Aim # 9: How are acids and bases formed from salts? H.W. # 9 Study pp. 686-691(sec. 14.8) Ans. ques. p. 706 # 113,119,123,129 # 126 (Hint: find Kb, then Ka, etc.)

I The relationship between Ka and Kb For any weak acid, (1) HA(aq) H+(aq) + A-(aq) Ka For the salt of its conjugate base, (2) NaA(aq) → Na+(aq) + A-(aq) and (3) A-(aq) + H2O(ℓ) HA(aq) + OH-(aq) Kb

Adding reactions (1) and (3) HA(aq) H+(aq) + A-(aq) A-(aq) + H2O(ℓ) HA(aq) + OH-(aq) H2O(ℓ) H+(aq) + OH-(aq) Kw Ka = [H+][A-] Kb = [HA][OH-] [HA] [A-] Ka x Kb = [H+][A-][HA][OH-] [HA][A-] Ka x Kb = [H+][OH-] = Kw

For any acid (base) and its conjugate base (acid) Ka x Kb = Kw = 1.0 x 10-14 Problem: Calculate the value of Kb for acetate ion, C2H3O2-. Ans: For acetic acid, HC2H3O2, Ka = 1.8 x 10-5 KaKb = Kw 1.0 x 10-14 Kb = Kw = 1.0 x 10-14 = 5.6 x 10-10 Ka 1.8 x 10-5 II Some salts in aqueous solution react with water to produce an acidic solution; Some salts react with water to produce a basic solution. The process by which either forms is called hydrolysis.

Other salts do not react, and produce a neutral solution. III Salts that produce basic solutions sodium acetate, NaC2H3O2, is the salt of a strong base (NaOH) and a weak acid (HC2H3O2). 1. NaC2H3O2(s) + H2O(ℓ) → Na+(aq) + C2H3O2-(aq) hydrolysis 2. C2H3O2-(aq) + H2O(ℓ) HC2H3O2(aq) + OH-(aq) Salts formed from a strong base and a weak acid produce basic solutions (pH >7).

Problem: Calculate the pH of a 0 Problem: Calculate the pH of a 0.20 M sodium acetate (NaC2H3O2) solution. Ans: NaC2H3O2(s) + H2O(ℓ) → Na+(aq) + C2H3O2-(aq) C2H3O2-(aq) + H2O(ℓ) → HC2H3O2(aq) + OH-(aq) Kb = [HC2H3O2][OH-] = (x)(x) ͘ [C2H3O2] 0.20 – x Ka for acetic acid, HC2H3O2, = 1.8 x 10-5 Kb = Kw = 1.0 x 10-14 = 5.55 x 10-10 Ka 1.8 x 10-5 5.55 x 10-10 = x2 ≈ x2 ͘ 0.20 – x 0.20

x2 = 1.11 x 10-10 [OH-] = x = 1.05 x 10-5 M pOH = -log(1.05 x 10-5) = 4.95 pH = 14.00 – 4.98 = 9.02 check: 1.05 x 10-5 x 100 = .005% 0.20 IV Salts that produce acidic solutions pyridine hydrochloride is the salt of weak base (pyridine, C5H5N) and a strong acid (HCl). 1. C5H5NHCl(s) + H2O(ℓ) → C5H5NH+(aq) + Cl-(aq) hydrolysis 2. C5H5NH+(aq) C5H5N(aq) + H+(aq) Salts formed from a weak base and a strong acid produce acidic solutions (pH < 7)

Problem: Calculate the pH of a 0 Problem: Calculate the pH of a 0.30 M pyridine hydrochloride (C5H5NHCl) solution. Ans: C5H5NHCl(s) + H2O(ℓ) → C5H5NH+(aq) + Cl-(aq) C5H5NH+(aq) C5H5N(aq) + H+(aq) Ka = [C5H5N][H+] = (x)(x) ͘ [C5H5NH+] 0.30 – x Kb for pyridine, C5H5N, = 1.7 x 10-9 Ka = Kw = 1.0 x 10-14 = 5.88 x 10-6 Kb 1.7 x 10-9 5.88 x 10-6 = (x)2 ≈ x2͘ ͘ 0.30 – x 0.30 x2 = 1.76 x 10-6 [H+] = x = 1.32 x 10-3

Check: 1.3 x 10-3 x 100 = .43% .30 pH = -log(1.32 x 10-3) = 2.88 V Salts that produce neutral solutions sodium chloride is the salt of a strong base (NaOH) and a strong acid (HCl) 1. NaCl(s) + HCl(ℓ) → Na+(aq) + Cl-(aq) 2. Cl-(aq) + H2O(aq) x Hydrolysis does not occur because HCl is a strong acid that dissociates completely. The solution is neutral (pH = 7)

VI Salts formed from a weak base and a weak acid ammonium cyanide is the salt of the weak base ammonia (NH3) and the weak acid hydrocyanic acid (HCN) 1. NH3(aq) + HCN(g) → NH4CH(s) 2. NH4CN(s) + H2O(ℓ) → NH4+(aq) + CN-(aq) acidic ion basic ion If Ka of the acidic ion > Kb of the basic ion, the solution will be acidic. If Kb of the basic ion > Ka of the acidic ion, the solution will be basic.

For NH4CN, Ka NH4+ = Kw = 1.0 x 10-14 = 5.6 x 10-10 KbNH3 1.8 x 10-5 KbCN- = Kw = 1.0 x 10-14 = 2.0 x 10-5 KaHCN 4.9 x 10-10 KbCN- > KaNH4+ therefore the solution will be basic.

Kb pyridine = 1.7 x 10-9 Ka nitrous acid = 4.5 x 10-4 Problem: Will a salt solution of the salt pyridinium nitrite be acidic or basic? Kb pyridine = 1.7 x 10-9 Ka nitrous acid = 4.5 x 10-4 Ans: Ka = Kw = 1.0 x 10-14 = 5.9 x 10-6 Kb 1.7 x 10-9 Kb = Kw = 1.0 x 10-14 = 2.2 x 10-11 Ka 4.5 x 10-4 Ka>Kb therefore, the salt solution will be acidic.

Problem: Will a salt solution of NaH2PO4 be acidic or basic? Ans: NaH2PO4(s) + H2O(ℓ) → Na+(aq) + H2PO4-(aq) H2PO4-(aq) ↔ HPO42-(aq) + H+(aq) Ka2 = 6.2 x 10-8 H2PO4-(aq) + H2O(ℓ) ↔ H3PO4(aq) + OH-(aq) Kb = Kw = 1.0 x 10-14 = 1.3 x 10-12 Ka 7.5 x 10-3 Because Ka>Kb , the solution will be acidic.

Practice Problems Zumdahl (8th ed.) p. 693 # 126,118,120