acids and bases review/equilibrium reversible reaction: r p and r p acid dissociation is a...

Acids and Bases Review/Equilibrium reversible reaction: R P and R P Acid dissociation is a reversible reaction and is said to be in equilibrium. H 2 SO 4 2 H + + SO 4 2

< 7 Acids and Bases pH taste ______ react with ______ proton (H + ) donor Both are electrolytes: turn litmus lots of H + /H 3 O + react w/metals pH taste ______ react with ______ proton (H + ) acceptor turn litmus lots of OH dont react w/metals sour bases red > 7 bitter acids blue they conduct electricity in soln litmus paper Aliens: Acid Blood Robot Chicken: Alien Blood

H + and H 3 O + are used interchangeably closer to reality; hydronium ion faster to write; hydrogen ion/ proton H3O+H3O+ H + H + versus H 3 O +

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ACIDBASE NEUTRAL pH scale: measures acidity/basicity Each step on pH scale represents a factor of ___. pH 5vs. pH 6 ___X more acidic pH 3 vs. pH 5:_______X different pH 8 vs. pH 13:_______X different 10 10 100 100,000 pH = log [H 3 O + ]

Measuring pH pH meter acid-base indicators: e.g., pH paper is paper impregnated with mixtures of various indicators. a pair of electrodes chemicals whose color depends on pH Litmus 7 Phenolphthalein 8.2 bromthymol blue 7.6 methyl red 6.2 4 5 6 7 8 9 10 R O Y G B I V

Common Acids Strong Acids stomach acid; (dissociate ~100%) hydrochloric acid: HCl H + + Cl pickling: cleaning metals w /conc. HCl sulfuric acid: H 2 SO 4 2 H + + SO 4 2 #1 chemical;(auto) battery acid explosives; nitric acid: HNO 3 H + + NO 3 fertilizer

Common Acids (cont.) Weak Acids (dissociate very little) acetic acid: CH 3 COOH H + + CH 3 COO hydrofluoric acid: HF H + + F citric acid: H 3 C 6 H 5 O 7 ascorbic acid: H 2 C 6 H 6 O 6 lactic acid: CH 3 CHOHCOOH vinegar; naturally made by apples used to etch glass; extremely caustic lemons or limes; sour candy vitamin C waste product of muscular exertion

LSD Acid Lysergic acid diethylamide, C 20 H 25 N 3 O LSD from the German :"Lysergsure- diethylamid Ironically, not acidic, but slightly basic Derived from ergot, a grain fungus

carbonic acid: H 2 CO 3 carbonated beverages CO 2 + H 2 O H 2 CO 3 dissolves limestone (CaCO 3 ) rainwater in air H 2 CO 3 : cave formation H 2 CO 3 : natural acidity of lakes H 2 CO 3 : beverage carbonation

Acid Attacks (primarily on women) Developing Nations Around the World UK Model Katie Piper Katie Piper before her acid attack and after.

Common Bases Strong Bases Lye: used to make soap; clogged drain cleaner (dissociate ~100%) sodium hydroxide: NaOH Na + + OH ammonia: NH 3 + H 2 O NH 4 + + OH common household cleaning soln (Windex); hair dye household bleach; pool chlorine sodium hypochlorite: NaClO + H 2 O HClO + OH Slaked lime: limestone plus water; mortar/plaster calcium hydroxide: Ca(OH) 2 Ca 2+ + 2 OH Weak Bases (dissociate very little)

Strong Acids and Bases these are strong electrolytes that exist entirely as ions in aqueous solution memorize the names and formulas of the seven strong acids......and the eight strong, hydroxide bases... the hydroxides of... strong base cations Li, Na, K, Rb, Cs, Ca, Sr, Ba hydrochloric, HCl hydrobromic, HBr hydroiodic, HI chloric, HClO 3 perchloric, HClO 4 nitric, HNO 3 sulfuric, H 2 SO 4 Polyatomic Ion Sheet Halogens

Molarity (M) Review molarity (M) = moles of solute L of soln used most often in this class mol L M Na + How many mol solute are reqd to make 1.35 L of 2.50 M soln? What mass sodium hydroxide is this? mol L M mol = M L= 2.50 M (1.35 L ) = 3.38 mol 3.38 mol= 135 g NaOH OH NaOH

