Chapter 33 Acid-Base Equilibria 1 Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved. Solutions of a Weak Acid or Base • The simplest acid-base equilibria are those in which a single acid or base solute reacts with water. – we will first look at solutions of pure weak acids or bases. – Next, we will also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water. Salts can be neutral, acidic, or basic. 2 33.1 Acid-Ionization Equilibria • Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion (hydrolysis). HC 2 H 3O 2 (aq ) H 2O(l ) H 3O (aq) C2 H 3O 2 (aq) <<100% K<<1 – Because acetic acid is a weak electrolyte, it ionizes to a small extent in water, hence double arrow; not 100% like strong acid which wrote single arrow. 3 Acid-Ionization Equilibria • For a weak acid, the equilibrium concentrations of ions in solution are determined by the acidionization constant (also called the aciddissociation constant, Ka). – Consider the generic monoprotic acid, HA. HA(aq ) H 2O(l ) H 3O (aq ) A (aq ) 4 Acid-Ionization Equilibria HA(aq) H 2O(l ) H 3O (aq) A (aq) – Recall that the concentrations of pure solids and pure liquids are not included in the K expression. – The corresponding equilibrium expression is: [ H 3O ][ A ] K Ka [ HA] – however, since we are referring to acid hydrolysis (acid + H2O) the constant is known as Ka , the acid-ionization constant. 5 Experimental Determination of Ka • Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. – It is important to realize that the solution was made 0.012 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.012 M, just a small amount less. 7 A Problem To Consider – Let x be the moles per liter of product formed. HC6 H 4 NO2 (aq) H 2O(l ) Initial[] Change[] Equilibrium[] H 3O (aq) C6 H 4 NO2 (aq) 0.012 - x 0.012 - x 0 0 +x +x x x [ H 3O ][C6 H 4 NO2 ] Ka [ HC6 H 4 NO2 ] 8 A Problem To Consider – We can obtain the value of x from the given pH. x [ H 3O ] anti log( pH ) 10 pH 9 A Problem To Consider – Substitute this value of x in our equilibrium expression. [ H 3O ][C6 H 4 NO2 ] Ka [ HC6 H 4 NO2 ] ionization <<100% K<<1 x2 (0.00041) 2 5 Ka 1.4 10 (0.012 x) (0.01159) – Note that HW 28 (0.012 x) (0.012 0.00041) 0.01159 0.012 the concentration of unionized acid remains virtually unchanged. Important point to remember for later. code: ka 10 33.1.1 Percent Ionzation • The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions. – Electrical conductivity or some other colligative property can be measured to determine the degree of ionization. – With weak acids, the pH can be used to determine the equilibrium composition of ions in the solution because it gives the concentration of hydronium ions at equil. pH of solution [H O+] amount ionized 3 eq 11 – To obtain the degree of ionization: % ionization [ H ]eq [acid ]o x 100% note: % is unit 0.00041 % ionization x100% 3.4% 0.012 – weak acids typically less than 5%. The lower the %ionized, less H+ formed, the lower Ka, weaker the acid. Ka is measure of strength of acid or Kb if base. Strength based on ionization not pH directly. Lower pH more acidic, does not mean stronger acid 12 Ka is a measure of strength of acid. It indicates the amount ionized to form hydronium ions. The more hydronium ions formed, naturally the more acidic and lower pH. However, strength of acid is measure of amount ionized, not conc of acid. Conc can vary which ultimately will vary the pH of solution. To determine the strength of an acid you must compare Ka values not pH of solution. pH is dependent on Ka as well as conc. of solution. Must compare apples and apples. 