Acids, Bases and the Common Ion Effect

Transcription

1 cids, Bases and the Common Ion Effect Consider the following acid equilibrium of a weak acid: HF + H O H 3 O + + F By LeChatelier s principle, we predict the HF dissociation should be driven left, suppressing the dissociation. H + + Cl HCl (aq) K a = [H 3 O + ][F ] [HF] What happens when we add some strong acid to the mixture? HCl completely dissociates, adding free H 3 O + to solution. This is a common ion to the weak acid equilibrium. More quantitative a) Determine [F ] in a solution of 0.500M HF. b) Determine [F ] in a solution of 0.500M HF and 0.10M HCl. K HF = 6.6x10 4. a) solution of a weak acid. Let s use the quadratic. HF HF HF HF + K K K C [ H ] = [ F ] = [F ] = M

2 Continued b) Use ICE table. Let HF and HCl dissociate separately. Initial HF + H O H 3 O + + F weak acid strong acid Change x + x + x Equilib 0.50x x x K a = [H 3 O + ][F ] [HF] K a = (0.10 +x)(x) / (0.50x) ssumption: ionization of HF is suppressed. ssume x<<0.1; x<<0.5 K a ~ (0.10)(x) / (0.50) Continued Rearrange: x = [F ] = (0.50M)K a / 0.10M = M Comparison a) no common ion, [F ] = M b) common ion, [F ] = M 5 X less! This is one example of the common ion effect. When a strong acid supplies the common ion, H 3 O +, the ionization of a weak acid is suppressed. When a strong base supplies the common ion, OH, the ionization of a weak base is suppressed.

3 Buffering and titrations! 16_1 ph Pink Phenolphthalein Colorless Equivalence point Blue Yellow Bromcresol green Volume of NaOH added (ml) Volume NaOH added (ml) ph Chemical buffer a chemical system that resists change Example: any chemical system in equilibrium a + bb + cc dd + ee + ff K eq d e f [D] [E] [F] = [] [B] [C] a b c cid/base buffer a system that resists changes to ph caused by addition of excess base or acid.

4 cid Base Buffer Systems mixture of a weak acid and it s conjugate base O O O Benzoic acid O OH H O benzoate O O dd as sodium salts for example CH 3 CH 3 acetic acid acetate mixture of a weak base and it s conjugate acid N H H 3 C CH + 3 dd as N CH H 3 C CH 3 3 CH 3 chloride trimethylamine trimethylammonium salt for example n acid/base buffer system What happens if we add acid and conjugate base in equal concentrations? i.e. equal moles of H and Na Initial H + H O H 3 O + + weak acid conj base Change x +x + x Equilib 0.10x +x 0.10+x K [H ][ ] [H] a = + K a = + [H ][ x] [0.10 x] But due to presence of, ionization of H is suppressed more than usual. x<<0.10 [H + ] = K a ph = pk a!

5 K HendersonHasselbalch Equation [H ][ ] [H] a = + Solve for [H + ] + K a [H] [H ] = [ ] Take log of both sides log [H + ] = log K a log [H] + log [ ] log [H + ] = log K a + log [ ]/[H] From previous ICE table, we see: [H] = C H x [ ] = C + x But in general, C H = C. If x<<c H ph = pk log [ ] a + ph = pk log C a + [H] C H Buffer ratios For correct buffering, there should be significant amounts of H and. Just as importantly, the ratio should be such that: 0.10 C C H 10.0 Substituting this constraint into the HH equation ph = pk ± 1 buffer a

6 How to prepare a buffer 1. What ph buffer do you require?. Find an acid/conj base system from tables that has pk a within 1 unit of the required ph. 3. Use the HH equation to determine the exact ratio of the acid and the base you will need. 4. What buffer capacity do you need 0.1M, 0.01M, 0.001M? nswer depends on how much acid or base could potentially impact the system that you are trying to buffer. 5. Prepare solutions taking into account 3) and 4). Example Prepare 1L of buffer with ph = 3.5, with a total concentration, [ ] + [H] = 0.01M. ). Pick buffer system. cid pk a Citric cid 3.13 Benzoic acid 4.0 cetic acid 4.77 Carbonic acid 6.36 mmonium ion 9.5 Phenol 9.89

7 ph = pk log C a + C log C C C C C C H H H = ph pka [ ] [H] =10 Continued 3) Determine ratio of [ ]/[H] needed. Use either HH equation or equilibrium expression. H ( ) (ph pka) = 10 =.34 Cont d 5) Prepare solution (i.e. What masses or volumes are needed?) C /C H =.34 (1) lso C + C H = 0.01M () from (1) C =.34 C H,substitute this into ().34 C H + C H = 0.01M C H = M C =.34 C H C = 0.007M What substances, and quantities, are needed to make these solutions?

8 cont d Citric acid, C 6 H 8 O 7 MW = 19.1g/mole monosodium citrate NaC 6 H 7 O 7 MW = 14.1g/mole Since we are preparing 1L, we need to add moles of citric acid and moles of monosodium citrate. mass H = mol x 19.1 g/mol = 0.576g mass Na = mol x 14.1 g/mol = 1.50g Dissolve to make 1L of solution. OR Dissolve 0.01mol x 19.1g/mol = 1.9g of citric acid in about 1L of water, and neutralize with NaOH until ph = 3.5! (Total volume must be 1L.) Examples of Buffers in Nature ph of blood is ph is maintained by a series of buffer systems, H CO 3 /HCO 3, phosphate and proteins. ph regulation in the body is critical since functioning of enzymes is highly ph dependent. lkalosis raising of ph results from hyperventilation, or exposure to high elevations (altitude sickness). cidosis lowering of ph in blood by organ failure, diabetes or long term protein diet.