Dissociation and Ion Concentration Strong acids or bases dissociate ~100%. HNO 3 H + + NO 3 NaOH Na + + OH For strongs, we often use two arrows of differing length OR just a single arrow, H + NO 3 + H+H+ 1 2 100 1000/L 0.0058 M 1 2 100 1000/L 0.0058 M 1 2 100 1000/L 0.0058 M + + + + +

HClH + + Cl 4.0 M + monoprotic acid H 2 SO 4 2 H + + SO 4 2 2.3 M4.6 M2.3 M+ SO 4 2 H+H+ H+H+ H+H+ H+H+ + diprotic acid Ca(OH) 2 Ca 2+ 2 OH + 0.025 M 0.050 M+

pH Calculations Recall that the hydronium ion (H 3 O + ) is the species formed when hydrogen ion (H + ) attaches to water (H 2 O). OH is the hydroxide ion. For right now, in any aqueous soln, [ H 3 O + ] [ OH ] = 1 x 10 14 ( or [ H + ] [ OH ] = 1 x 10 14 ) so whether were counting front wheels (i.e., H + ) or big wheels (i.e., H 3 O + ) doesnt much matter. H+H+ H3O+H3O+ The number of front wheels is the same as the number of big wheels

At 25 o C, calculate the hydrogen ion concentration if the hydroxide ion concentration is 2.7 x 10 4 M. Is this solution an acid or a base? [H + ] [OH ] = 1.0 x 10 14 [H + ] (2.7 x 10 4 ) = 1.0 x 10 14 [H + ] = 3.7 x 10 11 M [H + ] < [OH ] base

Given:Find: A. [ OH ] = 5.25 x 10 6 M[ H + ] = 1.90 x 10 9 M B. [ OH ] = 3.8 x 10 11 M [ H 3 O + ] = 2.6 x 10 4 M C. [ H 3 O + ] = 1.8 x 10 3 M [ OH ] = 5.6 x 10 12 M D. [ H + ] = 7.3 x 10 12 M [ H 3 O + ] = 7.3 x 10 12 M Find the pH of each soln above. Remember pH = log [ H 3 O + ]( or pH = log [ H + ] ) A. B.C.D. 3.592.7411.13 8.72 pH = log [ H 3 O + ]= log [1.90 x 10 9 M ] log 19EE9= . [ H 3 O + ] [ OH ] = 1 x 10 14

A few last equations pOH = log [ OH ] pH + pOH = 14 [ OH ] = 10 pOH [ H 3 O + ] = 10 pH ( or [ H + ] = 10 pH ) pOH pH [ OH ] [ H 3 O + ] pH + pOH = 14 [ H 3 O + ] [ OH ] = 1 x 10 14 [ H 3 O + ] = 10 pH pH = log [ H 3 O + ] [ OH ] = 10 pOH pOH = log [ OH ]

1. If pH = 4.87, find [ H 3 O + ]. pOH pH [ H 3 O 1+ ] pH + pOH = 14 pOH pH [ OH ] [ H 3 O + ] pH + pOH = 14 [ H 3 O + ] [ OH ] = 1 x 10 14 [ H 3 O + ] = 10 pH pH = log [ H 3 O + ] [ OH ] = 10 pOH pOH = log [ OH ] [ H 3 O + ] = 10 pH = 10 4.87 On a graphing calculator log .8=74 2 nd 10 x [ H 3 O + ] = 1.35 x 10 5 M

For the following problems, assume 100% dissociation. 2. Find pH of a 0.00057 M nitric acid (HNO 3 ) soln. HNO 3 0.00057 M (Who cares?)(GIVEN)(affects pH) H + + NO 3 pH = log [ H 3 O + ] = log (0.00057) = 3.24 pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ]

3. Find pH of a soln with 3.65 g HCl in 2.00 dm 3 of soln. HCl H + + Cl [ HCl ] 0.05 M pH = log [ H + ] = log (0.05) = 1.3 pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] (space) = 0.05 M HCl = M HCl 3.65 g = 2.00 L

4. Find the concentration of an H 2 SO 4 soln w /pH 3.38. H 2 SO 4 2 H + + SO 4 2 [ H + ] = 10 pH = 10 3.38 = 4.2 x 10 4 M pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] (space) 4.2 x 10 4 M(Who cares?)X M 2.1 x 10 4 M [ H 2 SO 4 ] = 2.1 x 10 4 M