0.1M HC2H3O2 Ka = 1.8 x 10-5 pH = 2.87 0.1M HF Ka = 6.9x10-4 pH = 2.08 10M HC2H3O2 Ka = 1.8 x 10-5 pH = 1.87 0.1M HF Ka = 6.9x10-4 pH = 2.08 Which stronger acid? HF, larger Ka 13 33.1.2 pH of Weak Acids • Once you know the value of Ka, you can calculate the equilibrium concentrations of species HA, A-, and H3O+ for solutions of different molarities. – The general method for doing this is same way we did Kc or Kp except we can simplify the math because of percent ionize is so small compared to initial conc of acid; remember 0.012-x=0.012. This wasn't the case for Kc or Kp problems so can't neglect in those type problems 14 Calculations With Ka • Note that in our previous example, the degree of ionization was very small as compared to the concentration of nicotinic acid. • It is the small value of the degree of ionization that allows us to ignore the subtracted x in the denominator of our equilibrium expressions to simplify our calculations. • The degree of ionization of a weak acid depends on both the Ka and the concentration of the acid solution 15 Calculations With Ka • How do you know when you can use this simplifying assumption? – It can be shown that if the acid concentration, Ca, divided by the Ka exceeds 100, that is, if [acid] Ka 100 – then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5% otherwise must do quadratic formula. Another way look at it factor of two - three difference in power of 10 is needed between Ka and conc of acid. 16 ex. Calculate the pH in a 1.00 M solution of nitrous acid, HNO2, for which Ka = 6.0 x 10-4 at 25oC. acid, base, or salt? acid strong or weak? weak HNO2 + H2O Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [HNO2] 1.00 -x 1.00 - x H3O+ + NO2- [H3O+] 0 +x x [NO2-] 0 +x x = Does answer make sense? Yes, acid pH 17 A Problem To Consider • Really two equil: Kw as well but neglect most of the time as long as [H+] is greater than 10-6 M from the acid. • Really [H+]tot = [H+]acid +[H+]H2O (water so small neglect) HNO2 + H2O H2O + H 2O H3O+ + NO2H3O+ + OH- [H+]tot = [H+]HNO2 +[H+]H2O [H+]tot = 0.024 M + <10-7M = 0.024 M 18 ex. Calculate the pH of 10.00 mL of 1.00 M solution of nitrous acid , HNO2, diluted to 100.0 mL with water for which Ka = 6.0 x 10-4 at 25oC. acid, base, or salt? acid H3O+ + NO2- HNO2 + H2O strong or weak? weak [HNO2] [H3O+] [NO2-] Initial, [ ]o 0.100 0 0 Change, D[ ] -x +x +x x x Equilibrium, [ ]eq 0.100 - x Does answer make sense? Yes, acid pH 19 A Problem To Consider • What is the pH at 25°C of a 3.6 x 10-3 M solution of acetyl salicylic acid (aspirin), HC9H7O4? The acid is monoprotic and • Ka=3.3 x 10-4 at 25°C. acid, base, or salt? acid strong or weak? weak 20 A Problem To Consider HC9 H 7O4 (aq ) H 2O(l ) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq 3.6 x 10-3 -x 3.6 x 10-3 - x H 3O (aq) C9 H 7O4 (aq) 0 0 +x +x x x 21 A Problem To Consider – Note that which is less than 100 or 10-3 vs 10-4 not factor 3 difference; therefore, won't work, so we must solve the equilibrium equation exactly . 22 – If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get HC9 H 7O4 (aq ) H 2O(l ) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq 3.6 x 10-3 -x 3.6 x 10-3 – x H 3O (aq) C9 H 7O4 (aq) 0 +x x 0 +x x 23 – You can solve this equation exactly by using the quadratic formula. – Rearranging the equation to put it in the form ax2 + bx 4 + c 6 =0 x ( 3.3 10 )x (1.2 10 ) 0 2 b b 4ac x 2a 2 (3.3 10 4 ) (3.3 10 4 ) 2 4(1)(1.2 10 6 ) x 2(1) x 9.4 10 4 M x 1.3 10 3 M 24 A Problem To Consider – Taking the positive value and neglect water 4 x [ H 3O ] 9.4 10 M [C9 H 7O4 ] – Now we can calculate the pH. 4 pH log(9.4 10 ) 3.03 Does answer make sense? Yes, acid pH HW 29 code: weak 25 Calculate the pH of a solution that has a concentration of 1.0 x 10-8 M NaOH. NaOH Na+ + OH1.0 x 10-8 M 1.0 x 10-8 M pOH = - log [OH-] = - log 1.0 x 10-8 = 8.00 pH = 14.00 – pOH = 14.00 – 8.00 = 6.00 Does the answer make sense? No, it is a base - what’s wrong? The principle reaction is the water not the NaOH; therefore, you can’t neglect the water. Must rework problem with water equilibrium and include the additional base from NaOH (pH = 7.02). 