reversible reaction: R P and P R Acid dissociation is a reversible reaction. H 2 SO 4 2 H SO 4

Transcription

1 Unit : Equilibrium / Acids and Bases reversible reaction: R P and P R Acid dissociation is a reversible reaction. H 2 SO 4 2 H + + SO 4 Rate at which equilibrium: R P = Rate at which P R -- looks like nothing is happening, however -- system is dynamic, NOT static Le Chatelier s principle: When a system at equilibrium is disturbed, it shifts to a new equilibrium that counteracts the disturbance. N 2 (g) + H 2 (g) 2 NH (g) Disturbance Equilibrium Shift Add more N 2.. H 2.. NH Remove NH.. Add a catalyst no shift Increase pressure.

2 Light-Darkening Eyeglasses AgCl + energy Ag o + Cl o (clear) (dark) Go outside Sunlight more intense than inside light; energy shift to a new equilibrium: GLASSES DARKEN Then go inside energy shift to a new equilibrium: GLASSES LIGHTEN In a chicken CaO + CO 2 CaCO (eggshells) In summer, [ CO 2 in a chicken s blood due to panting. -- shift ; eggshells are thinner How could we increase eggshell thickness in summer? -- give chickens carbonated water [ CO 2, shift -- put CaO additives in chicken feed [ CaO, shift

3 Acids and Bases ph < 7 ph > 7 taste sour react w /bases proton (H + ) donor turn litmus red taste bitter react w /acids proton (H + ) acceptor turn litmus blue lots of H + /H O + lots of OH react w /metals don t react w /metals Both are electrolytes. ph scale: measures acidity/basicity ACID D BASE NEUTRAL Each step on ph scale represents a factor of 0. ph 5 vs. ph 6 (0X more acidic) ph vs. ph 5 ph 8 vs. ph (00X different) (00,000X different)

4 Common Acids Strong Acids hydrochloric acid: HCl H + + Cl -- stomach acid; pickling: cleaning metals w /conc. HCl sulfuric acid: H 2 SO 4 2 H SO 4 -- # chemical; (auto) battery acid nitric acid: HNO H + + NO -- explosives; fertilizer Weak Acids acetic acid: CH COOH CH COO + H + -- vinegar; naturally made by apples hydrofluoric acid: HF H + + F -- used to etch glass citric acid, H C 6 H 5 O 7 -- citrus fruits; sour candy ascorbic acid, H 2 C 6 H 6 O 6 -- vitamin C lactic acid, CH CHOHCOOH -- waste product of muscular exertion carbonic acid, H 2 CO -- carbonated beverages -- H 2 O + CO 2 H 2 CO (dissolves limestone, CaCO )

5 Acid Nomenclature binary acids: acids w /H and one other element Binary Acid Nomenclature. Write hydro. 2. Write prefix of the other element, followed by -ic acid. HF hydrofluoric acid HCl hydrochloric acid HBr hydrobromic acid hydroiodic acid HI hydrosulfuric acid H 2 S oxyacids: acids containing H, O, and one other element Common oxyanions (polyatomic ions that contain oxygen) that combine with H to make oxyacids: BrO NO 2 CO PO 4 ClO 2 SO 4 IO

6 Oxyacid Nomenclature Write prefix of oxyanion, followed by -ic acid. HBrO bromic acid HClO chloric acid H 2 CO carbonic acid sulfuric acid H 2 SO 4 phosphoric acid H PO 4 Above examples show most common forms of the oxyacids. If an oxyacid differs from the above by the # of O atoms, the name changes are as follows: one more O = per ic acid most common # of O = ic acid one less O = ous acid two fewer O = hypo ous acid HClO 4 perchloric acid HClO chloric acid HClO 2 chlorous acid HClO hypochlorous acid phosphorous acid H PO hypobromous acid HBrO persulfuric acid H 2 SO 5

7 Various Definitions of Acids and Bases Arrhenius acid: yields H + in sol n e.g., HNO H + + NO Arrhenius base: yields OH in sol n e.g., Ba(OH) 2 Ba OH Lewis acid: Lewis base: e pair acceptor e pair donor TOPIC FOR FUTURE CHEM. COURSES Bronsted-Lowry acid: proton (i.e., H + ) donor Bronsted-Lowry base: proton (i.e., H + ) acceptor B-L theory is based on conjugate acid-base pairs. **Conjugate acid has extra H + ; conjugate base doesn t.** HCl + H 2 O H O + + Cl C-B C-A C-A C-B NH + H 2 O NH 4 + C-B C-A C-A + OH C-B CH COOH + H 2 O CH COO + H O + C-A C-B C-B C-A

