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Chem. 1B – 9/13 Lecture. Announcements First Mastering Homework due Tonight Starting Lab #7 on Wednesday Today’s Lecture – cont. –Le Châtelier’s Principle.

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Presentation on theme: "Chem. 1B – 9/13 Lecture. Announcements First Mastering Homework due Tonight Starting Lab #7 on Wednesday Today’s Lecture – cont. –Le Châtelier’s Principle."— Presentation transcript:

1 Chem. 1B – 9/13 Lecture

2 Announcements First Mastering Homework due Tonight Starting Lab #7 on Wednesday Today’s Lecture – cont. –Le Châtelier’s Principle (Stresses resulting in equilibrium shifts) –Chapter 15: Definition of Acids/Bases (3 types of definitions) –Acid Strengths and K a –Autoprotolysis and pH –[H + ] and pH in strong acid and weak acid solutions

3 Chem 1B - Equilibrium Le Châtelier’s Principle Changes in Conditions – Types: –Changes in reactant or product concentrations – covered last time –Effect of a change in volume (compression/expansion or dilution/concentration) –Change in temperature

4 Chem 1B - Equilibrium Le Châtelier’s Principle – Volume Stress Example: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) –Mathematical explanation: –Initially at equilibrium K C = 10 5 and [NO] = 0.0010 M, [O 2 ] = 0.0010 M and [NO 2 ] = 0.010 M –Now we reduce the volume from 10.0 to 1.00L –New concentrations: [NO] = 0.010 M, [O 2 ] = 0.010 M and [NO 2 ] = 0.10 M (same number of moles in 1/10 th the volume so 10X more concentrated) –Q = (0.10 M) 2 /[(0.010 M) 2 (0.010 M)] = 10 4 < K C, so products favored

5 Chem 1B - Equilibrium Le Châtelier’s Principle – Volume Stress Note: in aqueous solutions, dilution works in the same way (increase in space due to dilution favors side with more moles) a 1 M HC 2 H 3 O 2 (acetic acid) solution is diluted by adding an equal volume of water. How does this reaction change? HC 2 H 3 O 2 (aq) ↔ H + (aq) + C 2 H 3 O 2 - (aq)

6 Chem 1B - Equilibrium Le Châtelier’s Principle – Temperature Stress Note: change in T changes K (while initial K becomes Q) If ΔH>0, as T increases, products favored - this also means K increases with T If ΔH<0, as T increases, reactants favored Easiest to remember by considering heat a reactant or product Example: OH - + H + ↔ H 2 O(l) + heat (reaction  H < 0) Increase in T

7 Chem 1B - Equilibrium Le Châtelier’s Principle Looking at the reaction below, that is initially at equilibrium, AgCl(s) ↔ Ag + (aq) + Cl - (aq) (ΔH°>0) determine the direction (toward products or reactants) each of the following changes will result in a)increasing the temperature b)addition of water (dilution of system) c)addition of AgCl(s) d)addition of NaCl

8 Chem 1B – Aqueous Chemistry Acid Definitions Arrhenius Definition (most narrow definition, but most familiar to us) –An acid releases H + (essentially the same as H 3 O + ) when dissolved in water –A base releases OH - upon dissolution in water –Acid Example: HClO 4 can write reaction as: HClO 4 (aq) + H 2 O(l) → H 3 O + (aq) + ClO 4 - (aq) or as: HClO 4 (aq) → H + (aq) + ClO 4 - (aq) (simplified view) –Base Example: KOH KOH(aq) → K + (aq) + OH - (aq) makes Arrhenius acid an acid makes KOH a base

9 Chem 1B – Aqueous Chemistry Acid Definitions Brønsted-Lowry (broader definition because it applies to non-aqueous solutions) –An acid is a proton (H + ) donor –A base is a proton acceptor (normally must have an available electron pair) –Acid-base reactions will have both an acid and a base (different than Arrhenius definition) Example: acetic acid in water HC 2 H 3 O 2 (aq) + H 2 O(l) ↔ H 3 O + (aq) + ClO 4 - (aq) acid base C 2 H 3 O 2 -H + :O-H ↔ H-O + -H + C 2 H 3 O 2 - H H

10 Chem 1B – Aqueous Chemistry Acid Definitions Examples –In the following examples, indicate Arrhenius and Brønsted-Lowry acids and bases: 1)NH 3 (aq) + H 2 O(l) ↔ OH - (aq) + NH 4 + (aq) 2)HNO 2 (aq) + H 2 O(l) ↔ H 3 O + (aq) + NO 2 - (aq) 3)HCl(aq) → H + (aq) + Cl - (aq) 4)NH 4 + (aq) ↔ H + (aq) + NH 3 (aq) 5)Al(H 2 O) 6 3+ (aq) ↔ Al(H 2 O) 5 OH 2+ (aq) + H + (aq) 6)H 2 SO 4 (sol’n) + HC 2 H 3 O 2 (l) ↔ HSO 4 - (sol’n) + H 2 C 2 H 3 O 2 + (sol’n)

