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1 The common-ion effect Chapter 17 Additional Aspects of Acid-Base Equilibria Dr. Peter Warburton Say we make two acid solutions: M HCl (a strong acid) and M CH COOH (a weak acid). HO H O HCl Cl A M HCl solution by itself would have a ph 1.0 ([H O ] = M) since the reaction goes to completion. The common-ion effect The common-ion effect CHCOOH HO HO CH COO (all in mol/l) CH COOH (aq) H O (l) H O (aq) CH COO - (aq) Initial conc N/A Conc. change -x N/A x x Equil. conc x N/A x x A M CH COOH solution ( a = 1.8 x 10-5 ) by itself would have a ph.8 since an equilibrium is established where the equilibrium concentration of acetic acid M x = [H O ] = [CH COO - ] = 1. x 10 - M 4 1

2 The common-ion effect Say we put together a solution that is BOTH M HCl (a strong acid) and M CH COOH (a weak acid). The two reactions of the acids with water take place in the same container at the same time! The common-ion effect HCl H O H O Cl CHCOOH HO HO CHCOO Both reactions are a source of H O and so we could expect that [H O ] = M 1. x 10 - M [H O ] = M and ph 0.99 which appears to be true 5 6 The common-ion effect HCl H O CH COOH H O H O H O Cl CH COO However, both reactions share the common ion H O, and so they cannot be treated as independent reactions. Something that affects one reaction must also affect the other reaction. Le Chatalier s Principle! The common-ion effect CHCOOH H O HO CHCOO Imagine we start with the M CH COOH and then we add M H O by adding M HCl. Le Chatalier Principle tells us our reaction will shift back towards reactants! 7 8

3 The common-ion effect H O as a common-ion CHCOOH HO HO CHCOO (all in mol/l) CH COOH (aq) H O (l) H O (aq) CH COO - (aq) [HCl] initial N/A [CH COOH] initial N/A Conc. changes x N/A -x x Equil. conc x N/A x x If we assume x is much smaller than M we will quickly find that x = a x = [CH COO - ] = 1.8 x 10-5 M The value of x has decreased because of the added H O! 9 10 OH - as a common-ion Salts as source of a basic common-ion 11 1

4 Salts as source of a basic common-ion Salts as source of an acidic common-ion On the left is a solution of M CH COOH while on the right is a solution that is both M CH COOH and M CH COONa which is a source of the basic common ion CH COO -. The reaction has shifted back towards reactants! 1 14 Salts as source of an acidic common-ion On the left is a solution of M NH while on the right is a solution that is both M NH and M NH 4 Cl which is a source of the acidic common ion NH 4. The reaction has shifted back towards reactants! Buffer solutions Solutions that contain both a weak acid and its conjugate base are buffer solutions. These solutions are resistant to changes in ph

5 Buffer solutions The system has enough of the original acid and conjugate base molecules in the solution to react with added acid or added base, so the new equilibrium mixture will be very close in composition to the original equilibrium mixture. Buffer solutions A 0.10 mol L -1 acetic acid 0.10 mol L -1 acetate mixture has a ph of 4.74 and is a buffer solution! CH COOH (aq) H O (l) H O (aq) CH COO - (aq) (all in mol/l) CH COOH (aq) H O (l) H O (aq) CH COO - (aq) Initial 0.10 N/A Conc. changes -x N/A x x Equil. conc x N/A x 0.10 x a = 1.8x10 5 = [HO ][CHCOO [CH COOH] ] Buffer solutions Buffer solutions If we rearrange the a expression a[chcooh] [H O ] = = 1.8x10 [CH COO ] and make the assumption that x is much less than 0.10, we see If [CH COOH] = [CH COO - ], then [H O ] = 1.8 x 10-5 M = a and ph = p a = 4.74 [CHCOOH] [CH COO ]

