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Acids and Bases Review/Equilibrium

H 2 SO 4 2 H + + SO 4 2–. 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. . Acids and Bases. litmus paper. < 7. > 7. pH . pH . sour. bitter.

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Acids and Bases Review/Equilibrium

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  1. H2SO4 2 H+ + SO42– 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.

  2. Acids andBases litmus paper < 7 > 7 pH pH sour bitter taste ______ taste ______ bases acids react with ______ react with ______ proton (H+) donor proton (H+) acceptor red blue turn litmus turn litmus lots of H+/H3O+ lots of OH– react w/metals don’t react w/metals Both are electrolytes: they conduct electricity in sol’n

  3. ACID BASE 0 1 2 3 4 5 6 78 9 10 11 12 13 14 NEUTRAL pH = –log [H3O+] pH scale: measures acidity/basicity 10 Each step on pH scale represents a factor of ___. pH 5 vs. pH 6 10 ___X more acidic pH 3 vs. pH 5: _______X different 100 100,000 pH 8 vs. pH 13: _______X different

  4. a pair of electrodes Measuring pH • pH meter • acid-base indicators: chemicals whose color depends on pH e.g., Litmus<7 >7 Phenolphthalein<8.2 >8.2 bromthymolblue<6 >7.6 methylred<4.4 >6.2 • pH paper is paper impregnated with mixtures of various indicators. 4 5 6 7 8 9 10 ROYGBIV

  5. hydrochloric acid: HCl H++ Cl– sulfuric acid: H2SO42 H++ SO42– nitric acid: HNO3H++ NO3– Common Acids (dissociate ~100%) Strong Acids stomach acid; pickling: cleaning metals w/conc. HCl #1 chemical; (auto) battery acid explosives; fertilizer

  6. acetic acid: CH3COOHH++ CH3COO– hydrofluoric acid: HF H++ F– citric acid: H3C6H5O7 ascorbic acid: H2C6H6O6 lactic acid: CH3CHOHCOOH Common Acids (cont.) Weak Acids (dissociate very little) vinegar; naturally made by apples used to etch glass; extremely caustic lemons or limes; sour candy vitamin C waste product of muscular exertion

  7. H2CO3: beverage carbonation rainwater dissolves limestone (CaCO3) in air CO2 + H2O H2CO3 H2CO3: cave formation H2CO3: natural acidity of lakes carbonic acid: H2CO3 • carbonated beverages

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

  9. sodium hydroxide: NaOH Na++ OH– calcium hydroxide: Ca(OH)2 Ca2++ 2OH– ammonia: NH3 + H2O NH4+ + OH– sodium hypochlorite: NaClO + H2O HClO + OH– Common Bases (dissociate ~100%) Strong Bases Lye: used to make soap; clogged drain cleaner “Slaked lime”: limestone plus water; mortar/plaster Weak Bases (dissociate very little) common household cleaning sol’n (Windex); hair dye household bleach; pool “chlorine”

  10. hydrochloric, HCl hydrobromic, HBr hydroiodic, HI chloric, HClO3 perchloric, HClO4 nitric, HNO3 sulfuric, H2SO4 • these are strong electro- • lytes that exist entirely as • ions in aqueous solution Strong Acids and Bases • memorize the names • and formulas of the • seven strong acids... ...and the eight strong, hydroxide bases... the hydroxides of... Halogens Li, Na, K, Rb, Cs, Ca, Sr, Ba Polyatomic Ion Sheet “strong base cations”

  11. mol M L Molarity (M) Review mol molarity (M) = moles of solute L of sol’n M L • used most often in this class How many mol solute are req’d to make 1.35 L of 2.50 M sol’n? mol = M L = 2.50 M (1.35 L ) = 3.38 mol What mass sodium hydroxide is this? Na+ OH– NaOH 3.38 mol = 135 g NaOH

  12. HNO3 H+ + NO3– NaOH Na+ + OH– NO3– + NO3– H+ H+ 1 2 100 1000/L 0.0058 M Dissociation and Ion Concentration Strong acids or bases dissociate ~100%. For “strongs,” we often use two arrows of differing length OR just a single arrow, 1 + 1 2 + 2 100 + 100 1000/L + 1000/L 0.0058 M + 0.0058 M

