the [H+] and pH of 0.001 N HCl. [H+] = 0.001 M pH = - log 1 × 10-3 = 3 4-Calculate the [OH-] and pH of 0.01 N NaOH. [OH-] = 0.01 M pOH = - log 1.0 × 10-2 = 2 pH = 14 - 2 = 12 Solutions of weak acids and weak bases a-Calculation of the pH of a solution of a weak monoprotic acid HA ↔ H+ + A- Ka = [H+][A-] [HA] If acid is pure, the ionization gives [H+] = [A-]. Ka = [H+]2 Ca Therefore, [H+]= √ K xC a a and: pH = ½ (pKa + pCa) Examples 1-Calculate the pH of 0.1M HAc (pKa = 4.76) pH = ½ (4.76 + 1) = 2.88 b-Calculation of pH of solution of a weak monoequivalent base B + H2O ↔ BH+ + OH- Kb = [BH+][OH-] [OH-]= √ Kb x Cb [B] pOH = ½ (pKb + pCb) and , pH = 14 - ½ (pKb + pCb) Example Calculate pH, [H+ ] and [OH- ] of 0.001 M solution of cocaine base (pKb = 5.59) pCb = ‒ log 0.001 = 3 pH = pKw ‒ ½ (pKb + pCb )= 14 ‒ ½ (5.59 + 3) = 9.705 [H+ ] = 10-pH = 10-9.705 = 1.97 x 10-10 M Calculation of pH of salts: Most salts are completely ionized , HgCl2 is unionized. All salts are partially hydrolyzed NaCl + H2O NaOH + HCl pH of salts of a strong acid and a strong base: The pH of salts of strong acid and strong base is equal to 7 regardless the conc. of the salt NaCl, KCl, NaNO3 Solutions of salts of weak acids and weak bases: Salts of weak acids or weak bases react with water to give basic or acidic solutions, respectively. 1-Salts of weak acids and strong bases: Ex: sodium acetate, It is a strong electrolyte, The anion of the salt is a Brönsted base: H2O + CH3COO- ↔ CH3COOH + OH- acid1 base2 acid2 base1 pOH = ½ (pKb + pCs) pH = pKw - ½ (pKb + pCs) 2-Salts of weak bases and strong acids Ex: NH4Cl , which is a strong electrolyte and is completely ionized. Its hydrolytic reaction is: NH4+ + H2O ↔ NH4OH + H+ pH = ½ (pKa + pCs) 3-Solutions of acid salts: Solution of weak acid salts of the type HA- can ionize as acids and as bases, e.g. HCO3- , H2PO4- which can react in 2 ways: - Ionization as an acid: HCO3- ↔ H+ + CO32- Ionization as a base: HCO3- + H2O ↔ H2CO3 + OH- The pH is independent on the conc. of the salt: pH = ½ (pK1 + pK2)