26 33.1.3 pH of Polyprotic Acids • Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. – Sulfuric acid, for example, can lose two protons in aqueous solution. H 2SO 4 (aq ) H 2O(l ) H 3O (aq ) HSO4 (aq ) HSO4 (aq ) H 2O(l ) H 2O(l ) H 2O(l ) 2 H 3O (aq ) SO 4 (aq ) H 3O (aq ) OH (aq ) [H3O+]tot = [H3O+]H2SO4 + [H3O+]HSO4- + [H3O+]H2O Typically worry about principle rxn only 27 Polyprotic Acids – For a weak diprotic acid like carbonic acid, H2CO3, several simultaneous equilibria must be considered. < 5% H 2CO 3 (aq ) H 2O(l ) HCO3 (aq ) H 2O(l ) H 2O(l ) H 2O(l ) H 3O (aq ) HCO 3 (aq ) << 5% 2 H 3O (aq ) CO 3 (aq ) H 3O (aq ) OH (aq ) [H3O+]tot = [H3O+]H2CO3 + [H3O+]HCO3- + [H3O+]H2O Once again, typically worry about principle rxn only. i.e. NaHCO3 would use the second equation. 28 Polyprotic Acids – Each equilibrium has an associated acid-ionization constant.– For the loss of the first proton H 2CO 3 (aq ) H 2O(l ) H 3O (aq ) HCO 3 (aq ) [H 3O ][HCO 3 ] K a1 4.3 107 [H 2CO 3 ] – For the loss of the second proton 2 HCO3 (aq ) H 2O(l ) H 3O (aq ) CO 3 (aq ) 2 [H 3O ][CO 3 ] 11 K a2 4 . 8 10 [HCO 3 ] 29 Polyprotic Acids – In general, the second ionization constant, Ka2, for a polyprotic acid is smaller than the first ionization constant, Ka1. Harder to pull proton off charged species. Total H+ is sum of all equil H+ but usually neglect all but the principal rxn. – In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2. HPO42Ka3 H2PO4- << Ka2 H3PO4 << Ka1 30 Polyprotic Acids – When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions. – However, reasonable assumptions can be made that simplify these calculations as we show in the next example. 31 A Problem To Consider • Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.100 M solution? Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12. H 2C6 H 6O6 (aq) H 2O(l ) HC6 H 6O6 (aq) H 2O(l ) H 3O (aq) HC6 H 6O6 (aq) 2 H 3O (aq) C6 H 6O6 (aq) – For diprotic acids, Ka2 is so much smaller than Ka1 that the smaller amount of hydronium ion produced in the second reaction can be neglected. 32 A Problem To Consider H 2C6 H 6O6 (aq) H 2O(l ) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq H 3O (aq) HC6 H 6O6 (aq) 0.100 -x 0.100 - x 0 0 +x +x x x [ H 3O ][ HC6 H 6O6 ] K a1 [ H 2C6 H 6O6 ] 33 A Problem To Consider 2 x 5 [ H 3O ][ HC6 H 6O6 ] 7 . 9 10 K a1 (0.100 x) [ H 2C6 H 6O6 ] – Assuming that x is much smaller than 0.10 (1265>100), you get x 2 (7.9 10 5 ) (0.100) 7.9 10 6 3 x 2.8 10 M 0.0028 M [ H 3O ] [ HC6 H 6O6 ] pH log(0.0028) 2.55 34 A Problem To Consider • Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- in previous example? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12. – The ascorbate ion, C6H6O62-, which we will call y, is produced only in the second ionization of H2C6H6O6. HC6 H 6O6 (aq) H 2O(l ) from 1st equil 2 H 3O (aq) C6 H 6O6 (aq) some from 1st eq none to start 35 H 2C6 H 6O6 (aq) H 2O(l ) H 3O (aq) HC6 H 6O6 (aq) 0.0028 M from 1st equil 0.0028 M – Assume the starting concentrations for HC6H6O6- and H3O+ to be those from the first equilibrium. HC6 H 6O6 (aq) H 2O(l ) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq 2 H 3O (aq) C6 H 6O6 (aq) 0.0028 0.0028 0 -y +y +y 0.0028-y 0.0028+y y 2 Ka2 [ H 3O ][C6 H 6O6 ] [ HC 6 H 6O6 ] 36 – Substituting into the equilibrium expression (0.0028 y )( y ) 12 1.6 10 (0.0028 y ) – Assuming y is much smaller than 0.0028 (1.75 x 109 > 100), the equation simplifies to (0.0028)( y ) 12 1.6 10 (0.0028) – Hence, 2 y [C6 H 6 O 6 ] 1.6 10 – typically, the concentration of the anion ion equals Ka2 if starting with reactants of first equil. 