5. If [ OH ] = 5.6 x 10 11 M, find pH. pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] Find [ H 3 O + ] Find pOH = 1.79 x 10 4 M Then find pH pH = 3.75 = 10.25

6. Find pH of a 3.2 x 10 5 M barium hydroxide (Ba(OH) 2 ) soln. Ba(OH) 2 3.2 x 10 5 M 6.4 x 10 5 M (Who cares?)(GIVEN)(affects pH) Ba 2+ + 2 OH pOH = log [ OH ] = log (6.4 x 10 5 ) = 4.19 pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pH = 9.81

7. What mass of Al(OH) 3 is reqd to make 15.6 L of a soln with a pH of 10.72? Al(OH) 3 Al 3+ + 3 OH pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pOH pH [ OH 1 ] [ H 3 O 1+ ] pH + pOH = 14 [ H 3 O 1+ ] [ OH 1 ] = 1 x 10 14 [ H 3 O 1+ ] = 10 pH pH = log [ H 3 O 1+ ] [ OH 1 ] = 10 pOH pOH = log [ OH 1 ] pOH = 3.28 = 10 3.28 = 5.25 x 10 4 M (space) [ OH ] = 10 pOH 5.25 x 10 4 M(w.c.?)1.75 x 10 4 M mol = 1.75 x 10 4 (15.6) = 0.213 g Al(OH) 3 mol Al(OH) = M L 3 = 0.00273 mol Al(OH) 3

Amphoteric substances can be acids or bases, depending on the reaction conditions (e.g. H 2 O) HCl(aq) + H 2 O(l) NH 3 (aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) NH 4 + (aq) + OH (aq) NH 3 is another example. [] + [] NH 3 (aq) + H 2 O(l) NH 2 - (aq) + H 3 O + (aq) HCO 3 - (aq) + H 2 O(l) OH (aq) + H 2 CO 3 (aq)

The two substances in a conjugate acid-base pair differ by a H +... Strong acids / bases easily / ____ H +. Weak acids / bases do NOT easily / ____ H +. and the acid has the extra H +. In acid-base equilibria, protons are donated in forward and reverse reactions. HNO 2 (aq) + H 2 O(l)NO 2 (aq) + H 3 O + (aq) ACID CONJ. BASE base conj. acid donate donate accept accept The stronger a/n acid / base, the weaker its conj. base / acid.

__HCl + __NaOH ________ + ______ __H 3 PO 4 + __KOH ________ + ______ __H 2 SO 4 + __NaOH ________ + ______ __HClO 3 + __Al(OH) 3 ________ + ______ ________ + ________ __AlCl 3 + ______ ________ + ________ __Fe 2 (SO 4 ) 3 + ______ 1 Neutralization Reaction ACID + BASE SALT + WATER NaClH2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O 1 1 1 13 12 31 K 3 PO 4 3 1 Na 2 SO 4 2 Al(ClO 3 ) 3 3 1 HClAl(OH) 3 31 H 2 SO 4 Fe(OH) 3 32 1 6 31 1

Titration If an acid and a base are mixed together in the right amounts, the resulting solution will be perfectly neutralized and have a pH of 7. -- For pH = 7.. then mol = M L In a titration, the above equation helps us to use a KNOWN conc. of acid (or base) to determine an UNKNOWN conc. of base (or acid). mol H 3 O 1+ = mol OH 1

2.42 L of 0.32 M HCl are used to titrate 1.22 L of an unknown conc. of KOH. Find the molarity of the KOH. HCl H + + Cl KOH K + + OH 0.32 M X M 0.32 M(2.42 L)=(1.22 L) 1.22 L = M KOH = 0.63 M [ H 3 O + ] V A = [ OH ] V B [ OH ] Acid Base Fill up flask with acid Then, titrate with base

458 mL of HNO 3 ( w /pH = 2.87) are neutralized w /661 mL of Ba(OH) 2. What is the pH of the base? [ H 3 O + ] = 10 pH = 10 2.87 = 1.35 x 10 3 M [ H 3 O + ] V A = [ OH ] V B (1.35 x 10 3 )(458 mL) = [ OH ] (661 mL) [ OH ] = 9.35 x 10 4 M pOH = log (9.35 x 10 4 )= 3.03 pH = 10.97 OK If we find this, we can find the bases pH. OK

How many L of 0.872 M sodium hydroxide will titrate 1.382 L of 0.315 M sulfuric acid? H 2 SO 4 2 H + + SO 4 2 NaOH Na + + OH 0.872 M (1.382 L) [ H 3 O + ] V A = [ OH ] V B 0.630 M=(V B )0.872 M 0.315 M0.630 M ?? V B = 0.998 L (H 2 SO 4 ) (NaOH)

Partial Neutralization 1.55 L of 0.26 M KOH 2.15 L of 0.22 M HCl pH = ? Procedure: 1. Calc. mol of substance, then mol H + and mol OH . 2. Subtract smaller from larger. 3. Find [ ] of whats left over, and calc. pH.