6 Buffer solutions What happens if we add 0.01 mol of NaOH (strong base) to 1.00 L of the acetic acid acetate buffer solution? CH COOH (aq) OH - (aq) H O (l) CH COO - (aq) This goes to completion and keeps occurring until we run out of the limiting reagent OH - (all in moles) CH COOH (aq) OH - (aq) H O (l) CH COO - (aq) Initial N/A 0.10 Change -x -x N/A x Final (where x = 0.01 due to limiting OH - ) 0.10 x = x = 0.00 N/A 0.10 x = 0.11 New [CH COOH] = 0.09 M and new [CH COO - ] = 0.11 M Buffer solutions (all in mol/l) CH COOH (aq) H O (l) H O (aq) CH COO - (aq) Initial conc N/A Conc. change -x N/A x x Equil. conc x N/A x 0.11 x 5 [HO ][CHCOO ] 5 (x)(0.11 x) a = 1.8x10 = so 1.8x10 = With the assumption [CH COOH] that x is much smaller (0.09than x) 0.09 mol (an assumption we always need to a[chcooh] [HO ] = = 1.8x10 = 1.5x10 M check after [CH COO calculations ] are done!), 0.11 we find Note we ve made the assumption that x << 0.09! ph = - log [H O ] ph = - log 1.5 x 10-5 ph = Buffer solutions Adding 0.01 mol of OH - to 1.00 L of water would have given us a ph of 1.0! There is no significant amount of acid in water for the base to react with. Buffer solutions What happens if we add 0.01 mol of HCl (strong acid) to 1.00 L of the acetic acid acetate buffer solution? CH COO - (aq) H O (aq) H O (l) CH COOH (aq) This goes to completion and keeps occurring until we run out of the limiting reagent H O (all in moles) CH COO - (aq) H O (aq) H O (l) CH COOH (aq) Initial N/A 0.10 Change -x -x N/A x Final (where x = 0.01 due limiting H O ) 0.10 x = x = 0.00 N/A 0.10 x = 0.11 New [CH COOH] = 0.11 M and new [CH COO - ] = 0.09 M 4 6

7 Buffer solutions (all in mol/l) CH COOH (aq) H O (l) H O (aq) CH COO - (aq) Initial conc N/A Conc. change -x N/A x x Equil. conc x N/A x 0.09 x 5 [HO ][CHCOO ] 5 (x)(0.09 x) a = 1.8x10 = so 1.8x10 = With the assumption [CH that COOH] x is much smaller than ( mol x) (an assumption we always need to check after a[chcooh] [H calculations are done!), we find O ] = = 1.8x10 =.x10 M [CHCOO ] 0.09 Note we ve made the assumption that x << 0.09! ph = - log [H O ] ph = - log. x 10-5 ph = 4.66 Buffer solutions Adding 0.01 mol of H O to 1.00 L of water would have given us a ph of.0! There is no significant amount of base in water for the acid to react with. 5 6 Adding acid or base to a buffer Problem Calculate the ph of a L buffer solution that is 0.5 mol/l in HF and 0.50 mol/l in NaF. (all in mol/l) HF (aq) H O (l) H O (aq) F - (aq) Initial conc. 0.5 N/A Conc. change -x N/A x x Equil. conc. 0.5 x N/A x 0.50 x a =.5x10 4 [HO ][F ] = [HF] so.5x10 4 (x)(0.50 x) = (0.5 x) With the assumption that x is much smaller than 0.5 mol (an assumption we always need to check after calculations are done!), we find 7 8 7

8 Problem a[hf] [HO ] = =.5x10 [F ] = ph = - log [H O ] ph = - log x 10-4 ph =.76 5 x10 4 M Problem a) What is the change in ph on addition of 0.00 mol of HNO? (all in moles) F - (aq) H O (aq) H O (l) HF (aq) Initial N/A 0.05 Change -x -x N/A x Final (where x = x 0.00 x N/A 0.05 x due to limiting H O ) = = 0.00 = 0.07 a New [HF] = 0.7 M and new [F - ] = 0.48 M (all in mol/l) HF (aq) H O (l) H O (aq) F - (aq) Initial conc. 0.7 N/A Conc. change -x N/A x x Equil. conc. 0.7 x N/A x 0.48 x =.5x10 4 [HO ][F = [HF] ] so.5x10 4 (x)(0.48 x) = (0.7 x) 9 0 Problem a[hf] [HO ] = =.5x10 [F ] = x10 Notice we ve made the assumption that x << 0.7. We should check this! ph = - log [H O ] ph = - log x 10-4 ph =.71 4 M Problem b) What is the change in ph on addition of mol of OH? (all in moles) HF (aq) OH - (aq) H O (l) F - (aq) Initial N/A 0.10 Change -x -x N/A -x Final (where x = x x N/A x due to limiting OH - ) = 0.01 = 0.00 = New [HF] = 0.1 M and new [F - ] = 0.54 M (all in mol/l) HF (aq) H O (l) H O (aq) F - (aq) Initial conc. 0.1 N/A Conc. change -x N/A x x Equil. conc. 0.1 x N/A x 0.54 x a =.5x10 4 [HO ][F = [HF] ] so.5x10 4 (x)(0.54 x) = (0.1 x) 1 8