9 in the lungs CO (g) CO (aq) + H O The bicarbonate buffer is essential for controlling blood ph t high elevation, P drops, hyperventilation! H CO 3(aq) What causes alkalosis at high elevation? a multiple equilibrium hypothesis. in the blood H CO 3(aq) + H O H 3 O + + HCO 3 pk a = 6.4 [HCO 3 ] [H CO ] ~ OH H O ph increases! cid Base Titrations cid base titrations are examples of volumetric techniques used to analyze the quantity of acid or base in an unknown sample. cid + Base H O + salt This is done by detecting the point at which we have added an equal number of equivalents of base to the acid. This is the equivalence point. For neutralization of an unknown monoprotic acid () with a base (B), we have at equivalence: moles of acid = moles of base C V = C B V B CBV C = V equiv We detect the equivalence point with a ph meter or by identifying the endpoint with an acid base indicator.

10 cid Base Titrations some terminology equivalence point: the point at which moles of acid = moles of base endpoint: the experimental approximation of the equivalence point titration error: the difference between the equivalence point and the endpoint titrant: the solution that is added in a measured quantity standard solution: a solution of known concentration Indicators cidbase indicators are highly coloured weak acids or bases. HIndic Indic + H + colour 1 colour Colour transition occurs for 0.1< [Indic ]/[HIndic]<10 ind [In ] ind phtransition = pk a + log = pk a ± 1 [HIn] They may have more than one colour transition. Example. Thymol blue Red Yellow Blue One of the forms may be colourless phenolphthalein (colourless to pink)

11 Selection of an indicator. Indicators The colour transition range for an indicator is pk a + 1. Choose an indicator that has a pk a close to the ph at the equivalence point. phenolphthalein Indicator Examples bromthymol blue methyl red

12 16_11 ph cid Base Titrations Titration of strong acid with strong base Pink Colorless Blue Yellow Phenolphthalein Equivalence point Bromcresol green Volume of NaOH added (ml) Volume NaOH added (ml) ph Low initial ph, large ph change at equivalence point Example What is the ph at 0mL, 10mL, 5.0mL and 35mL in titration of 5mL of M strong acid (HCl) with 0.100M NaOH? 0 ml [H + ] = C = 0.100M ph = log (0.1) ph = mL (all calculated in moles) HCl (aq) + NaOH (aq) H O + NaCl (aq) Initial strong acid strong base Change Equilib ~ [H + ] = mol H + /V T = /0.035L = ph = 1.37

13 Cont d In general, before the equivalence point: [H + ] = (C V C B V B ) / (V +V B ) 5.0 ml moles acid = moles base (i.e. this is the equivalence point) all acid is neutralized no excess base We are left with pure NaCl (aq) solution, ph = mL ph is determined by amount of excess base [OH ] = (C B V B C V ) / (V +V B ) = {(.1M)(.035L)(.1M)(.05L)}/{ L} = M ph = 14 poh = 14 (log [OH ]) = ph = 1. 16_1 Titration of Weak cid with Strong Base 1 ph Yellow 3 Colorless Blue Pink Bromcresol green Phenolphthalein Equivalence point Volume of NaOH added (ml) moderate initial ph, buffer region, basic equivalence point, change in ph not as great as with strong acid strong base 4 Volume NaOH added (ml) ph

14 Titration of Weak cid with Strong Base 1. Initial Point calculate ph of solution of a weak acid. (Use approximation for weak acid or solve quadratic.). Buffer Region Use HH equation to determine ph. moles H = initial moles H moles OH added moles = moles OH added 3. Equivalence Point a pure solution of a weak base, moles = moles H we started with (Use approximation for weak base or solve quadratic.) 4. Excess base (region 4) ph is determine solely by the amount of excess strong base added. (See Strong cid/strong Base example.) Example What is the ph at the equivalence point in the titration of 50 ml of a 0.1M solution of acetic acid with 0.1M NaOH? K a (acetic acid) = 1.8 x10 5 What is the volume of NaOH at the equivalence point? For a monoprotic acid, at the equivalence point, moles of acid = moles of base C V = C NaOH V NaOH V equiv =V NaOH =C V /C NaOH =(0.100M)(0.050L)/(0.100M) = 50mL t equivalence point, [conj. base]= C V /V TOTL = [ ] [ ], = (0.100M)(0.050L)/( L) = 0.05M

15 Cont d So, at the equivalence point, what is the ph of a 0.05M solution of CH 3 COO Na +? 14 6 C K (0.05M) (1x10 ) [OH ] = CBKB = B w = = 5.7x10 K 5 1.8x10 ph = 14 poh = 14 (log (5.7 x 10 6 )) ph = _8 Indicator name Methyl violet yellow 0 ph range for color change violet Which indicator would we use? Thymol blue (acidic range) Bromphenol blue Methyl orange red yellow yellow blue red yellow pk a(indicator) = ph equivalence + 1 Bromcresol green yellow blue Methyl red red yellow Bromthymol blue yellow blue Thymol blue (basic range) Phenolphthalein yellow colorless blue pink lizarin yellow R yellow red