Resources:

Transcription

1 I. A/B General A. Definitions 1. Arrhenius 2. B-L 3. Lewis *H+=H3O+ B. Properties 1. Acid 2. Base II. ACIDS A. Types 1. Binary 2. Ternary 3. Acid Anhydride B. Protic 1. mono 2. di 3. tri C. Strengths 1. Determining Factors a. Degree of Dissociation b.molecular Shape and composition 2. Strong Acids 3. Weak Acids 4. Ka, Kb, Kw III. Bases A. Types of bases 1. hydroxy bases (-OH) 2. carbonates (-CO3) 3. molecular bases (NH3) 4. Strong Bases IV. Amphoteric Substances V. Acid Base Reactions A. Arrhenius (Neutralization) STRONG + STRONG B. B-L (Conjugates) STRONG + Weak C. B-L Incomplete Neutralization D. Carbonates + Acid VI. ph and poh A. Definition, Kw B. Formula C. ph + poh = 14 D. calculating [H3O+] and [OH-] from ph or poh VII. Titrations A. 1:1 B. Unbalanced Resources:

2 %20Notes%20(answers).pdf %20Notes/notes14.PDF -Henderson Hesselbach Equation

3 Acids, Bases and Salts I. Definitions A. Definitions: Acid Base 1. Arrhenius A substance which A substance which increases [H+] increases [OH-] 2. Bronsted-Lowry A substance that A substance that donates [H+] accepts [H+] 3. Lewis An e - pair acceptor An e - pair donor ***** H + (aq) = H 3 O + hydrogen ion = a proton = hydronium ion B. Properties 1. Acids A. dilute acids taste sour B. acids react with most metals to liberate H 2 gas (S.R.rxn) C. are neutralized by strong bases to form: Acid + Base(-OH) a salt + HOH Acid + Base(-CO 3 ) a salt + HOH + CO 2 D. cause color changes in indicators Table M E. are electrolytes in proportion to their [ ] (keep in mind [ ], solubility (K sp ) and degree of dissociation (K a ) 2. Bases A. bases taste bitter B. bases feel slippery C. are neutralized by strong acids. D. cause color changes in indicators Table M E. are electrolytes in proportion to their [ ] (keep in mind [ ], solubility (K sp ) and degree of dissociation (K b )

4 II. ACIDS A. Types 1. Binary two element acid, (H + nm) A. Naming Rule: hydro ic acid B. Examples: 1. HCl hydrochloric acid 2. HBr hydrobromic acid 3. H 2 S hydrosulfuric acid 2. Ternary three or more element acid, (H + PAI) TABLE E!!!!! A. Naming Rule: ate - ic acid ite ous acid B. Examples: 1. H 2 SO 3 sulfurous acid 2. H 2 sulfuric acid 3. HNO 3 nitric acid 4. H 3 PO 4 phosphoric acid B. Protic- indicates the # of protons which can potentially be donated 1. monoprotic 1 H + examples: HCl, HNO 3 2. diprotic - 2 H + examples: H 2 S, H 2 3. triprotic - 3 H + examples: H 3 PO 4 C. Strengths 1. Determining Factors a. Degree of Dissociation 1. The more an acid dissociates the stronger it is. This is determined through K a values. 2. Acids that dissociate less are weaker. b. Inorganic acids are stronger than organic (carbon- based)

5 2. STRONG acids. A. The following acids are considered strong because they have a large degree of dissociation: 3. WEAK Acids A. Acids that do not dissociate readily are considered weak. B. Examples: HF, HC 2 H 3 O 2 aka CH 3 COOH (vinegar), HC 6 H 7 O 7 (citric) 4. Kw values a. K w = dissociation constant for water = 1 x H 2 O (l) H + (aq) + OH - (aq) III. Bases A. Types 1. hydroxy bases (m-oh) -STRONGEST bases b/c they completely dissociate and increase the [OH-]. They can completely neutralize strong acids. Examples: NaOH, KOH, Mg(OH)2, Ca(OH)2 2. carbonates (m-co 3 ) - milder bases, largely insoluble Examples: Na2CO3, K2CO3, MgCO3, CaCO3

