Acids Bases Salts I Properties of Acids Bases
Acids, Bases, & Salts
I. Properties of Acids & Bases A. Properties of Acids 1. Aqueous solutions have a sour taste 2. Acids change the color of acid-base indicators 3. Some acids react with active metals to release hydrogen Zn(s) + H 2 SO 4(aq) Zn. SO 4(aq) + H 2(g) 4. Acids react with bases to produce salts and water HCl(aq) + Na. OH(aq) Na. Cl(aq) + H 2 O(l) 5. Acids conduct electric current
B. Properties of Bases 1. 2. 3. 4. 5. Aqueous solutions of bases have a bitter taste Bases change the color of acid-base indicators Dilute aqueous solutions of bases feel slippery Bases react with acids to produce salts and water Bases conduct electric current
II. Arrhenius Acids and Bases - Svante Arrhenius, Swedish chemist (1859 -1927) A. Arrhenius Acid – A chemical compound that increases the concentration of hydrogen ions, H+, in aqueous solution [H+] > [OH-] B. Arrhenius Base – A substance that increases the concentration of hydroxide ions, OH-, in aqueous solution [H+] < [OH-]
C. Aqueous Solutions of Acids 1. Acids are molecular compounds that ionize in solution HNO 3 + H 2 O H 3 O+ + NO 3 H 2 SO 4 + H 2 O H 3 O+ + HSO 4 H 2 O + HCl H 3 O+ + Cl-
D. Strength of Acids 1. Strong acids ionize completely in solution 2. Weak acids ionize only slightly and are weak electrolytes Strong Acids Weak Acids H 2 SO 4 HSO 4 - HCl. O 4 H 3 PO 4 HCl HF HNO 3 CH 3 COOH HBr H 2 CO 3 HI H 2 S HCN HCO 3 -
E. Aqueous Solutions of Bases 1. Ionic bases dissociate to some extent when placed in water Na. OH (s) Na+ (aq) + OH − (aq) 1. Basic solutions are referred to as “alkaline” 2. Molecular bases react with water to produce hydroxide ions NH 3 (g) + H 2 O (l) ↔ NH 4+ (aq) + OH- (aq)
F. Strength of Bases 1. Strength of ionic bases is linked to solubility a) High solubility = strong base b) Low solubility = weak base 2. Molecular bases tend to be weak regardless of solubility
III. Bronsted-Lowry Acids & Bases A. Bronsted-Lowry Acid – molecule or ion that donates a proton (H+) B. Bronsted-Lowry Base – molecule or ion that accepts protons 1. Hydroxide ions (OH-) are acceptor of ionic bases
IV. Conjugate Acids & Bases A. Conjugate Base 1. The species that remains after an acid has given up a proton H 3 PO 4 (aq) + H 2 O (l) ↔ H 3 O+ (aq) + H 2 PO 4 -(aq) acid conjugate base 2. The stronger an acid, the weaker its conjugate base
B. Conjugate Acid 1. The species that is formed when a base gains a proton H 3 PO 4 (aq) + H 2 O (l) ↔ H 3 O+ (aq) + H 2 PO 4 - (aq) base conjugate acid 2. The stronger a base, the weaker its conjugate acid
IV. Ionization Constants A. Ionization constant = Ka B. Compares to relative strength of acids Ka = stronger acid) (high
B. p. H and p. OH 1. H 2 O(l) + H 2 O(l) ↔ H 3 O+ + OH 2. At 25 o. C, [H 3 O+] and [OH-] = 1. 0 x 10 -7 mol/L, and remains constant in pure water and dilute aqueous solutions. 3. This constant, Kw, is called the ionization constant of water. Kw = [H 3 O+][OH-] = (1. 0 x 10 -7) = 1. 0 x 10 -14 mol/L
V. p. H Scale & [H 3 O+] [OH-] A. p. H Scale 1. p. H < 7 : acid [H 3 O+] > [OH-] 2. p. H = 7 : neutral [H 3 O+] = [OH-] 3. p. H > 7 : basic/alkaline [H 3 O+] < [OH-]
B. Formulas for p. H, p. OH, [H 3 O+], & [OH-] p. H = - log [H 3 O+] p. OH = - log [OH-] p. H + p. OH = 14 Calculating p. H: [H 3 O+] must be in scientific notation! Calculate the log of the [H 3 O+] Multiply by -1 If [H 3 O+] = 0. 0261, then p. H = 0. 0261 = 2. 61 x 10 -2 (log 2. 61 x 10 -2) x -1 = 1. 58 (In calculators type: 2. 61, EE, 2, +/-, log, +/-)
Calculating p. OH [OH-] must be in scientific notation! Calculate the log of the [OH-] Multiply by -1 If [OH-] = 0. 000048, then p. OH = 0. 000048 = 4. 8 x 10 -9 log 4. 8 x 10 -9 x-1 = 8. 3 (In calculators type: 4. 8, EE, 9, +/-, log, +/-)
Calculating [H 3 O+] and [OH-] • From p. H & [OH-] from [H 3 O+] (and vice versa) [H 3 O+] = antilog (-p. H) If p. H = 7. 52, what are the [H 3 O+] and [OH-] concentrations? a) [H 3 O+] = antilog (-7. 52) (In calculators type: 7. 52, +/-, 2 nd, 10 x or 7. 52, +/-, 2 nd, log) b) [H 3 O+] = c) [H 3 O+] [OH-] = 1. 0 x 10 -14 d) [OH-] = 1. 0 x 10 -14 = 3. 3 x 10 -7 M OH[H 3 O+]
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