Chem Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation

Transcription

1 Chem Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation 3//011 1

2 The net ionic equation for the reaction of KOH(aq) and HNO 3 (aq) is 1. K + + NO - 3 KNO 3 (aq). K + + OH - + H 3 O + H O(l) + K + 3. H 3 O + + OH - H O(l) 4. HNO 3 + KOH H O(l) + KNO 3 (aq) 3//011

3 The net ionic equation for the reaction of KOH(aq) and HNO 3 (aq) is 1. K + + NO - 3 KNO 3 (aq). K + + OH- + H 3 O + H O(l) + K + 3. H 3 O + + OH- H O(l) 4. HNO 3 + KOH H O(l) + KNO 3 (aq) ANSWER=(3) You need to be able to recognize KOH as a strong base, and HNO 3 as a strong acid. The neutralization occurs with K + and NO 3 - as spectator ions. 3//011 3

4 Section 18.1 Common Ion Effect In acid-base, and other equilibria involving ionic components, you can manipulate the position of the equilibrium by adding or removing one ionic reactant or product. We can use an ICE table to quantify this effect. Or, qualitatively, the effect is predicted by L Chatelier. ClCH CO H + H O ClCH CO - + H 3 O + 3//011 4

5 Review. We have seen what happens when, say, 0.10 M chloroacetic acid is dissolved in water (K a = 1.35 x 10-3 ): ClCH CO H + H O ClCH CO - + H 3 O + I 0.10 M 0 0 C -x x x E x x x 3//011 K a [ HCO [ H 3 O [ HCO H x 10 3 x x x x ( 1 x ( x ( 3 ph st nd rd approx ) approx ) log( ) approx ) (Use quadratic or method of successive approximations if C o < 100 x K a. Here (100)K a = M. 5

6 Now let s see what happens when we add M sodium chloroacetate (NaClCH CO ) to 0.10 M chloroacetic acid: ClCH CO H + H O ClCH CO - + H 3 O + I 0.10 M C -x x x E x x x //011 K a. 35 [ HCO [ H 3 O [ HCO H 3 ( x ) x x x 3 ( ) x x x ( 1 x ( x ( 3 ph st nd rd approx ) approx ) log( ) approx ). 5 6

7 Now, think about this in terms of L Chatelier s Principle: If you add more chloroacetate ClCH CO -, the reaction will go to the LEFT to use up the excess ClCH CO -. ClCH CO H + H O ClCH CO - + H 3 O + Therefore [H 3 O + decreases, and ph goes UP. 3//011 7

8 3//011 9

9 Section 18. Buffers 3//011 10

10 Definition A buffer is an aqueous solution containing a mixture of a weak acid and its conjugate base, (or a weak base and its conjugate acid). The function of a buffer is to absorb H + or OH - ions, minimizing the change in ph that might otherwise occur. HA + H O A - + H 3 O + B + H O BH + + OH - (Notice that either equilibrium can absorb H + or OH - ions) 3//011 11

11 The Henderson-Hasselbalch Equation is used to describe buffer solutions (these have appreciable amounts of both HA and A -, like at least 5% of the total concentration as one or the other). 3//011 1

12 Henderson-Hasselbalch Equation Example buffer solutions, which contain a weak acid and conjugate base of the weak acid (or weak base and conjugate acid of the weak base). CH 3 COOH + H O CH 3 COO - + H 3 O + Acetic acid acetate HNO + H O NO - + H 3 O + Nitrous acid nitrite NH H O NH 3 + H 3 O + Ammonium ammonia F- + H O HF + OH - Fluoride hydrofluoric acid 3//011 13

13 CH 3 COOH + H O CH 3 COO - + H 3 O + Acetic acid acetate (HOAc) (OAc - ) equilibrium constant definition K a products [ OAc [ H 3 O reactants [ HOAc Take log 10 of both sides log of product = sum of logs Mult both sides by -1 px = -log 10 (x) Solve for ph log K log K a a log K [ OAc [ H 3O log [ HOAc [ OAc log[ H log 3O [ HOAc a log[ H [ OAc pk a ph log [ HOAc [ OAc ph pk log a [ HOAc 3 O [ OAc log [ HOAc 3//011 14

14 Logic of the H-H equation ph pk a [ A log [ HA Please remember this equation and know how to use it. Say 1/5 [A -, 4/5 [HA Then log(1/4) = ph = pk a Say [A - = [HA Then log(1) = 0 ph = pk a + 0 = pk a Say /3 [A - and 1/3 [HA Then log(/1) = 0.30 ph = pk a or [H 3 O + > K a or [H 3 O + = K a or [H 3 O + < K a 3//011 15

15 Most buffers are made up with equal or near-equal concentrations of acid and conjugate base. This has maximum ability to absorb H + or OH - ions. Buffers can be prepared 3 ways. 1) Start with the acid form, and add ½ mole NaOH per mole acid. ) Start with the conjugate base, and add ½ mole HCl per mole base. 3) Add equal amounts of acid and conjugate base. 3//011 16

