Acids, Bases and the Common Ion Effect Consider the following acid equilibrium of a weak acid: HF + H O H 3 O + + F - K a = [H 3 O + ][F - ] [HF] By LeChatelier s principle, we predict the HF dissociation should be driven left, suppressing the dissociation. H + + Cl - HCl (aq) 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.5m HF. b) Determine [F - ] in a solution of.5m HF and.m HCl. K HF = 6.6x -. a) A solution of a weak acid. Let s use the quadratic. HF HF HF HF + K K K C [ H ] = [ F ] = + + [F - ] =.178M Continued b) Use ICE table. Let HF and HCl dissociate separately. HF + H O H 3 O + + F - Initial weak acid.5 strong acid. Change -x + x + x Equilib.5-x. +x x K a = [H 3 O + ][F - ] [HF] K a = (. +x)(x) / (.5-x) Assumption: ionization of HF is suppressed. Assume x<<.1; x<<.5 K a ~ (.)(x) / (.5)
Continued Rearrange: x = [F - ]= (.5M)K a /.M =.33 M Comparison a) no common ion, [F - ] =.178 M b) common ion, [F - ] =.33 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. Buffering and titrations! 16_1 1 Volume NaOH added (ml) ph ph 1 8 6 Pink Phenolphthalein Colorless Equivalence point Blue Yellow Bromcresol green 5 15 5 3 35 5 5 Volume of NaOH added (ml) 1 3 5 15 1 3 5 6 7 8 9 3 35 5 5.9 3.7 3.79 3.98.13.5.67 5.3 5.5 5.57 5.7 5.91 6.3 8.78 11.9 11.59 11.75 11.87 11.96 1. 1.36 1.6 1.5 Chemical buffer a chemical system that resists change Example: any chemical system in equilibrium aa + bb + cc dd + ee + ff K eq d e f [D] [E] [F] = [A] [B] [C] a b c Acid/Base buffer a system that resists changes to ph caused by addition of excess base or acid.
Acid Base Buffer Systems A mixture of a weak acid and it s conjugate base O O H Benzoic acid O OH CH 3 acetic acid O O benzoate O O CH 3 acetate Add as sodium salts for example A mixture of a weak base and it s conjugate acid H N + Add as H C CH CH 3 3 3 H 3 C CH 3 CH 3 N chloride trimethylamine trimethylammonium salt for example An acid/base buffer system What happens if we add acid and conjugate base in equal concentrations? i.e. equal moles of HA and NaA HA + H O H 3 O + + A - Initial weak acid. conj base. Change -x +x + x Equilib.-x +x.+x K a = + [H ][A ] [HA] K a = + [H ][. + x] [.- x] But due to presence of A -, ionization of HA is suppressed more than usual. x<<. [H + ] = K a ph = pk a! Henderson-Hasselbalch Equation + [H ][A ] Ka = Solve for [H [HA] + + K a [HA] ] [H ] = [A ] Take log of both sides -log [H + ] = -log K a log [HA] + log [A - ] -log [H + ] = -log K a + log [A - ]/[HA] From previous ICE table, we see: [HA] = C HA x [A - ] = C A- + x But in general, C HA = C A-. If x<<c HA ph pk log [A - = ] a + ph pk log C - = A a + [HA] CHA
Buffer ratios For correct buffering, there should be significant amounts of HA and A -. Just as importantly, the ratio should be such that:. C C HA. A Substituting this constraint into the H-H equation ph = pk ± 1 buffer a 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 H-H equation to determine the exact ratio of the acid and the base you will need.. What buffer capacity do you need.1m,.1m,.1m? Answer 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 ). Example Prepare 1L of buffer with ph = 3.5, with a total concentration, [A - ] + [HA] =.1M. ). Pick buffer system. Acid pk a Citric Acid 3.13 Benzoic acid. Acetic acid.77 Carbonic acid 6.36 Ammonium ion 9.5 Phenol 9.89
Continued 3) Determine ratio of [A - ]/[HA] needed. Use either H-H equation or equilibrium expression. ph pk log C - = a + A CHA log C - A = ph pka CHA C - - A [A ] (ph pka) = CHA [HA] C - A (3.5 3.13) = =.3 CHA Cont d 5) Prepare solution (i.e. What masses or volumes are needed?) C A- /C HA =.3 (1) Also C A- + C HA =.1M () from (1) C A- =.3 C HA,substitute this into ().3 C HA + C HA =.1M C HA =.3 M C A- =.3 C HA C A- =.7M What substances, and quantities, are needed to make these solutions? cont d Citric acid, C 6 H 8 O 7 MW = 19.1g/mole monosodium citrate NaC 6 H 7 O 7 MW = 1.1g/mole Since we are preparing 1L, we need to add.3 moles of citric acid and.7 moles of monosodium citrate. mass HA =.3 mol x 19.1 g/mol =.576g mass NaA =.7 mol x 1.1 g/mol = 1.5g Dissolve to make 1L of solution. OR Dissolve.1mol 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 7.+.5. 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. Alkalosis raising of ph results from hyperventilation, or exposure to high elevations (altitude sickness). Acidosis lowering of ph in blood by organ failure, diabetes or long term protein diet. in the lungs CO (g) CO (aq) + H O The bicarbonate buffer is essential for controlling blood ph At 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. [HCO 3 ] [H CO ] ~ 3 + OH - H O ph increases! Acid Base Titrations Acid base titrations are examples of volumetric techniques used to analyze the quantity of acid or base in an unknown sample. Acid + 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 (A) with a base (B), we have at equivalence: moles of acid = moles of base C A V A = C B V B CBVequiv C A = VA We detect the equivalence point with a ph meter or by identifying the end-point with an acid base indicator.