12 M HW 30 [H3O+]tot = 0.0028 MKa1 + 1.6 x 10-12 Ka2 + <1 x10-7 H2O = 0.0028 M code: multiple 37 33.2 Base-Ionization Equilibria • Equilibria involving weak bases are treated similarly to those for weak acids. – Ammonia, for example, ionizes in water as follows. NH 3 (aq ) H 2O(l ) NH 4 (aq ) OH (aq ) 38 Base-Ionization Equilibria NH 3 (aq ) H 2O(l ) NH 4 (aq ) OH (aq ) – Recall that the concentrations of pure solids and pure liquids are not included in the K expression. – The corresponding equilibrium expression is: 4 [ NH ][OH ] Kb [ NH 3 ] – however, since we are referring to base hydrolysis (base + H2O) the constant is known as Kb , the base-ionization constant. 39 Base-Ionization Equilibria • Equilibria involving weak bases are treated similarly to those for weak acids. Larger Kb , the stronger base, ionize more OH-, more basic. – In general, a weak base B with the base ionization HB (aq ) OH (aq ) B(aq ) H 2O(l ) has a base ionization constant equal to HW 31 [HB ][OH ] K b [B ] code: plate 40 A Problem To Consider • What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. acid, base, or salt? base strong or weak? weak – As before, we will follow the three steps in solving an equilibrium. 1. Write the equation and make a table of concentrations (ICE). 2. Set up the equilibrium constant expression. 3. Solve for x = [OH-]. 41 A Problem To Consider C5 H 5 NH (aq ) OH (aq ) C5 H 5 N(aq ) H 2O(l ) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [C5H5NH+] 0 [OH-] 0 - x +x +x 0.20 - x x x [C5H5N] 0.20 42 A Problem To Consider C5 H 5 NH (aq ) OH (aq ) C5 H 5 N(aq ) H 2O(l ) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [C5H5NH+] 0 [OH-] 0 - x +x +x 0.20 - x x x [C5H5N] 0.20 43 A Problem To Consider – Note that Cb Kb 0.20 1.4 10 9 1.4 10 8 which is much greater than 100 (also factor of 8 difference between the powers of 10 of Kb [2.0 x 10-1] and concentration of base [1.4 x 10-9]), so we may use the simplifying assumption that (0.20-x) (0.20). 44 A Problem To Consider C5 H 5 NH (aq ) OH (aq ) C5 H 5 N(aq ) H 2O(l ) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [C5H5NH+] 0 [OH-] 0 - x +x +x 0.20 - x x x [C5H5N] 0.20 45 A Problem To Consider – Solving for pOH 5 pOH log[OH ] log(1.7 10 ) 4.77 – Since pH + pOH = 14.00 pH 14.00 pOH 14.00 4.77 9.23 Does answer make sense? Yes, base pH or combine equations into one: HW 32 code: base 46 33.3 Acid-Base Properties of Ions and Salts • One of the successes of the BrønstedLowry concept of acids and bases was in pointing out that some ions can act as acids or bases. – Consider a solution of sodium cyanide, NaCN. NaCN(s ) Na (aq ) CN (aq ) H 2O Cations: neutral or acidic Anion: neutral or basic – A 0.1 M solution has a pH of 11.1 and is therefore fairly basic. 47 Acid-Base Properties of a Salt Solution – Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH- making it a basic solution CN (aq ) H 2O(l ) HCN(aq ) OH (aq ) – From the Brønsted-Lowry point of view, the CN- ion acts as a base, because it accepts a proton from H2O. 48 Acid-Base Properties of a Salt Solution – You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic. CN (aq ) H 2O(l ) HCN(aq ) OH (aq ) NaCN is a basic salt. – The reaction of the CN- ion with water is referred to as the hydrolysis of CN-. 49 Acid-Base Properties of a Salt Solution • The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. – The CN- ion hydrolyzes to give the conjugate acid and hydroxide. CN (aq ) H 2O(l ) HCN(aq ) OH (aq ) 50 Acid-Base Properties of a Salt Solution – The hydrolysis reaction for CN- has the form of a base ionization so you write the Kb expression for it. CN (aq ) H 2O(l ) HCN(aq ) OH (aq ) [ HCN ][OH ] K b [CN ] Note: need Kb of CN- but probably given Ka HCN (ionization constant tables contain molecular species), will need to convert. 