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General properties ACIDS Taste sour Turn litmus React with active metals – Fe, Zn React with bases BASES Taste bitter Turn litmus Feel soapy or slippery (react with fats to make soap) React with acids blue to redred to blue
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Acid Nomenclature Review Binary Ternary An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”
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Acid Nomenclature Flowchart
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HBr (aq)HBr (aq) H 2 CO 3H 2 CO 3 H 2 SO 3H 2 SO 3 hydrobromic acid carbonic acid sulfurous acid Acid Nomenclature Review
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Name ‘Em! HI (aq)HI (aq) HCl (aq)HCl (aq) H 2 SO 4H 2 SO 4 HNO 3HNO 3 H 3 PO 4H 3 PO 4
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Definitions Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent used in organic chemistry, wider range of substances
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Examples Arrhenius Bronsted-Lowry Lewis HCl NaOH HClNH 3 :NH 3 BF 3 HCN The hydrogen ion in aqueous solution H + + H 2 O H 3 O + (hydronium ion)
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Neutralization In general: Acid + Base Salt + Water All neutralization reactions are double replacement reactions. HCl + NaOH NaCl + HOH HCl + Mg(OH) 2 H 2 SO 4 + NaHCO 3
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HA Let’s examine the behavior of an acid, HA, in aqueous solution. What happens to the HA molecules in solution?
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HA H+H+ A-A- Strong Acid 100% dissociation of HA Would the solution be conductive?
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HA H+H+ A-A- Weak Acid Partial dissociation of HA Would the solution be conductive?
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HA H+H+ A-A- Weak Acid HA H + + A - At any one time, only a fraction of the molecules are dissociated.
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Strong and Weak Acids/Bases Strong acids/bases – 100% dissociation into ions HClNaOH HNO 3 KOH H 2 SO 4 Weak acids/bases – partial dissociation, both ions and molecules CH 3 COOHNH 3
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pH 23456789101112 neutral @ 25 o C (H + ) = (OH - ) distilled water acidic (H + ) > (OH - ) basic or alkaline (H + ) < (OH - ) natural waters pH = 6.5 - 8.5 normal rain (CO 2 ) pH = 5.3 – 5.7 acid rain (NO x, SO x ) pH of 4.2 - 4.4 in Washington DC area 0-14 scale for the chemists fish populations drop off pH < 6 and to zero pH < 5
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You are here! http://nadp.sws.uiuc.edu/isopleths pH of Rainwater across United States in 2001 Increasing acidity Why is the eastern US more acidic? air masses
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What is acid rain? CO 2 (g) + H 2 O H 2 CO 3 H + + HCO 3 - Dissolved carbon dioxide lowers the pH Atmospheric pollutants from combustion NO, NO 2 + H 2 O … HNO 3 SO 2, SO 3 + H 2 O … H 2 SO 4 both strong acids pH < 5.3
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Effects of Acid Rain on Marble (calcium carbonate) George Washington: BEFORE George Washington: AFTER
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pH 123456891011 The biological view in the human body gastric juice urine saliva cerebrospinal fluid blood pancreatic juice bile acidicbasic/alkaline 7 Tortora & Grabowski, Prin. of Anatomy & Physiology, 10 th ed., Wiley (2003)
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Chapter 16.2 “Concentration of Solutions”
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Concentration is... a measure of the amount of solute dissolved in a given quantity of solvent A concentrated solution has a large amount of solute A dilute solution has a small amount of solute –These are qualitative descriptions But, there are ways to express solution concentration quantitatively
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Concentrated vs. Dilute
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Molarity Molarity = moles of solute liters of solution Abbreviated with a capital M, such as 6.0 M This is the most widely used concentration unit used in chemistry.
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- Page 481
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Practice Problem A sample of 0.0341 mol iron(III) chloride, Fe Cl 3, was dissolved in water to give 25. 0 mL of solution. What is the molarity of the solution? Since molarity = Moles of FeCl 3 Liters of solution 0.0341 moles of FeCl 3 0.0250 Liters of solution Since molarity = = 1.36 M FeCl 3
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Making solutions 1)Pour in a small amount of the solvent, maybe about one-half 2)Then add the pre-massed solute (and mix by swirling to dissolve it) 3)Carefully fill to final volume. –Fig. 16.8, page 481 Can also solve: moles = M x L Sample Problem 16.3, page 482
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Dilution Adding water to a solution will reduce the number of moles of solute per unit volume but the overall number of moles remains the same! Think of taking an aspirin with a small glass of water vs. a large glass of water You still have one aspirin in your body, regardless of the amount of water you drank, but a larger amount of water makes it more diluted.
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Dilution water (solvent)solute concentrated, M initial diluted, M final adding water lowers the solute concentration moles of solute remain constant V initial V final moles initial = moles final M final x V final = M initial x V initial
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Dilution The number of moles of solute in solution doesn’t change if you add more solvent! The # moles before = the # moles after Formula for dilution: M 1 x V 1 = M 2 x V 2 M 1 and V 1 are the starting concentration and volume; M 2 and V 2 are the final concentration and volume. Stock solutions are pre-made solutions of known Molarity. Sample 16.4, p.484
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Percent solutions can be expressed by a) volume or b) mass Percent means parts per 100, so Percent by volume: = Volume of solute x 100% Volume of solution indicated %(v/v) Sample Problem 16.5, page 485
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Percent solutions Percent by mass: = Mass of solute(g) x 100% Volume of solution (mL) Indicated %(m/v) More commonly used 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution? How many grams of salt are there in 52 mL of a 6.3 % solution?
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Titration Calculation HCl + NaOH NaCl + HOH at equivalence point: mole HCl = mole NaOH moles = M x V L M acid x V initial acid = M base x V buret A way to analyze solutions! indicator PGCC CHM 101 Sinex If the mole to mole ratio = 1:1
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Titration Calculation HCl + NaOH NaCl + HOH at equivalence point: mole HCl = mole NaOH moles = M x V L A way to analyze solutions! indicator PGCC CHM 101 Sinex
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M acid x V initial acid = M base x V buret If the mole to mole ratio = 1:1 If the mole to mole ratio is not 1:1 you need to add a mole factor to the equation above based on the coefficients in the balanced equation. Example: 2 HCl + Sr(OH) 2 SrCl 2 + 2H 2 O Mole to mole ratio = 2 acid : 1 base
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The number of moles of acid is twice the number of moles of base, therefore it will take twice as much base to neutralize the acid. moles acid = 2(moles base ) The '2' in this formula is called the mole factor. Substituting into equation (1) we have: M A V A = 2 M B V B
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Sample Problem: If 25.0 mL of a standard 0.05 M HCl solution is required to neutralize 20.0 mL of a solution of Sr(OH) 2, what is the concentration of the base? (Use the equation in the example above.) M A V A = 2 M B V B (0.05 M)(25.0 mL) = 2(X)(20.0 mL) X = 0.0313
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35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.
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PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!
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