1.55 L of 0.26 M KOH 2.15 L of 0.22 M HCl mol KOH = = 0.403 mol OH 0.26 M (1.55 L)= 0.403 mol KOH mol HCl = = 0.473 mol H + 0.22 M (2.15 L)= 0.473 mol HCl [ H 1+ ] = = 0.070 mol H + = 0.0189 M H + 0.070 mol H + 1.55 L + 2.15 L LEFT OVER = log (0.0189) pH = log [ H + ] = 1.72 mol L M

4.25 L of 0.35 M hydrochloric acid is mixed w /3.80 L of 0.39 M sodium hydroxide. Find final pH. Assume 100% dissociation. (HCl) (NaOH) mol HCl = = 1.4875 mol H 1+ 0.35 M (4.25 L)= 1.4875 mol HCl mol NaOH = = 1.4820 mol OH 1 0.39 M (3.80 L)= 1.4820 mol NaOH [ H 1+ ] = = 0.0055 mol H 1+ = 6.83 x 10 4 M H 1+ 0.0055 mol H 1+ 4.25 L + 3.80 L LEFT OVER = log (6.83 x 10 4 )pH = log [ H 1+ ]= 3.17

5.74 L of 0.29 M sulfuric acid is mixed w /3.21 L of 0.35 M aluminum hydroxide. Find final pH. Assume 100% dissociation. (H 2 SO 4 ) (Al(OH) 3 ) mol H 2 SO 4 = = 3.3292 mol H + 0.29 M (5.74 L)= 1.6646 mol H 2 SO 4 mol Al(OH) 3 = = 3.3705 mol OH 0.35 M (3.21 L)= 1.1235 mol Al(OH) 3 = 0.0413 mol OH LEFT OVER [ OH 1 ] == 0.00461 M OH 0.0413 mol OH 5.74 L + 3.21 L pOH = log (0.00461)= 2.34 pH = 11.66

A. 0.038 g HNO 3 in 450 mL of soln. Find pH. = 0.45 L 0.038 g HNO 3 63 g 1 mol () = 6.03 x 10 4 mol HNO 3 1000 mL 1 L () = 6.03 x 10 4 mol H 1+ [ H 1+ ] == 1.34 x 10 3 M H 1+ 6.03 x 10 4 mol H 1+ 0.45 L = log (1.34 x 10 3 )pH = log [ H 1+ ]= 2.87 mol L M

B. 0.044 g Ba(OH) 2 in 560 mL of soln. Find pH. = 0.56 L 1000 mL 1 L () 0.044 g Ba(OH) 2 171.3 g 1 mol () = 2.57 x 10 4 mol Ba(OH) 2 = 5.14 x 10 4 mol OH 1 [ OH 1 ] == 9.18 x 10 4 M OH 1 5.14 x 10 4 mol OH 1 0.56 L pOH = log (9.18 x 10 4 )= 3.04 pH = 10.96 mol L M

C. Mix them. Find pH of resulting soln. From acid 6.03 x 10 4 mol H 1+ From base 5.14 x 10 4 mol OH 1 = 8.90 x 10 5 mol H 1+ LEFT OVER [ H 1+ ] == 8.81 x 10 5 M H 1+ 8.90 x 10 5 mol H 1+ 0.45 L + 0.56 L = log (8.81 x 10 5 )pH = log [ H 1+ ]= 4.05

Chemical Equilibrium

Chemical equilibrium is reached when reaction rates become equal, and the concentrations of Rs and Ps no longer change reactants products system must be closed equilibrium is a dynamic process (although it might look static) rate at which R P rate at which P R = amt. of R = amt. of P

law of mass action: expresses the relationship between [ ]s of R and P in any reaction For a system at equilibrium with the balanced equationaA + bB pP + qQ the equilibrium-constant expression is: i.e., PRODUCTS REACTANTS )( You can only include reaction species in the gaseous or aqueous phase. Do NOT include solids/liquids.