8 Dissociation and Ion Concentration Strong acids or bases dissociate ~00%. HNO H + + NO HNO H + + NO H + NO NO H /L 000/L + 000/L M M M HCl H + + Cl 4.0 M 4.0 M M monoprotic acid H 2 SO 4 2 H + + SO SO 4 + H + H + H + H + SO 4 2 diprotic acid 2. M 4.6 M + 2. M Ca(OH) 2 Ca OH M M M

9 ph Calculations Recall that the hydronium ion (H O + ) is the species formed when hydrogen ion (H + ) attaches to water (H 2 O). OH is the hydroxide ion. For this class, in any aqueous sol n, [ H O + [ OH = x 0 4 ( or [ H + [ OH = x 0 4 ) If hydronium ion concentration = 4.5 x 0 9 M, find hydroxide ion concentration. [ H O + [ OH = x x 0 x 0 [ OH -9 [ H O 4.5 x 0 M 0 x y x = 2.2 x 0 6 M = M M

10 Given: Find: A. [ OH = 5.25 x 0 6 M [ H +.90 x 0 9 M B. [ OH =.8 x 0 M [ H O x 0 4 M C. [ H O + =.8 x 0 M [ OH 5.6 x 0 2 M D. [ H + = 7. x 0 2 M [ H O + 7. x 0 2 M Find the ph of each sol n above. ph = log [ H O + ( or ph = log [ H + ) A. ph = log [ H O + = log [.90 x 0 9 M On a graphing calculator ph = 8.72 B..6 C. 2.7 D.. A few last equations poh = log [ OH ph + poh = 4 [ H O + = 0 ph ( or [ H + = 0 ph ) [ OH = 0 poh

11 ph [ H O + = 0 ph ph = log [ H O + [ H O + ph + poh = 4 [ H O + [ OH = x 0 4 poh [ OH = 0 poh poh = log [ OH [ OH If ph = 4.87, find [ H O +. [ H O + = 0 ph = On a graphing calculator [ H O + =.5 x 0 5 M If [ OH = 5.6 x 0 M, find ph. Find [ H O + =.8 x 0 4 M Find poh = 0. Then find ph Then find ph ph =.7

12 For the following problems, assume 00% dissociation. Find ph of a M nitric acid (HNO ) sol n. HNO H + + NO M M M GIVEN ph = log [ H O + = log [ =.24 Find ph of.2 x 0 5 M barium hydroxide (Ba(OH) 2 ) sol n. Ba(OH) 2 Ba OH.2 x 0 5 M.2 x 0 5 M x 0 5 M GIVEN poh = log [ OH = log [ 6.4 x 0 5 = 4.9 ph = 9.8 Find the concentration of an H 2 SO 4 sol n w /ph.8. H 2 SO 4 2 H + + SO 4 2 [ H O + = [ H + = 0 ph = 0.8 = 4.2 x 0 4 M H 2 SO 4 2 H SO 4 X 4.2 x 0 4 M + (Who cares?) X = [ H 2 SO 4 = 2. x 0 4 M

13 Find ph of a sol n with.65 g HCl in 2.00 dm of sol n. HCl H + + Cl mol L.65 g mol HCl 6.5 ghcl M 2.00 dm 0.05 MHCl HCl H + + Cl 0.05 M 0.05 M M ph = log [ H O + = log [ 0.05 =. What mass of Al(OH) is req d to make 5.6 L of a sol n with a ph of 0.72? Assume 00% dissociation. M Al(OH) Al + + OH poh =.28 [ OH = 0 poh = 0.28 = 5.25 x 0 4 M Al(OH) Al + + OH.75 x 0 4 M (Who cares?) x 0 4 M mol L mol mol ML Al(OH) mol Al(OH) 78 g Al(OH).75 x 0-4 M(5.6 L) 0.2 g Al(OH) mol Al(OH)

14 Acid-Dissociation Constant, K a For the generic reaction in sol n: K a [ PRODUCTS [ REACTANTS A + B C + D K a [ C [ D [ A [ B For strong acids, e.g., HCl HCl H + + Cl K a [ H [ Cl [ HCl - "BIG." Assume 00% dissociation; K a not applicable for strong acids. For weak acids, e.g., HF HF H + + F - [ H [ F K a "SMALL" 6.8 x 0 [ HF -4 Other K a s for weak acids: CH COOH CH COO + H + K a =.8 x 0 5 HC H 5 O H + + C H 5 O K a =.4 x 0 4 HNO 2 H + + NO 2 K a = 4.5 x 0 4 The weaker the acid, the smaller the K a. stronger, larger.