16 Say you prepared a buffer using equal moles of sodium nitrite NaNO and nitrous acid HNO (K a = 3. x 10-4 ). What is the ph of the resulting solution? (Try using pencil and paper, and not a calculator ) //011 17

17 Say you prepared a buffer using equal moles of sodium nitrite NaNO and nitrous acid HNO (K a = 3. x 10-4 ). What is the ph of the resulting solution? (Try using pencil and paper, and not a calculator ) ph = pk a + log(a-/ha) = pk a ph = -log(3. x 10-4 ) ph = -[log3. + log(10-4 ) ph = -(~0.5-4) ph = -(-3.5) = 3.5 3//011 18

18 1. Calculate the stoichiometry (moles) first mol HCN L HCN mol HCN L HCN. Strong bases react completely with weak acids. HCN + OH - CN- 1 mol CN mol NaOH mol CN mol NaOH + H O HCN + OH - CN - I C E ~ [ CN [ HCN mol L mol L + H O M M 3//011 19

19 3//011 0

20 ph ph pk a [ CN log [ HCN log(4.0 x10 10 ) log ph ph ph log( ) ( ) Notice that if [HA > [A -, the ph must be LESS THAN the pk a. 3//011 1

21 3//011

22 Demo Vernier ph electrode/loggerpro software 30. ml water 3.0 mmoles sodium acetate (0.5 g) 3.0 M HCl (aq) ( 3.0 mmol/ml) Pipetman 1000 (delivers ml max) 3//011 3

23 Observations: 1.00 ml 3.0 M HCl ml water: ph ~ 1.5 (This is about 0.10 M strong acid, which should give ph = 1.0.) 0.4 g NaCH 3 CO (sodium acetate) + 30 ml water: ph ~ 7.8. (This should be about 8.7 if you do the calculation.) Adding 0.50 ml of 3.0 M HCl to the sodium acetate solution. ph ~ 4.8 (This should be 4.74 since it is a buffer with about 0.05 M acetate and 0.05 M acetic acid where ph = pka). Adding another 0.50 ml of 3.0 M HCl to the above solution gives ph ~ 3.0. (This is now essentially a solution of 0.10 M in acetic acid, and the calculated ph =.87). Why is the ph not the same as the first dilution?? Because the acetate reacts with, and holds, the H + in its weak acid structure. 3//011 4

24 HClO + OH- ClO- + H O 0.35 x 0 -x -x x 0.35 x ~0 x ph pk a We know these 3//011 5 [ A log [ HA [ HA [ A M

25 3//011 6 M A HA [ [ OH mol L L mol OH [ [ [ [ [ [ log [ [ log A HA HA A HA A pka ph HA A pka ph pka ph (Substituting real numbers gives less writing.) M A M M A M A M A A 0.09 [ [ [ 0.35 [ [

26 3//011 7

27 18.3 Titration Calculations Adding OH- of a known concentration to acid solution until moles OH - = moles H + ( neutralization reaction ) 3//011 8

28 HI(aq) + Ca(OH) (aq) CaI (aq) + H O(l) L Ca ( OH )? L HI L Ca ( OH ) 0. 9 mol Ca ( OH ) L Ca ( OH ) mol HI mol Ca ( OH ) 1 L HI mol HI L HI 3//011 9

29 L Ca ( OH )? L HI L Ca ( OH ) L Ca ( OH ) 1 L HI? L HI L Ca ( OH ) mol Ca ( OH ) L Ca ( OH ) 1 L HI mol HI? L HI L Ca ( OH ) mol Ca ( OH ) L Ca ( OH ) mol Ca ( OH ) 1 L HI mol HI? L HI L Ca ( OH ) mol Ca ( OH ) L Ca ( OH ) mol HI mol Ca ( OH ) 1 L HI mol HI? L HI L Ca ( OH ) 0. 9 mol Ca ( OH ) L Ca ( OH ) mol HI mol Ca ( OH ) 1 L HI mol HI L HI 3//011 30

30 3//011 31

31 This shows the equation for the entire titration curve (ph vs. volume of OH- titrant) for 5 different acids. This equation will NOT be on the next exam. However, we have studied several sections of the typical titration curve: 1. Acid by itself: For a strong acid, [H 3 O + = molarity of acid. For a weak acid use the ICE table.. Acid + conjugate base: Use the Henderson-Hasselbalch Equation. 3. At the equivalence point: ph = 7.00 for strong acid vs. OH - titration. Use ICE table for hydrolysis of the conjugate base of a weak acid. 4. Past the equivalence point: This is now just dilute OH -, and so ph = -log(k w /[OH - ). 3//011 3

32 ph Titrating a weak acid (1.0 M, K a = 1 x 10-6 ) with strong base NaOH HA + OH - A - + H O 14 Acid ph of NaOH solutions buffer of weak acid & its conjugate base ph of conjugate base ph of weak acid [weak acid=[conjugate base ph = pk a Mol OH-/mol HA titration equivalence pt 3//011 33