  13. monoprotic acid diprotic acid H+ H+ SO42– H+ H+ HCl H+ + Cl– 4.0 M 4.0 M + 4.0 M H2SO4 2 H+ + SO42– + SO42– 2.3 M 4.6 M + 2.3 M Ca(OH)2 Ca2+ + 2 OH– 0.025 M 0.025 M + 0.050 M

  14. The number of front wheels is the same as the number of big wheels… pH Calculations Recall that the hydronium ion (H3O+) is the species formed when hydrogen ion (H+) attaches to water (H2O). OH– is the hydroxide ion. H3O+ H+ …so whether we’re counting front wheels (i.e., H+) or big wheels (i.e., H3O+) doesn’t much matter. For right now, in any aqueous sol’n, [ H3O+ ] [ OH– ] = 1 x 10–14 ( or [ H+ ] [ OH– ] = 1 x 10–14 )

  15. At 25oC, 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

  16. – – 1 9 . EE 9 = log [ H3O+ ] [ OH– ] = 1 x 10–14 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 [ H3O+ ] = 2.6 x 10–4 M C. [ H3O+ ] = 1.8 x 10–3 M [ OH– ] = 5.6 x 10–12 M D. [ H+ ] = 7.3 x 10–12 M [ H3O+ ] = 7.3 x 10–12 M Find the pH of each sol’n above. Remember… pH = –log [ H3O+ ] ( or pH = –log [ H+ ] ) A. pH = –log [ H3O+ ] = –log [1.90 x 10–9 M ] 8.72 B. 3.59 C. 2.74 D. 11.13

  17. [ H3O+ ] = 10–pH pH [ H3O+ ] pH = –log [ H3O+ ] pH + pOH = 14 [ H3O+ ] [ OH– ] = 1 x 10–14 [ OH– ] = 10–pOH pOH [ OH– ] pOH = –log [ OH– ] A few last equations… [ H3O+ ] = 10–pH ( or [ H+ ] = 10–pH ) pOH = –log [ OH– ] [ OH– ] = 10–pOH pH + pOH = 14

  18. [ H3O+ ] = 10–pH pH pH [ H3O1+ ] [ H3O+ ] pH = –log [ H3O+ ] [ H3O+ ] [ OH– ] = 1 x 10–14 pH + pOH = 14 pH + pOH = 14 – 4 . 7 = 2nd log 8 10x [ OH– ] = 10–pOH pOH pOH [ OH– ] pOH = –log [ OH– ] 1. If pH = 4.87, find [ H3O+ ]. [ H3O+ ] = 10–pH = 10–4.87 On a graphing calculator… [ H3O+ ] = 1.35 x 10–5 M

  19. [ H3O1+ ] = 10–pH [ H3O1+ ] = 10–pH pH pH [ H3O1+ ] [ H3O1+ ] pH = –log [ H3O1+ ] pH = –log [ H3O1+ ] pH + pOH = 14 pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ OH1– ] = 10–pOH [ OH1– ] = 10–pOH pOH pOH [ OH1– ] [ OH1–] pOH = –log [ OH1– ] pOH = –log [ OH1– ] For the following problems, assume 100% dissociation. 2. Find pH of a 0.00057 M nitric acid (HNO3) sol’n. H+ + NO3– HNO3 0.00057 M 0.00057 M 0.00057 M (GIVEN) (affects pH) (“Who cares?”) pH = –log [ H3O+ ] = –log (0.00057) = 3.24

  20. (space) [ H3O1+ ] = 10–pH [ H3O1+ ] = 10–pH pH pH [ H3O1+ ] [ H3O1+ ] pH = –log [ H3O1+ ] pH = –log [ H3O1+ ] 2.00 L pH + pOH = 14 pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ OH1– ] = 10–pOH [ OH1– ] = 10–pOH pOH pOH [ OH1– ] [ OH1– ] pOH = –log [ OH1– ] pOH = –log [ OH1– ] 3. Find pH of a sol’n with 3.65 g HCl in 2.00 dm3 of sol’n. H+ + Cl– HCl 0.05 M 0.05 M 0.05 M 3.65 g [ HCl ] = MHCl = = 0.05 M HCl pH = –log [ H+ ] = –log (0.05) = 1.3

