Lecture 152/25/04
Quiz 5 Given a 0.25 M solution of NH 3, calculate the pH and the equilibrium concentrations of NH 3, NH 4 +, and OH -. K b = 1.8 x Equations: K w = K a K b = [H 3 O + ][OH - ]
Binary Acids Within group: HA bond strength decreases (acid becomes stronger) as A increases with size Acid strength: HF < HCl < HBr < HI Across Row: Stronger acid (more polar - weaker bond) as you move left to right Acid strength: H 2 O < HF Molecular structure and acid strength
H-O-Z- Acid strength increases with increases electronegativity HOClHOBrHOI KaKa 3.5 x x x electronegativityCl (3.0)Br (2.8)I (2.5) Oxoacids with different central atoms
Acid strength increases as the number of O atoms attached to Z increases HOClHClO 2 HClO 3 HClO 4 KaKa 3.5 x x ~ 10 3 ~ 10 8 HNO 2 vs. HNO 3 H 2 SO 3 vs. H 2 SO 4 Molecular structure and acid strength Oxoacids with the same central atom
Acid strength related to R The more electronegative R, the stronger the acid C 5 H 11 -COOHCH 3 -COOHCH 2 Cl-COOHCCl 3 -COOH KaKa 1.4 x x x x hexanoic acidAcetic acid Chloroacetic acid trichloroacetic acid Carboxylic Acids (R-COOH)
Assign pH: (≤ 2, 2-6, 6-8, 8-12, ≥ 12) No calculators 0.1 M solutions HNO 2 NH 4 Cl NaF Mg(CH 3 COO - ) 2 BaO KHSO 4 NaHCO 3 BaCl 2
Put in order high pH to low (0.1 M solutions) KNO 3 Na 2 SO 4 CaO HBr C 6 H 5 COOH NaH
Common NameFormulaConstantpKa arsenic acidH 3 AsO 4 K 1 = 5.65 x H 2 AsO 4 - K 2 = 1.75 x HAsO 4 2- K 3 = 2.54 x carbonic acid H 2 CO 3 K 1 = 4.35 x HCO 3 - K 2 = 4.69 x chromic acid H 2 CrO 4 K 1 = HCrO 4 - K 2 = 3.36 x hydrogen sulfideH2SH2SK 1 = 1.02 x HS - K 2 = 1.22 x phosphoric acidH 3 PO 4 K 1 = 7.11 x H 2 PO 4 - K 2 = 6.23 x HPO 4 2- K 3 = 4.55 x sulfuric acidH 2 SO 4 K 1 > 1Negative HSO4-HSO4- K 2 = 1.01 x sulfurous acidH 2 SO 3 K 1 = 1.71 x HSO 3 - K 2 = 5.98 x
Acid–Base Properties of Salt Solution 1. Strong acid + strong base 2. Weak acid + strong base 3. Strong acid + weak base 4. Weak acid + weak base