9 Problem a[hf] [HO ] = =.5x10 [F ] = x10 Notice we ve made the assumption that x << 0.1. We should check this! ph = - log [H O ] ph = - log 1. 6 x 10-4 ph =.87 4 M Predicting whether a solution is a buffer Any solution that becomes a mixture of a conjugate acid-base pair will be a buffer. 1) Weak acid-base conjugate pairs like CH COOH and CH COO - or NH 4 and NH. ) Weak acid reacting with small amounts of strong base like CH COOH and NaOH. ) Weak base reacting with small amounts of strong acid like NH and HCl. 4 Problem Describe how a mixture of a strong acid such as HCl and a salt of a weak acid such as CH COONa can be a buffer solution. Problem What is the ph of a buffer solution prepared by dissolving.1 g of NaCHO (molar mass is g mol -1 ) in a sufficient volume of 0.4 M HCHO to make ml of the buffer? a of formic acid is 1.8 x 10-4 Answer: ph =

10 The Henderson-Hasselbalch equation We ve seen that, for solutions with both members of a conjugate acid-base pair, that [acid] [H O ] = a [base] [acid] [acid] log[ho ] = log a = loga log [base] [base] ph = p a log [base] / [acid] This is called the Henderson- Hasselbalch Equation. The Henderson-Hasselbalch Equation If we have a buffer solution of a conjugate acid-base pair, then the ph of the solution will be close to the p a of the acid. This p a value is modified by the logarithm of ratio of the concentrations of the base and acid in the solution to give the actual ph. 7 8 The Henderson-Hasselbalch equation ph = p a log [base] / [acid] ALWAYS remember when you use the H-H eqn that there is the assumption that the equilibrium concentrations of acid and base are relatively unchanged from the initial concentrations. That is we have assumed x is very small compared to the initial concentrations! 9 The Henderson-Hasselbalch equation ph = p a log [base] / [acid] Generally this assumption is valid as long as we know 0.10 < [base] / [acid] < 10 AND [base] / a > 100 AND [acid] / a >

11 Alternate Henderson-Hasselbalch equation We can always look at a buffer solution as a base combined with its conjugate acid B (aq) H O (l) OH - (aq) BH (aq) [BH ][OH ] [B] b = so [OH ] = b = [B] [BH ] log[oh ] = log b [base] = log [acid] poh = p b log [acid] / [base] b b [base] [acid] [base] log [acid] Alternate Henderson-Hasselbalch equation If we have a buffer solution of a conjugate acid-base pair, then the poh of the solution will be close to the p b of the base. This p b value is modified by the logarithm of ratio of the concentrations of the acid and base in the solution to give the actual ph Problem Use the Henderson-Hasselbalch equation to calculate the ph of a buffer solution prepared by mixing equal volumes of 0.0 mol/l NaHCO and 0.10 mol/l Na CO. a of HCO - = 4.7 x (see Table 16.4) We should also check the validity of using H-H at the end to be sure! 4 Problem answer If we mix equal volumes, the total volume is TWICE the volume for the original acid or base solutions. Since the number of moles of acid or base DON T CHANGE on mixing, the concentrations will be half the given values. ph = p a log [base] / [acid] ph = (-log 4.7 x ) log (0.05) / (0.10) ph = ph =

12 Preparing buffer solutions How do we adjust the concentrations? If we want to create a buffer solution of a specific ph, the H-H equation tells us we need to pick a conjugate acid-base pair with a p a for the acid close to the ph we want, and then we adjust the amounts of the conjugate acid and base Problem How many grams of (NH 4 ) SO 4 (molar mass is 1.141g mol -1 ) must be dissolved in L of 0.5 M NH to produce a solution with ph = 9.00? Assume that the solution volume remains at L. b for ammonia is 1.8 x Answer: 1 g Buffer capacity Buffer capacity is the measure of the ability of a buffer to absorb acid or base without significant change in ph. Larger volumes of buffer solutions have a higher buffer capacity, and buffer solutions of higher initial concentrations of the conjugate acid-base pair have a larger buffer capacity

13 Buffer range We ve seen that as long as 0.10 < [base] / [acid] < 10 then the assumption that the Henderson-Hasselbalch equation is based upon (x << [base] and x << [acid]) is likely to be valid. Buffer range ph = p a log [base] / [acid] ph = p a log 10 ph = p a 1.0 OR ph = p a log [base] / [acid] ph = p a log 0.10 ph = p a Buffer range Application of buffers In general buffer solutions have a useful range of ph that is p a ± 1.0 For instance an acetic acid - acetate buffer has a useful ph range of about.7 to 5.7 since p a is 4.7 ( a = 1.8 x 10-5 ) Many biological processes can only occur at very specific ph values (usually between ph = 6 and ph = 8). The reactions often take place in buffered environments (e.g. human blood is buffered to ph = 7.4 see text pg. 74)