6 3. molecular (NH 3 ) covalently bonded. Typically have unused electron pairs. IV. Amphoteric Substances!!!!!!!!!!!!! 1. Definition: Substances that can act as either an acid or a base!!! 2. Examples: H2O, HSO4-, H2PO4-2 (any partially neutralized acid) V. Acid/Base Reactions A. Complete Neutralization (a salt + H 2 O) 1. HCl + NaOH --> 2. H 2 + Mg(OH) 2 --> 3. H 2 + KOH --> 4. HClO 4 + Ca(OH) 2 --> B. Incomplete Neutralization 1. H 2 + (1 mole)naoh --> 2. H 3 PO 4 + (1 mole) KOH --> C. B-L Conjugate Acid/Base Pairs 1. HNO 3 + H 2 O --> 2. HC 2 H 3 O 2 + H 2 O --> 3. HC 2 H 3 O 2 + NH 3 --> 4. HClO 2 + H 2 O --> D. Carbonates (a salt, H 2 O, CO 2 ) 1. HC 2 H 3 O 2 + NaHCO 3 --> 2. HCl + CaCO 3 -->

7 VI. Kw, H +, OH -, ph, poh - CLICK HERE for a cool animation A. Kw = [H + ] [OH - ] 1. Kw = 1x10-14 = [H + ] [OH - ] acidic [H + ] > [OH - ] basic [H + ] < [OH - ] neutral [H + ] = [OH - ] Therefore in a neutral solution 1x10-14 = (1x10-7 )(1x10-7 ) 2. Calculating [H + ] or [OH - ] [H + ] = Kw / [OH - ] [OH - ] = Kw / [H + ] example: 1. Determine the hydronium ion concentration if the hydroxide concentration =. 0001M [H + ] = Kw / [OH - ] [H + ] = 1x10-14 / [1x10-4 ] [H + ] = 1x Determine [OH - ] if [H 3 O + ] = 4.8x10-11? [OH - ] = Kw / [H + ] [OH - ] = 1x10-14 / [4.8x10-11 ] [OH - ] = 2.08x10-4

8 3. ph = -log [H 3 O + ] Example: a. What is the ph of a m HCl sln? ph = -log (1x10-8 ) ph = 8 b. What is the ph of a m H 2 sln? ph = -log (4.82 x 10-4 ) ph = 3.32 c. What is the ph of a.00078m NaOH sln? We need to first determine the [H3O+] from [OH-] 4. poh calcs [H 3 O + ] = Kw / [OH - ] [H 3 O + ] = 1x10-14 / [7.8x10-4 ] [H 3 O + ] = 1.28x10-11 ph = -log [H 3 O + ] ph = -log 1.28x10-11 ph = poh = - log [OH - ] or 14 = ph + poh COMPLETING THE CALCULATION CIRCUIT!!!!!!!!!! 5. [H 3 O + ] = inv log -ph [OH - ] = inv log -poh

9 VII. Titrations A. Definition: The laboratory procedure where a known concentration of an acid or base is used to neutralize the other to and end point in order to determine the unknown's [ ]. Typically NaOH is added to the acid and an indicator like phth is used. B. In a titration; moles acid = moles base following formula: Assuming that the # H + = #OH - available then you can use the Ma x Va = Mb x Vb Ma = molarity of acid Va = volume of acid Mb = molarity of base Vb = volume of base 1. Examples a. 40 mls. of a 0.50M HCl are titrated with 20 mls of NaOH. What is the [NaOH]? MaVa=MbVb HCl and NaOH are 1:1 so use formula directly. Mb = MaVa / Vb Mb = (0.50M x 40mls) / 20 mls Mb = 1.0M NaOH

10 b. 60 mls of HNO3 are neutralized by 20 mls of 0.1M NaOH. What is the [HNO 3 ]? MaVa = MbVb H + :OH - = 1:1 ratio Ma = MbVb / Va Ma = (0.1M x 20 mls) / 60 mls Ma = 0.033M HNO3 c. 10 mls H 2 are neutralized by 25 mls of 0.04M Ca(OH) 2. What is the [H 2 ]? MaVa = MbVb H + :OH - = 2:2 or 1:1 ratio Ma = MbVb / Va Ma = (0.04M x 25mls) / 10mls Ma = 0.1M H 2 2. If the H + :OH - is not 1:1 you need to introduce the equivalents factor into the equation and then solve!!!! a. 10 mls of 0.1M H 2 are neutralized by 20 mls of NaOH. What is the [NaOH]? MaVa = MbVb H + :OH - = 2:1 So we need to introduce those #'s to the equation 2 MaVa = MbVb Mb = 2 MaVa / Vb Mb = (2 x 0.1M x 10 mls) / 20 mls Mb = 0.1M NaOH b. 10 mls of 0.1M H 3 PO 4 are neutralized by 40 mls of Ca(OH) 2. What is the [Ca(OH) 2 ]? MaVa = MbVb equation 3MaVa = 2MbVb Mb = 3MaVa / 2Vb Mb = (3 x 0.1M x 10 mls) / (2 x 40 mls) Mb = M Ca(OH) 2 H + :OH - = 3:2 So we need to introduce those #'s to the