11 Chem 1B – Aqueous Chemistry Acid Definitions Brønsted-Lowry – Conjugate acids and bases –When a Brønsted-Lowry acid loses its H +, the remainder is called a conjugate base –When a Brønsted-Lowry base extracts a H +, the remainder is called a conjugate acid Examples HNO 2 (aq) + H 2 O(l) ↔ H 3 O + (aq) + NO 2 - (aq) NH 3 (aq) + H 2 O(l) ↔ OH - (aq) + NH 4 + (aq) Conjugate base Conjugate acid

12 Chem 1B – Aqueous Chemistry Acid Definitions Lewis Acids and Bases (15.10) –A Lewis Acid is an electron pair acceptor –A Lewis Base is an electron pair donor –Brønsted-Lowry acids are also Lewis acids (but not always visa versa) –Examples: :NH 3 (aq) + H 2 O(l) ↔ OH - (aq) + NH 4 + (aq) Electron pair on N makes NH 3 a base Ag + + 2NH 3 (aq) ↔ Ag(NH 3 ) 2 + (aq) (formation of metal- ligand complex) Top example H 2 O(l) is both a Brønsted-Lowry and also a Lewis acid, while Ag + in bottom is only a Lewis acid

13 Chem 1B – Aqueous Chemistry Acid Strength In water, extent of formation of H + gives strength of acid Generic Acid: HA HA(aq) ↔ H + (aq) + A - (aq) The more strongly the above reaction favors the products, the stronger the acid This is given by the equilibrium constant – called K a in this example where: K a = [H + (aq)][A - (aq)]/[HA(aq)]

14 Chem 1B – Aqueous Chemistry Acid Strength – Strong Acids These are characterized by a K a >>1 in which the products dominate to an extent that no reactant is expected Also said to fully dissociate Example: HCl HCl(aq) → H + (aq) + Cl - (aq) No HCl(aq) expected Other strong acids: HBr, HI HNO 3 HClO 4 H 2 SO 4 (only first loss of H + )

15 Chem 1B – Aqueous Chemistry Acid Strength – Weak Acids Partially dissociate in water (so for HA, it will exist to some extent as both HA(aq) and A - (aq) These are characterized by 0.01 > K a > 10 -12 Examples: –HC 2 H 3 O 2 – acetic acid (in vinegar) –HCHO 2 – formic acid (used by some ants) –HClO – hypochlorous acid (bleach) Stronger weak acids have larger K a values Which is stronger: HCN (K a = 4.9 x 10 -10 ) or HC 2 H 3 O 2 (K a = 1.8 x 10 -5 )?

16 Chem 1B – Aqueous Chemistry Autoprotolysis of Water and pH Water, and some other “protic” solvents, reacts with itself as both an acid and a base 2H 2 O(l) ↔ H 3 O + (aq) + OH - (aq) or H 2 O(l) ↔ H + (aq) + OH - K = K w = [H + ][OH - ] = 1.0 x 10 -14 (at 25ºC) For pure water (using an ICE approach), we can show [H + ] = [OH - ] (= x) And K w = [H + ][OH - ] = [H + ] 2 or [H + ] = (K w ) 0.5 Or [H + ] = [OH - ] = 1.0 x 10 -7 M An acidic solution is where [H + ] > [OH - ] or [H + ] > 1.0 x 10 -7 M

17 Chem 1B – Aqueous Chemistry Autoprotolysis of Water and pH The pH scale is based on [H + ] but on a log scale pH = -log[H + ] (note: inverse relationship between acid conc. and pH) Very acidic solution (1 M HCl) pH = 0.0 Neutral solution [H + ] = [OH - ] = 10 -7 M: pH = 7.0 Very basic solution (1 M KOH) pH = 14.0

18 Chem 1B – Aqueous Chemistry Acid Strength and pH A Few Questions: 1)An unknown acid is dissolved in water so that [HA] o = 0.010 M. The pH is measured and found to be 3.90. Is this a strong or a weak acid? 2)ClO - is known as the: _________ __________(acid or base) of HClO? 3)At around 45ºC, K w = 4.0 x 10 -14. What is neutral pH at this K w value? 4)At a pH of a solution at equilibrium is 12.11 (and at 25ºC where K w = 1.0 x 10 -14 ), what is [OH - ]?

19 Chem 1B – Aqueous Chemistry Equilibrium Problems Involving Acids Strong Acids (large K a values) Example: Determine the pH of a 0.014 M HCl solution We could set up an ICE table, but with strong acids, we assume 100% dissociation HCl(aq) → H + (aq) + Cl - (aq) So [H + ] = [HCl(aq)] o = 0.014 M pH = -log[H + ] = 1.85 Note: this approach works as long as [H + ] from HA is greater than [H + ] from H 2 O

20 Chem 1B – Aqueous Chemistry Equilibrium Problems Involving Acids Weak Acids (K a values < 1) Example: Determine the pH of HCHO 2 (formic acid, K a = 1.8 x 10 -4 ) solution initially made to be 0.20 M.


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