Write the equilibrium-constant expressions for the following reactions. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) Fritz Haber (18681934) discovered a way to generate ammonia from hydrogen and nitrogen at high pressure. The ammonia was needed for Germanys munitions industry, which was cut off from the nitrate sources of South America by the British blockade during WWI.

Write the equilibrium-constant expression for CaCO 3 (s) CaO(s) + CO 2 (g). Write expressions for K c. CO 2 (g) + H 2 (g) CO(g) + H 2 O(l) SnO 2 (s) + 2 CO(g) Sn(s) + 2 CO 2 (g)

Type of Equilibrium Constants (K) There are lots of different Ks they are all the same concept, just different types of equations. N 2 (g) + 3H 2 (g) 2NH 3 (g) HNO 2 (aq) + H 2 O(l) H 3 O + (aq) + NO 2 - (aq) NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) For use with acids For use with bases K conc K press

Calculating K H 2 (g) + I 2 (g) 2HI(g) The system is allowed to reach equilibrium and the following data is collected: 1.[H 2 ] = 4.953x10 -4, [I 2 ] =4.953x10 -4, [HI] = 3.655x10 -3 2. [H 2 ] = 1.141x10 -3, [I 2 ] = 1.141x10 -3, [HI] = 8.410x10 -3 3. [H 2 ] = 3.560x10 -3, [I 2 ] = 1.250x10 -3, [HI] = 1.559x10 -2

they are reciprocals The Magnitude of the Equilibrium Constant If K >> 1... If K

For the above equation, K c = If we exclude the pure liquid... This equation is taken to be valid for pure water and for dilute aqueous solutions. [ H + ] > [ OH ][ H + ] < [ OH ][ H + ] = [ OH ] H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) [H 3 O + ] [OH ] [H 2 O] 2 K w = [H 3 O + ] [OH ] = [H + ] [OH ] = 1.0 x 10 14 (@ 25 o C) BASENEUTRALACID

Acid-Dissociation Constant: K a For the generic reaction in soln:A + B C + D For strong acids, e.g., HCl HCl H + + Cl = BIG. Assume 100% dissociation; K a NOT applicable for strong acids. lots~0lots

large K a : Weak Acids Most acids are weak (i.e., only partially ionized) For a weak acid HX... HX(aq) H + (aq) + X (aq) acid-dissociation constant K a = [H + ] [X ] [HX] small K a : stronger acid weaker acid The % of a weak acid that is ionized is given by the equation: [H + ] at eq. x 100% ionization = [acid] orig. = small K a acetic acid = 1.8 x 10 5

If you know the concentrations of only some substances at equilibrium, make a chart and use reaction stoichiometry to figure out the other concentrations at equilibrium. THEN plug and chug. What kind of chart? Ice, ice, baby I = initial C = change E = equilibrium

Weak Acid Equilibrium: K a 1. Write the reaction for carbonic acid (H 2 CO 3 ) dissociating in water. H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 - (aq) a.Write the equilibrium expression for this reaction. b. If [H 2 CO 3 ] = 0.100M and it shows 0.212% ionization, what is the value of Ka? H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 - (aq) Initial Change Equilibrium 0.100M-------00 - 2.12x10 -4 ------- + 2.12x10 -4 + 2.12x10 -4 0.099788-------2.12x10 -4 2.12x10 -4

Weak Acid Equilibrium: K a Write the reaction for carbonic acid (H 2 CO 3 ) dissociating in water. H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 - (aq) c.If [H 2 CO 3 ] = 0.0500M and it shows 0.300% ionization, what is the value of Ka? H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 - (aq) ICEICE 0.0500M-------00 - 1.50x10 -4 ------- + 1.50x10 -4 + 1.50x10 -4 0.04985-------1.50x10 -4 1.50x10 -4

Weak Acid Equilibrium: K a Write the reaction for carbonic acid (H 2 CO 3 ) dissociating in water. H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 - (aq) c. Given the value of Ka, if [H 2 CO 3 ] = 0.0100M, what will the concentrations of the products? What is the % ionization? H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 - (aq) ICEICE 0.0100M-------00 - X------- + X + X 0.0100-X-------XX 0 X 2 = 4.50x10 -7 (0.01) X = (4.50x10 -9 ) X = 6.71x10 -5 M < 5.00% 5% Rule: at equilibrium, [H 2 CO 3 ] initial [H 2 CO 3 ] 0. Only good if X < 5% of the initial concentration!