  21. 4.2 x 10–4 M (space) [ H3O1+ ] = 10–pH [ H3O1+ ] = 10–pH pH pH [ H3O1+ ] [ H3O1+ ] pH = –log [ H3O1+ ] pH = –log [ H3O1+ ] pH + pOH = 14 pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ OH1– ] = 10–pOH [ OH1– ] = 10–pOH pOH pOH [ OH1– ] [ OH1– ] pOH = –log [ OH1– ] pOH = –log [ OH1– ] 4. Find the concentration of an H2SO4 sol’n w/pH 3.38. 2 H+ + SO42– H2SO4 X M (“Who cares?”) 2.1 x 10–4 M [ H+ ] = 10–pH = 10–3.38 = 4.2 x 10–4 M [ H2SO4 ] = 2.1 x 10–4 M

  22. [ H3O1+ ] = 10–pH [ H3O1+ ] = 10–pH pH pH [ H3O1+ ] [ H3O1+ ] pH = –log [ H3O1+ ] pH = –log [ H3O1+ ] pH + pOH = 14 pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ OH1– ] = 10–pOH [ OH1– ] = 10–pOH pOH pOH [ OH1– ] [ OH1–] pH = 3.75 pH = 3.75 pOH = –log [ OH1– ] pOH = –log [ OH1– ] 5. If [ OH– ] = 5.6 x 10–11 M, find pH. Find [ H3O+ ] = 1.79 x 10–4 M Find pOH = 10.25 Then find pH…

  23. [ H3O1+ ] = 10–pH [ H3O1+ ] = 10–pH pH pH [ H3O1+ ] [ H3O1+ ] pH = –log [ H3O1+ ] pH = –log [ H3O1+ ] pH + pOH = 14 pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ OH1– ] = 10–pOH [ OH1– ] = 10–pOH pOH pOH [ OH1– ] [ OH1– ] pOH = –log [ OH1– ] pOH = –log [ OH1– ] 6. Find pH of a 3.2 x 10–5 M barium hydroxide (Ba(OH)2) sol’n. Ba2+ + 2 OH– Ba(OH)2 3.2 x 10–5 M 3.2 x 10–5 M 6.4 x 10–5 M (GIVEN) (“Who cares?”) (affects pH) pOH = –log [ OH– ] = –log (6.4 x 10–5) = 4.19 pH = 9.81

  24. (space) [ H3O1+ ] = 10–pH [ H3O1+ ] = 10–pH pH pH [ H3O1+ ] [ H3O1+ ] pH = –log [ H3O1+ ] pH = –log [ H3O1+ ] [ OH– ] = 10–pOH molAl(OH) = M L pH + pOH = 14 pH + pOH = 14 3 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ H3O1+ ] [ OH1– ] = 1 x 10–14 [ OH1– ] = 10–pOH [ OH1– ] = 10–pOH pOH pOH [ OH1– ] [ OH1– ] pOH = –log [ OH1– ] pOH = –log [ OH1– ] 7. What mass of Al(OH)3 is req’d to make 15.6 L of a sol’n with a pH of 10.72? Al3+ + 3 OH– Al(OH)3 1.75 x 10–4 M (“w.c.?”) 5.25 x 10–4 M pOH = 3.28 = 10–3.28 = 5.25 x 10–4 M mol = 1.75 x 10–4(15.6) = 0.00273 mol Al(OH)3 = 0.213 g Al(OH)3

  25. __HCl + __NaOH________ + ______ __H3PO4 + __KOH________ + ______ __H2SO4 + __NaOH________ + ______ ACID + BASE SALT + WATER __HClO3 + __Al(OH)3________ + ______ ________ + ________ __AlCl3 + ______ ________ + ________ __Fe2(SO4)3 + ______ Neutralization Reaction 1 1 NaCl H2O 1 1 1 3 K3PO4 H2O 1 3 1 2 Na2SO4 H2O 1 2 1 3 Al(ClO3)3 H2O 3 1 H2O 3 3 1 HCl Al(OH)3 1 3 2 6 H2SO4 Fe(OH)3 H2O 1

  26. 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…………….. mol H3O1+ = mol OH1– 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).