14 Acid-base indicators We often measure the ph of a solution with a chemical acid-base indicator. Such indicators are weak acids in their own right (symbolized HIn) and indicate ph because the acid form has a different colour than the conjugate base form (In - ) Diprotic indicator! HIn (aq) H O (l) H O (aq) In - (aq) colour A colour B 5 54 Acid-base indicators HIn (aq) H O (l) H O (aq) In - (aq) colour A colour B If we increase the [H O ] we shift this reaction towards reactants. The colour will be that of HIn. If we decrease the [H O ] we shift this reaction towards products. The colour will be that of In -. A [H O ] in between these two extremes will give a colour that is a mixture of the two colours because both HIn and In - are present in significant amounts! Acid-base indicators HIn (aq) H O (l) H O (aq) In - (aq) colour A colour B More specifically, the HIn and In - form a buffer so the indicator works in a ph range of about ±1 around the p HIn of HIn

15 Since indicators work in a range of about ph units we often put several indicators that have different ranges in a single solution to give a universal indicator for ph range of about Universal indicator Applications of indicators Indicators are useful when we want a general idea of the value of the ph without using a ph meter. Pool chlorination (with Cl or NaOCl) is done to avoid algae growth. This works best at ph = 7.4, so we might need to add some acid or base

16 Neutralization reactions A reaction of an acid and a base often produces water and an aqueous salt as products. Such reactions are called neutralization reactions, and can be categorized by the strengths of the acid and base involved. Strong acid strong base neutralization The reaction of a strong acid such as HCl and a strong base such as NaOH becomes a reaction of H O and OH -, Therefore while the overall reaction is HCl (aq) NaOH (aq) H O (l) NaCl (aq) the actual net ionic equation is H O (aq) OH - (aq) H O (l) where the Na and Cl - ions are not involved (they are neutral spectator ions which don t react with water!) See slides 10, 10 and 106 of last chapter! 61 6 Strong acid strong base neutralization Strong acid strong base neutralization If we mix equal numbers of moles of HCl and NaOH, we will create twice as much water, while leaving an excess of NEITHER ion. In this case the [H O ] = [OH - ] = 1.0 x 10-7 M at 5 C (reaction goes to completion). H O (aq) OH - (aq) H O (l) is the reverse of the autoionization of water reaction so o n = = = = 1.0 x 10 at 5 C 14 H O OH w 1.0x10 [ ] [ ] Since the equilibrium constant is very large, we see the reaction goes to completion, and at the end, the equilibrium mixture will consist of water and an aqueous salt of ions that are neutral in character because they don t react with water. The ph after the reaction will be 7 (a neutral solution)

17 Strong acid titration by strong base Titration curve strong acid titrated by strong base Say we add the strong base to the strong acid solution drop by drop. Each drop of base (the titrant) will react with some of the acid to completion until the added base (the limiting reagent!) is all gone. If we measure the ph after we add each drop of base, we can plot a ph versus total volume of added base graph. This is called a titration curve End point versus equivalence point How do we know when we ve reached the end of our titration? We often will use an indicator to tell us when the titrated solution reaches a specific ph. This is the end point of the titration. The end point DEPENDS on the indicator we use! End point versus equivalence point The equivalence point is that point in the titration where we have added equal numbers of MOLES of acid and base. For strong acid strong base neutralizations the equivalence point occurs when the ph is 7 (a neutral solution)

18 Titration curve strong acid titrated by strong base Different end points depending on indicator End point versus equivalence point Bromothymol blue is a good indicator for strong acid-strong base titrations because the end point is very close to the equivalence point. Methyl red and phenolphtalein are pretty good choices too because the titration curve is very steep in their effective ph ranges Titration curve strong base titrated by strong acid Equivalence point The curve looks exactly the same, just flipped vertically! 71 Millimoles (mmol) Since we often set up titrations where our titrant concentration is less than 1 M and total titrant volume used is less than 50 ml this means we are adding about 5.00 x 10 - moles of acid or base in our titration. Because of this, we often do calculations in millimoles (1 mmol = 1 x 10 - mol) 7 18