2. The reaction for benzoic acid (HC 7 H 5 O 2 ) dissociating in water: If [HC 7 H 5 O 2 ] = 1.500M and it ionizes 0.651%, calculate the Ka for benzoic acid. HC 7 H 5 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 7 H 5 O 2 - (aq) Calculate the pH for this solution. Weak Acid Equilibrium: K a ICEICE 1.500M-------00 -.009765------- +.009765 +.009765 1.490------- +.009765 +.009765 [H 3 O + ] = 0.009765M pH = -log (0.009765) = 2.01

3. Write the reaction for boric acid (H 3 BO 3 ) dissociating in water. Given Ka = 5.4x10 -10, if [H 3 BO 3 ] = 0.500M, what will be the pH of the resulting solution? H 3 BO 3 (aq) + H 2 O(l) H 3 O + (aq) + H 2 BO 3 - (aq) ICEICE 0.500M-------00 - X------- + X + X 0.500-X-------XX 0 X 2 = 5.40x10 -10 (0.50) X = (2.70x10 -10 ) X = 1.64x10 -5 M X = [H 3 O + ] = 1.64x10 -5 M pH = -log (1.64x10 -5 ) = 4.78

The reaction for ammonia (NH 3 ) in water is as follows: If K b = 1.80x10 -5 and [NH 3 ] = 2.500M, calculate the pH of this solution. NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Calculate the pH for this solution. Weak Base Equilibrium: K b ICEICE 2.500M-------00 - X------- + X + X 2.500 - X------- + X + X X= [OH - ] = 0.00671M pOH = -log (0.00671M) = 2.17 11.83 0 X 2 = 1.80x10 -5 (2.50) X = (4.50x10 -5 ) X = 0.00671M

Le Chatliers Principle H 2 NH 3 N 2 orig. eq. new eq. H 2 added at this time system counteracting stress N 2 (g) + 3 H 2 (g) 2 NH 3 (g) As long as T stays the same, K stays the same! The changes keep K the same. When a system at equilibrium is disturbed, it shifts to a new equilibrium that counteracts the disturbance.

Le Chateliers principle: When a system at equilibrium is disturbed, it shifts to a new equilibrium that counteracts the disturbance. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Disturbance Equilibrium Shift Add more N 2 Add more H 2 Add more NH 3 . Add a catalyst.. Remove NH 3 Anything with a (s) or (l) no shift

Change in pressure for gaseous equilibrium systems 2 NO 2 (g)N 2 O 4 (g) If we decrease pressure, the system wants to... (goes to the side with more gas) increase it If we increase pressure, the system wants to... (goes to the side with less gas) decrease it SHIFT For H 2 (g) + I 2 (g) 2 HI(g), pressure changes result in... NO SHIFT

Changes in temperature These almost always result in... shifts in eq. AND changes in K. For exothermic reactions: R P + heat ( H is ____) as T increases... shift,K as T decreases... shift,K For endothermic reactions: R + heat P ( H is ____) as T increases... shift,K as T decreases... shift,K +

For PCl 5 (s) PCl 3 (g) + Cl 2 (g), H o = 87.9 kJ. Predict shifts for... (a) adding Cl 2 (b) increasing temperature (c) decreasing volume (d) adding PCl 5 (s) PCl 5 (s) + 87.9 kJ PCl 3 (g) + Cl 2 (g) We might want to rewrite the eq. as SHIFT NO SHIFT

AgCl + energy Ag o + Cl o shift to a new equilibrium: Then go inside shift to a new equilibrium: Light-Darkening Eyeglasses energy Go outside Sunlight more intense than inside light; GLASSES DARKEN (clear) (dark) energy GLASSES LIGHTEN

In a chicken CaO + CO 2 CaCO 3 In summer, [ CO 2 ] in a chickens blood due to panting. -- How could we increase eggshell thickness in summer? (eggshells) eggshells are thinner shift ; give chickens carbonated water put CaO additives in chicken feed [ CO 2 ], shift [ CaO ], shift I wish I had sweat glands.

acids and bases review/equilibrium reversible reaction: r p and r p acid dissociation is a...

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