CH 111, Week 5 Notes
CH 111, Week 5 Notes CH111
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This 2 page Class Notes was uploaded by Jordyn Meekma on Sunday October 2, 2016. The Class Notes belongs to CH111 at Northern Michigan University taught by Dr. Tom Getman in Fall 2016. Since its upload, it has received 4 views. For similar materials see General Chemistry 1 in Chemistry at Northern Michigan University.
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Date Created: 10/02/16
CH 111 Notes: Week 5 Electrical Conductivity o Electrolyte: a solution that can conduct electricity NaCl (aq) Na +(aq) Cl-(aq) o Nonelectrolyte: a solution that cannot conduct electricity Sugar solution o Strong Electrolyte: a substance that dissociates completely in solution (almost 100%) Dissociation: ionic compounds, such as salts, separate or split into their ions, usually in a reversible manner HCl (aq) H +(aq) Cl (aq) o Weak Electrolyte: a substance that barely dissociates in solution (<10%) CH CO H ←→ H + + CH CO - 3 2 (aq) (aq) 3 2 (aq) Bronsted-Lowry Acid/Base Theory o Acid: proton (H ) donor + o Base: proton (H ) acceptor o Amphiprotic: can act as either an acid or a base, depending on the situation Ex. water + + H is a shorthand notation for H O 3 Strong Acids: dissociate completely in solution to yield hydrogen ions o HCl hydrochloric acid o HNO ni3ric acid o H 2O 4 sulfuric acid (the first proton is a strong acid) o HClO 4 perchloric acid Weak Acids: dissociate very little *Memorize these acids and - o HSO 4 hydrogen sulfate bases* o CH C3 H 2 acetic acid o HF hydrofluoric acid Strong Bases: dissociate completely in solution to yield hydroxide ions o LiOH lithium hydroxide o NaOH sodium hydroxide o KOH potassium hydroxide Weak Bases: dissociate very little o NH 3 ammonia o Any metal hydroxide which is insoluble or slightly soluble Neutralization Reaction: a reaction between an acid and a base which results in a proton transfer and the formation of a salt o CH 3O H2 (aq) NaOH (aq) H 2 (l)aCH CO 3 2(aq) o HNO 3(aq) NH 3(aq) NH 4O 3(aq) Chemical Equation Methods 1. Molecular equation: write all the species in neutral form (normal chemical equations) CH 3O H2 (aq) NaOH (aq) H 2 (l)aCH CO 3 2(aq) 2. Complete ionic equation: write all the strong electrolytes in their ions (NaOH) CH 3O H 2 (aq)+ Na +(aq) OH (aq)→ H O2+(l) +(aq) CH C3 2 (aq) 3. Net ionic equation: cancel the spectator ions out of the complete ionic equation (Na ) + Spectator ions: ions that are not directly involved in the reaction CH 3O H 2 (aq)+ Na +(aq) OH (aq)→ H O2+(l) +(aq) CH C3 2 (aq) - - CH 3O H 2 (aq)+ OH (aq)→ H 2 +(l) CO 3 2 (aq) o HNO 3(aq)+ KOH (aq) KNO 3(aq)+ H 2 (l) Net ionic: H +(aq)+ OH -(aq) H O2*(l)s is always the net ionic equation for the reaction of a strong acid and a strong base* o H 2O 4(aq) 2LiOH (aq)→ Li 2O 4(aq)+ 2H O2 (l) Net ionic: H +(aq)+ HSO 4 (aq) 2OH (aq)→ 2H O2+ (l) 42(aq) o Sc(OH) 3(s)+ 3HClO 4(aq)→ 3H O2+(l)(ClO ) 4 3(aq) Net ionic: Sc(OH) + 3H + → 3H O + Sc 3+ 3(s) (aq) 2 (l) (aq) If 17.21 mL of 0.9256 M NaOH are needed to neutralize 25.00 mL of HNO , 3hat is the concentration of HNO ? 3 o HNO + NaOH → NaNO + H O 3(aq) (aq) 3(aq) 2 (l) o 0.9256 mol NaOH/1 L (0.01721 L)(1 mol HNO /1 mol NaOH)3= -2 1.59296x10 mol HNO 3 1.59296x10 mol HNO /0.025003L = 0.6372 M HNO 3 How many mL of 0.965 M HClO are needed to 4eutralize 38.2 mL of 0.0105 M Al(OH) ? 3 o Al(OH) 3(s) 3HClO 4(aq) 3H O2+ (l)ClO ) 4 3(aq) o 0.0105 mol Al(OH) /1 L3(0.0382 L)(3 mol HClO /1 mol Al(4H) )(1 L 3 HClO /4.965 mol) = 0.00125 L = 1.25 mL HClO 4 Titration: a method used to determine the concentration of a solute by a reaction with a standard solution (of a known concentration) Equivalence Point: when you’ve added the exact amount of known solution to completely react with the solute you’re analyzing End Point: when your indicator changes color Hydrolysis Reaction: a reaction between water and another reagent o NH 3(aq)+ H O2→(l) 4 (aq)+ OH -(aq) o CO 2(g)+ H O2→(l)CO 2 3(aq)carbonic acid) *this is how acids are formed in nature → nonmetal oxides + H O* 2
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