19 Millimoles (mmol) M = mol / L M = mmol / ml Problem For the titration of 5.00 ml of M HCl with 0.50 M NaOH, calculate a) the initial ph; b) the ph when the neutralization is 50.0% complete; c) the ph when the neutralization is 100% complete; and d) the ph when 1.00 ml of NaOH is added beyond the equivalence point Problem answers a) ph = 0.84 (V total = 5.00 ml) b) ph = 1.8 (V total =.50 ml) c) ph = 7.00 (V total = ml) d) ph = (V total = ml) Weak acid strong base Because a weak acid HA is largely undissociated in water, while the strong base is completely dissociated in water (becoming a source of OH - ), the neutralization reaction of a weak acid and strong base becomes that of HA (aq) OH - (aq) H O (l) A - (aq) Acetic acid is a weak acid, so when it reacts with NaOH, the net ionic equation is CH COOH (aq) OH - (aq) H O (l) CH COO - (aq) Now remember that net = 1 x x x x n

20 Weak Acid Strong Base CH COOH (aq) H O (aq) OH OH (aq) (aq) H O H O (l) _ CH COOH (aq) H O (l) H O (l) (aq) CH COO The equilibrium constant shows us the reaction goes to completion, so for equal numbers of moles of weak acid and strong base, we expect only water and the aqueous salt in the equilibrium mixture. CH COO (aq) = 1.8 x 10 a 1 n w (aq) = x a 5 = 1.0 x 10 1 w = 1.8 x Weak acid strong base CH COOH (aq) OH (aq) H O (l) CH COO However, the salt comprises of a neutral cation (Na in this case) and a weakly basic anion (CH COO - ), meaning the equilibrium mixture will be BASIC and have a ph greater than 7. See slide 107 of last chapter! In a weak acid titration by a strong base this means the equivalence point is NOT at ph = 7 but rather at ph > 7! (aq) = x n a 1 w = 1.8 x Titration curve weak acid titrated by strong base Problem. ph = p a when moles added OH - = ½ initial moles weak acid 4. buffer breaks! Titration of. creation of a buffer CH COOH with NaOH. 1. higher initial ph for a weak acid 6. Added strong base dominates weak base and determines ph 5. ph > 7 because of water hydrolysis by conjugate base For the titration of 0.00 ml of M HF with 0.50 M NaOH, calculate a) the initial ph; a = 6.6 x 10-4 b) the ph when the neutralization is 5.0% complete; c) the ph when the neutralization is 50.0% complete; and d) the ph when the neutralization is 100% complete

21 Problem answers a) ph =.00 (V total = 0.00 ml) b) ph =.70 (V total =.00 ml) c) ph =.18 (V total = 6.00 ml) d) ph = 8.08 (V total =.00 ml) Weak base strong acid Because a weak base B is largely undissociated in water, while the strong acid is completely dissociated in water (becoming a source of H O ), the neutralization reaction of a strong acid and weak base becomes that of H O (aq) B (aq) H O (l) BH (aq) Ammonia is a weak base, so when it reacts with HCl, the net ionic equation is H O (aq) NH (aq) H O (l) NH 4 (aq) Again net = 1 x x x x n 81 8 Weak base strong acid NH H O H O (aq) (aq) (aq) H O (l) OH NH (aq) (aq) NH NH 4 H O (l) 4 (aq) (aq) OH H O (l) Again the equilibrium constant shows us the reaction goes nearly to completion, so for equal number of reactant moles, we expect only water and the aqueous salt in the equilibrium mixture. (aq) b 1 w n = 1.8 x 10 = 1.0 x 10 = b x = 1.8 x w 9 8 Weak base strong acid H O (aq) NH (aq) NH 4 (aq) H O (l) However, the salt comprises of a neutral anion (Cl - in this case) and a weakly acidic cation (NH 4 ), meaning the equilibrium mixture will be ACIDIC and have a ph less than 7. In a weak base titration by a strong acid this means the equivalence point is NOT at ph = 7 but rather at ph < 7! n = b x 1 = 1.8 x 10 w

22 Titration curve weak base titrated by strong acid ph = p b when moles added H O = ½ initial moles weak base Problem For the titration of ml of M NH with 0.5 M HCl, calculate a) the initial ph; b = 1.8 x 10-5 b) the ph when the neutralization is 5.0% complete; c) the ph when the neutralization is 50.0% complete; and d) the ph when the neutralization is 100% complete Problem answers a) ph = (V total = ml) b) ph = 9.74 (V total = ml) c) ph = 9.6 (V total = ml) d) ph = 5.0 (V total = 7.55 ml) 87

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