51 Acid-Base Properties of a Salt Solution • An example of an cation that undergoes hydrolysis is the ammonium ion NH 4Cl NH 4 (aq) Cl (aq) – The NH4+ ion hydrolyzes to the conjugate base (NH3) and hydronium ion. NH 4 (aq ) H 2O(l ) NH 3 (aq ) H 3O (aq ) [ NH 3 ][H 3O ] Ka [ NH 4 ] – A 0.1 M solution has a pH of 5.1 and is therefore acidic. 52 33.3.1 Salts: Acidic, Basic, or Neutral • How can you predict whether a particular salt will be acidic, basic, or neutral and undergoes hydrolysis? – The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs. – Consequently, the anions of weak acids are good proton acceptors. – Anions of weak acids are basic. 53 Predicting Whether a Salt is Acidic, Basic, or Neutral – On the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze. Therefore, conjugate base of strong acids have do not undergo hydrolysis - neutral. – For example, the Cl- ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water. If did, strong acid breaks up 100% and hydroxide produced cancels with hydronium of strong acid and make neutral solution HCl (aq ) H 2O(l ) Cl H 3O Cl (aq ) H 2O(l ) HCl OH in essence, Cl H 3O Cl (aq ) H 2O(l ) no reaction 54 Predicting Whether a Salt is Acidic, Basic, or Neutral – Small highly charged (attract O more weakens O-H bond) cations have an attraction for the O on water and cause the production of H+ while large low charged cations do not. Low charged large ions are group I and II metals (cations of strong bases) and do not undergo hydrolysis - neutral. – For example, In summary, Na (aq ) H 2O(l ) no reaction If cation is that of the strong base (Na+, K+, Li+, Sr2+, Ba2+, Ca2+), it will have neutral hydrolysis; otherwise, cation will have acidic hydrolysis If anion of monoprotic strong acid (NO3-, Cl-, Br-, I-, ClO4-), it will have neutral hydrolysis; otherwise, anion will have basic hydrolysis 55 • Salts may be acidic/basic/neutral and are composed of cations (positive ions) and anions (negative ions) cation anion Acidic or neutral Cations of strong bases are neutral; otherwise, cation contributes acidity basic or neutral Anions of monoprotic strong acids are neutral; otherwise, anion contributes basicity Na+, K+, Li+, Ca2+, Sr2+, Ba2+ add neutral; rest of cations add acidity to salt Cl-, Br-, I-, NO3-, ClO4- add neutral; rest of anions add basicity to salt 56 Predicting Whether a Salt is Acidic, Basic, or Neutral • To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. Depending on the two components (cation/anion) the overall salt will be acidic/neutral/basic : nn n an a nb b ab depends on which is larger Ka or Kb 57 Predicting Whether a Salt is Acidic, Basic, or Neutral NH 4Cl ( s) NH 4 (aq) Cl (aq) neutral acidic salt pH < 7 acid NaCN ( s ) Na (aq ) CN (aq ) neutral basic basic salt pH > 7 KC2H3O2(s) K+(aq) + C2H3O2-(aq) neutral basic basic salt pH > 7 58 Predicting Whether a Salt is Acidic, Basic, or Neutral acid neutral acidic • AlCl 3 • Zn(NO3)2 acid neutral acidic • KClO4 neutral neutral neutral • Na3PO4 neutral basic basic • LiF neutral basic basic HW 33 code: important • NH4F (Ka = 5.6x10-10 > Kb = 1.4x10-11) acid basic acidic • NH4ClO (Ka = 5.6x10-10 < Kb = 3.6x10-7) acid basic basic • NH4C2H3O2 (Ka = 59 5.6x10-10 = Kb = 5.6x10-10) acid basic neutral 33.3.2 pH of Salt Solution • To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion. – The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species. – Thus the Kb for CN- is related to the Ka for HCN. 60 The pH of a Salt Solution • To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-. HCN(aq ) H 2O(l ) H 3O (aq ) CN (aq ) CN (aq ) H 2O(l ) HCN(aq ) OH (aq ) H 2O(l ) H 2O(l ) Ka Kb H 3O (aq) OH (aq) Kw 61 HCN(aq ) H 2O(l ) CN (aq ) H 2O(l ) H 2O(l ) H 2 O(l ) H 3O (aq ) CN (aq ) Ka HCN(aq ) OH (aq ) Kb H 3O (aq) OH (aq) Kw – When two reactions are added, their equilibrium constants are multiplied. – Therefore, K a K b K w 1.00 x1014 @ 25o C Calculate the Ka of NH4+ given the Kb of NH3 is 1.8 x 10-5 at 25oC code: kw HW 34 62 The pH of a Salt Solution • For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases. – The only difference is first must obtain the Ka or Kb for the ion that hydrolyzes. 