  27. 1.22 L 1.22 L 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. H+ + Cl– HCl 0.32 M 0.32 M K+ + OH– KOH X M X M [ H3O+ ] VA = [ OH– ] VB 0.32 M (2.42 L) = [ OH– ] (1.22 L) = MKOH = 0.63 M [ OH– ]

  28. 458 mL of HNO3 (w/pH = 2.87) are neutralized w/661 mL of Ba(OH)2. What is the pH of the base? [ H3O+ ] VA = [ OH– ] VB OK OK If we find this, we can find the base’s pH. [ H3O+ ] = 10–pH = 10–2.87 = 1.35 x 10–3 M (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

  29. (NaOH) How many L of 0.872 M sodium hydroxide will titrate 1.382 L of 0.315 M sulfuric acid? (H2SO4) [ H3O+ ] VA = [ OH– ] VB ? ? 0.630 M (1.382 L) = 0.872 M (VB) Na+ + OH– 2 H+ + SO42– NaOH H2SO4 0.315 M 0.630 M 0.872 M 0.872 M VB = 0.998 L

  30. Chemical Equilibrium

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

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

  33. N2 (g) + 3 H2 (g) 2 NH3 (g) 2 SO3 (g) 2 SO2 (g) + O2 (g) Fritz Haber (1868–1934) discovered a way to generate ammonia from hydrogen and nitrogen at high pressure. The ammonia was needed for Germany’s munitions industry, which was cut off from the nitrate sources of South America by the British blockade during WWI. Write the equilibrium-constant expressions for the following reactions.

  34. Write the equilibrium-constant expression for CaCO3(s) CaO(s) + CO2(g). CO2(g) + H2(g) CO(g) + H2O(l) SnO2(s) + 2 CO(g) Sn(s) + 2 CO2(g) Write expressions for Kc.

  35. Type of Equilibrium Constants (K) There are lots of different “K’s” they are all the same concept, just different types of equations. N2(g) + 3H2(g) ↔ 2NH3(g) HNO2(aq) + H2O(l) ↔ H3O+(aq) + NO2-(aq) NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq) Kconc≠ Kpress For use with acids… For use with bases…

  36. Calculating K • H2(g) + I2(g)  2HI(g) • The system is allowed to reach equilibrium and the following data is collected: • [H2] = 4.953x10-4, [I2] =4.953x10-4, [HI] = 3.655x10-3 • 2. [H2] = 1.141x10-3, [I2] = 1.141x10-3, [HI] = 8.410x10-3 • 3. [H2] = 3.560x10-3, [I2] = 1.250x10-3, [HI] = 1.559x10-2

  37. ( ) PRODUCTS REACTANTS The Magnitude of the Equilibrium Constant If K >> 1... products are favored. Eq. “lies to the right.” If K << 1... reactants are favored. Eq. “lies to the left.” The K for the forward and reverse reactions are NOT the same. • they are reciprocals • You must write out the reaction • and specify the temperature • when reporting a K.

  38. H2O(l) + H2O(l) H3O+(aq) + OH–(aq) The Autoionization of Water: Kw In ordinary water, we constantly have... • reaction is very rapid in each direction • at room temp., ~1 out • of a billion m’cules are • ionized, so pure water • is a poor conductor

  39. [H3O+] [OH–] H2O(l) + H2O(l) H3O+(aq) + OH–(aq) [H2O]2 [ H+ ] > [ OH– ] [ H+ ] < [ OH– ] [ H+ ] = [ OH– ] For the above equation, Kc = If we exclude the pure liquid... Kw = [H3O+] [OH–] = [H+] [OH–] = 1.0 x 10–14 (@ 25oC) This equation is taken to be valid for pure water and for dilute aqueous solutions. ACID BASE NEUTRAL

  40. For the generic reaction in sol’n: A + B C + D HCl H+ + Cl– Acid-Dissociation Constant: Ka For strong acids, e.g., HCl… ~0 lots lots = “BIG.” Assume 100% dissociation; KaNOT applicable for strong acids.