15 Find the ph of.75 M acetic acid (K a =.8 x 0 5 ). K a CH COOH CH COO + H +.8 x [ CHCOO [ H [ CH COOH [ CH COO = [ H + = x.8 x x.75 ph = log (0.0056) = 2.25 x M - [ CHCOO.75 [ H [ H Hypobromous acid has K a = 2.5 x 0 9. Find ph if 45 g of acid are in 50 L of sol n. HBrO H + + BrO [ HBrO 45 g 96.9 g 50 L 4.28 x [ H [ BrO x Ka 2.5 x 0 - [ HBrO 4.28 x 0 x = [ H + =.27 x 0 6 M ph = 5.5 If instead, 45 g of sulfuric acid were used [ H SO g 98.g 50 L x 0 [ H + = 8.44 x 0 M ph = M M For.75 M HCl, ph would be 0.24.

16 If nitrous acid has K a = 4.5 x 0 4, what mass is req d to make 85 L of a sol n with ph =.? [ H + = 0 ph = 0. = 7.94 x 0 4 M HNO 2 H + + NO 2 [ HNO x 0 4 M 7.94 x 0 4 M K a [ H [ NO2 [ HNO 2 - [ HNO 2 [ H [ NO K a - 2 [ HNO 2 (7.94 x x ) 2.40 x 0 - M mol HNO 2 = M L =.40 x 0 M (85 L) = 0.2 mol HNO 2 = 5.6 g HNO 2

17 Indicators chemicals that change color, depending on the ph Two examples, out of many: litmus red in acid, blue in base phenolphthalein..clear in acid, pink in base Measuring ph litmus paper Basically, ph < 7 or ph > 7 phenolphthalein ph paper -- contains a mixture of various indicators -- each type of paper measures a range of ph -- ph 0 to 4 universal indicator -- is a mixture of several indicators -- ph 4 to R O Y G B I V ph meter -- measures small voltages in solutions -- calibrated to convert voltages into ph -- precise measurement of ph

18 Neutralization Reaction ACID + BASE SALT + WATER HCl + NaOH + HCl + NaOH NaCl + H 2 O HCl + NaOH NaCl + H 2 O H PO 4 + KOH + H + PO 4 K + OH H PO 4 + KOH K PO 4 + H 2 O H PO 4 + KOH K PO 4 + H 2 O H 2 SO 4 + NaOH + H + 2 SO 4 Na + OH H 2 SO 4 + NaOH Na 2 SO 4 + H 2 O H 2 SO NaOH Na 2 SO H 2 O HClO + Al(OH) + HClO + Al(OH) Al(ClO ) + H 2 O + AlCl + HCl + Al(OH) AlCl + H 2 O + Fe 2 (SO 4 ) + H 2 SO Fe(OH) Fe 2 (SO 4 ) + 6 H 2 O

19 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 H O + = mol OH mol Since M, mol ML M L H O A M - L OH B L M - - VA M V H O OH B or [ HO VA [ OH VB In a titration, the above equation helps us to use a KNOWN conc. of acid (or base) to determine the UNKNOWN conc. of base (or acid) L of 0.2 M HCl are used to titrate.22 L of an unknown conc. of KOH. Find the molarity of the KOH. HCl H + + Cl and KOH K + + OH M H O V A M OH 0.2 M (2.42 L) M - M OH - M KOH OH - V B (.22 M) 0.6 M

20 458 ml of HNO (w/ph = 2.87) are neutralized w/66 ml of Ba(OH) 2. What is the ph of the base? [ H O + = 0 ph = =.5 x 0 M [ H O - VA [ OH VB (.5 x 0 - M ) (458 ml) [ OH - (66mL) [ OH = 9.5 x 0 4 M poh = log (9.5 x 0 4 ) =.0 ph = 0.97 How many L of M sodium hydroxide will titrate.82 L of 0.5 M sulfuric acid? H 2 SO 4 2 H SO 4 and NaOH Na + + OH 0.5 M 0.60 M M M.82 L X L [ H O - VA [ OH VB (0.60 M) (.82 L) (0.872 M) (X L) X = L NaOH

21 Example Titration with HNO and NaOH From a known [ HNO, find the unknown [ NaOH. HNO H + + NO NaOH Na + + OH 0.0 M 0.0 M? Buret Readings, in ml Trial Acid Base Initial Final Amt. Used [ H O - VA [ OH VB [ OH = [ NaOH = Buret Readings, in ml Trial 2 Acid Base Initial Final Amt. Used [ H O - VA [ OH VB [ OH = [ NaOH =