Acid Base Titrations - some terminology equivalence point: the point at which moles of acid = moles of base end-point: the experimental approximation of the equivalence point titration error: the difference between the equivalence point and the end-point titrant: the solution that is added in a measured quantity standard solution: a solution of known concentration Indicators Acid-base indicators are highly coloured weak acids or bases. HIndic Indic - + H + colour 1 colour Colour transition occurs for.1< [Indic - ]/[HIndic]< 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) Indicators phenolphthalein Selection of an indicator. 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.
Indicator Examples bromthymol blue methyl red ph Acid Base Titrations 16_11 1 Volume NaOH added (ml) 1 5 Pink 15 Phenolphthalein 1 Colorless 8 3 Equivalence point 5 6 Blue 6 7 Bromcresol green 8 Yellow 9 3 35 5 5 5 15 5 3 35 5 5 Volume of NaOH added (ml) Titration of strong acid with strong base Low initial ph, large ph change at equivalence point ph 1. 1.18 1.37 1.6 1.95.6..38.69 7. 11.9 11.59 11.75 11.87 11.96 1. 1.36 1.6 1.5 Example What is the ph at ml, ml, 5.mL and 35mL in titration of 5mL of. M strong acid (HCl) with.m NaOH? ml [H + ] = C A =.M ph = -log (.1) ph = 1. ml (all calculated in moles) HCl (aq) + NaOH (aq) H O + NaCl (aq) Initial strong acid.5 strong base.1 Change -. -.1.1.1 Equilib.15 ~.1.1 [H + ] = mol H + /V T =.15 /.35L =.9 ph = 1.37
Cont d In general, before the equivalence point: [H + ] = (C A V A C B V B ) / (V A +V B ) 5. 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 = 7. 35.mL ph is determined by amount of excess base [OH - ] = (C B V B C A V A ) / (V A +V B ) = {(.1M)(.35L)-(.1M)(.5L)}/{.5+.35L} =.167M ph = 1 poh = 1 (-log [OH - ]) = 1-1.78 ph = 1. 16_1 1 Titration of Weak Acid with Strong Base Volume NaOH added (ml) ph.9 1 3 Pink Phenolphthalein 1 3 5 3.7 3.79 3.98.13.5 ph 8 Colorless Equivalence point 15.67 5.3 5.5 6 Blue 1 5.57 5.7 1 Yellow Bromcresol green 3 5 6 7 5.91 6.3 8.78 11.9 11.59 8 11.75 9 11.87 5 15 5 3 35 5 5 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 3 35 5 5 11.96 1. 1.36 1.6 1.5 Titration of Weak Acid 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 H-H equation to determine ph. moles HA = initial moles HA moles OH - added moles A - = moles OH - added 3. Equivalence Point- a pure solution of a weak base, A - moles A - = moles HA we started with (Use approximation for weak base or solve quadratic.). Excess base (region ) ph is determine solely by the amount of excess strong base added. (See Strong Acid/Strong Base example.)
15_8 Example What is the ph at the equivalence point in the titration of 5 ml of a.1m solution of acetic acid with.1m NaOH? K a (acetic acid) = 1.8 x -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 A V A = C NaOH V NaOH V equiv =V NaOH =C A V A /C NaOH =(.M)(.5L)/(.M) = 5mL At equivalence point, [conj. base]= C A V A /V TOTAL = [A - ] [A - ], = (.M)(.5L)/(.5+.5L) =.5M Cont d So, at the equivalence point, what is the ph of a.5m solution of CH 3 COO - Na +? -1 C K (.5M) (1x ) [OH ] C K B w 6 = B B = = = 5.7x K -5 A 1.8x ph = 1 poh = 1 (-log (5.7 x -6 )) ph = 8.7 Indicator name Methyl violet ph range for color change 6 8 1 violet Which indicator would we use? Thymol blue (acidic range) Bromphenol blue Methyl orange red blue red pk a(indicator) = ph equivalence + 1 Bromcresol green blue Methyl red red Bromthymol blue blue Thymol blue (basic range) Phenolphthalein colorless blue pink Alizarin R red