  41. HCl(aq) + H2O(l) NH3(aq) + H2O(l) NH3(aq) + H2O(l) NH3(aq) + H2O(l) + – [ [ ] ] Amphoteric substances can be acids or bases, depending on the reaction conditions (e.g. H2O) H3O+(aq) + Cl–(aq) OH–(aq) + NH4+(aq) NH3 is another example. NH2-(aq) + H3O+(aq) NH4+(aq) + OH–(aq)

  42. HNO2(aq) + H2O(l) NO2–(aq) + H3O+(aq) conj. acid base • In acid-base equilibria, protons are donated in • forward and reverse reactions. • The two substances in a conjugate acid-base pair differ by a H+... and the acid has the extra H+. CONJ. BASE ACID • Strong acids / bases easily / ____ H+. accept donate • Weak acids / bases do NOT easily / ____ H+. accept donate The stronger a/n acid / base, the weaker its conj. base / acid.

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

  44. 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”

  45. Weak Acid Equilibrium: Ka • 1. Write the reaction for carbonic acid (H2CO3) dissociating in water. • H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) • Write the equilibrium expression for this reaction. • b. If [H2CO3] = 0.100M and it shows 0.212% ionization, what is the value of Ka? H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) 0.100M ------- 0 0 Initial Change Equilibrium - 2.12x10-4 ------- + 2.12x10-4 + 2.12x10-4 0.099788 ------- 2.12x10-4 2.12x10-4

  46. Weak Acid Equilibrium: Ka • Write the reaction for carbonic acid (H2CO3) dissociating in water. • H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) • If [H2CO3] = 0.0500M and it shows 0.300% ionization, what is the value of Ka?H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) 0.0500M ------- 0 0 I C E - 1.50x10-4 ------- + 1.50x10-4 + 1.50x10-4 0.04985 ------- 1.50x10-4 1.50x10-4

  47. Weak Acid Equilibrium: Ka 0 Write the reaction for carbonic acid (H2CO3) dissociating in water. H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) d. Given the value of Ka, if [H2CO3] = 0.0100M, what will the concentrations of the products? What is the % ionization? H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) 0.0100M ------- 0 0 I C E - X ------- + X + X 0.0100-X ------- X X X2 = 4.50x10-7(0.01) 5% Rule: at equilibrium, [H2CO3] ≈ initial [H2CO3]0. Only good if X < 5% of the initial concentration! < 5.00% X = √(4.50x10-9) X = 6.71x10-5M

  48. Weak Acid Equilibrium: Ka 2. The reaction for benzoic acid (HC7H5O2) dissociating in water: HC7H5O2(aq) + H2O(l) H3O+(aq) + C7H5O2 -(aq) If [HC7H5O2] = 1.500M and it ionizes 0.651%, calculate the Ka for benzoic acid. HC7H5O2(aq) + H2O(l) H3O+(aq) + C7H5O2 -(aq) Calculate the pH for this solution. 1.500M ------- 0 0 I C E - .009765 ------- + .009765 + .009765 1.490 ------- + .009765 + .009765 [H3O+] = 0.009765M pH = -log (0.009765) = 2.01

  49. Weak Acid Equilibrium: Ka 0 3. Write the reaction for boric acid (H3BO3) dissociating in water. H3BO3(aq) + H2O(l) H3O+(aq) + H2BO3-(aq) Given Ka = 5.4x10-10, if [H3BO3] = 0.500M, what will be the pH of the resulting solution? H3BO3(aq) + H2O(l) H3O+(aq) + H2BO3-(aq) 0.500M ------- 0 0 I C E - X ------- + X + X 0.500-X ------- X X X2 = 5.40x10-10(0.50) X = [H3O+] = 1.64x10-5M pH = -log (1.64x10-5) = 4.78 X = √(2.70x10-10) X = 1.64x10-5M

  50. Weak Base Equilibrium: Kb 0 The reaction for ammonia (NH3) in water is as follows: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) If Kb = 1.80x10-5 and [NH3] = 2.500M, calculate the pH of this solution. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Calculate the pH for this solution. 2.500M ------- 0 0 I C E - X ------- + X + X 2.500 - X ------- + X + X X2 = 1.80x10-5(2.50) X = √(4.50x10-5) X= [OH-] = 0.00671M pOH = -log (0.00671M) = 2.17  11.83 X = 0.00671M

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