22 Titration Using a Weak Acid 44.0 g of solid butanoic acid (HC 4 H 7 O 2, K a =.5 x 0 5 ) are dissolved in 0.6 L of sol n. If L NaOH neutralizes acid sol n, find conc. of NaOH. ** Find [ H + -, then use [ H VA [ OH VB.** [ HC 44 g 88 g 0.6 L 4H7O MHC4H7O2 K a.5 x 0-5 [ H [ C4H7O2 [ HC H O x = [ H + = 8.4 x 0 4 M x [ H V A [ OH - V (8.4 x 0 4 M) (0.6 L) = [ OH (0.590 L) B [ OH = [ NaOH = 0.05 M

23 Buffers chemicals that resist changes in ph Example: The ph of blood is 7.4. Many buffers are present to keep ph stable. H + + HCO H 2 CO H 2 O + CO 2 hyperventilating: CO 2 leaves blood too quickly [ CO 2 shift right [ H + ph (more basic) alkalosis: blood ph is too high (too basic) Remedy: Breathe into bag. [ CO 2 shift left [ H + ph (more acidic, closer to normal) acidosis: blood ph is too low (too acidic)

24 More on buffers: -- a combination of a weak acid and a salt -- together, these substances resist changes in ph (A) weak acid: CH COOH CH COO + H + (lots) (little) (little) (B) salt: NaCH COO Na + + CH COO (little) (lots) (lots) If you add acid (e.g., HCl H + + Cl ). large amt. of CH COO consumes extra H + as in (A) going 2. **Conclusion: ph remains relatively unchanged. If you add base (e.g., KOH K + + OH ). extra OH grabs H + from the large amt. of available CH COOH and forms CH COO and H 2 O 2. **Conclusion: ph remains relatively unchanged. Amphoteric Substances can act as acids OR bases e.g., H 2 O and NH NH 2 NH + NH H O + H 2 O OH

25 Partial Neutralization.55 L of 0.26 M KOH L of 0.22 M HCl Find ph..55 L of 0.26 M KOH 2.5 L of 0.22 M HCl ph =? Procedure:. Calc. mol of substance, then mol H + and mol OH. 2. Subtract smaller from larger.. Find [ of what s left over, and calc. ph. mol = M L mol KOH = 0.26 M (.55 L) = 0.40 mol KOH = 0.40 mol OH mol HCl = 0.22 M (2.5 L) = 0.47 mol HCl = 0.47 mol H + LEFT OVER mol H + [ H mol H M (.55 L 2.5 L) ph = log (0.089 M) =.72

26 4.25 L of 0.5 M hydrochloric acid is mixed w /.80 L of 0.9 M sodium hydroxide. Find final ph. Assume 00% dissociation. mol HCl = 0.5 M (4.25 L) =.4875 mol HCl =.4875 mol H + mol NaOH = 0.9 M (.80 L) =.4820 mol NaOH =.4820 mol OH LEFT OVER mol H + [ H mol H (4.25 L.80 L) 6.8 x 0 ph = log (6.8 x 0 4 M) =.7-4 M 5.74 L of 0.29 M sulfuric acid is mixed w /.2 L of 0.5 M aluminum hydroxide. Find final ph. Assume 00% dissociation. mol H 2 SO 4 = 0.29 M (5.74 L) =.6646 mol H 2 SO 4 =.292 mol H + mol Al(OH) = 0.5 M (.2 L) =.25 mol Al(OH) =.705 mol OH LEFT OVER 0.04 mol OH [ OH mol OH M (5.74 L.2L) poh = log ( M) = 2.4 ph =.66

27 A g HNO in 450 ml of sol n. Find ph. [ HNO 0.08 g 6 g L -.4 x 0 MHNO [ H + =.4 x 0 M ph = log [ H + = log (.4 x 0 M) = 2.87 B g Ba(OH) 2 in 560 ml of sol n. Find ph. [ Ba(OH) g 7. g L x 0 MBa(OH) 2 [ OH = 9.8 x 0 4 M poh = log [ OH = log (9.8 x 0 4 M) =.04 ph = 0.96 C. Mix them. Find ph of resulting sol n. **Governing equation: mol = M L mol H + =.4 x 0 M (0.450 L) = 6.0 x 0 4 mol H + mol OH = 9.8 x 0 4 M (0.560 L) = 5.4 x 0 4 mol OH LEFT OVER 8.90 x 0 5 mol H + [ H x 0 mol H (0.450 L L) 8.8x 0 ph = log (8.8 x 0 5 M) = M