63 Attack of hydrolysis problem When attacking pH (hydrolysis problems) always ask yourself 1) reaction or single solute 2) acid, base, or salt 3) if acid or base, is it strong (straight forward) or weak (Ka or Kb problem) if salt, write dissociation eq and examine cation/anions involved to determine if acid or base hydrolysis is available then Ka or Kb problem. 100% strong acid single solute weak Ka strong 100% neutral base weak soluble reaction neutralization cation SB Kb salt cation otherwise Ka neutral anion insoluble Ksp Anion of monoprotic SA otherwise Kb 64 A Problem To Consider • What is the pH of a 0.10 M NaCN solution at 25 °C? The Ka for HCN is 4.9 x 10-10. acid, base, or salt? basic salt soluble or insoluble? soluble NaCN(s) Na+(aq) + CN-(aq) 0.10 M 0.10 M CN- (aq) + H2O(l) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [CN-] 0.10 -x 0.10 - x [HCN] 0 +x x HCN(aq) + OH-(aq) [OH-] 0 +x x 65 A Problem To Consider Does answer make sense? Yes, basic salt 66 example: What is the pH of 0.10 M NH4I solution? acid, base, or salt? acidic salt -5 Kb NH3 = 1.8 x 10 soluble or insoluble? soluble NH4I(s) NH4+(aq) + I-(aq) 0.10 M 0.10 M NH4+ (aq) + H2O(l) H3O+(aq) + NH3 (aq) [NH4+] [H3O+] [NH3] Initial, [ ]o 0.10 0 0 Change, D[ ] -x +x +x Equilibrium, [ ]eq 0.10 - x x x HW 35 Does answer make sense? Yes, acidic salt code: salts 67 33.4 The Common Ion Effect HC 2 H 3O 2 (aq ) H 2O(l ) Add NaC2H3O2 H 3O (aq) C2 H 3O 2 (aq) Initial: 0 pH will increase 0 • The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. pH of acetic acid is lower than the pH of acetic acid/sodium acetate because of the common ion effect. 68 The Common Ion Effect NaC 2 H 3O2 Na (aq) C2 H 3O2 (aq ) HC 2 H 3O 2 (aq ) H 2O(l ) yM H 3O (aq) C2 H 3O 2 (aq) x pH x+ y M – The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased, meaning less hydronium formed and higher pH of solution as compared to pure acid. – This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect. Another source of ion coming from an additional solute added to the solution. 69 33.4.1 Buffers • When common ion effect involves weak acid/conjugate base (or vice versa), it is referred to as a buffer. Buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. – Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Typically conj is in form of salt. The components differ by one proton. HC2H3O2 / NaC2H3O2 HCN / NaCN H2CO3 / KHCO3 NH4Cl / NH3 H3PO4 / Na2HPO4 buffer? No, differ by 2 protons HCl / NaCl buffer? No, strong acid – Thus, a buffer contains both an acid species and a base species in equilibrium which allows it to absorb any added acid or base and keep pH of solution relatively constant 70 Buffers – Consider a buffer with equal molar amounts of HA and its conjugate base A-. NaA Na (aq ) A (aq ) HA(aq) H 2O(l ) H 3O (aq) A (aq) replace consumed A- and maintain pH (slight decrease in pH due to additional hydronium) If add small amount acid – react with acid or base component? base A (aq) H 3O (aq) H 2O(l ) HA(aq) neutralization If add small amount of base – reacts with acid component , produce A-, and shifts to left replace consumed HA and maintain pH (slight increase in pH due to decrease in hydronium) 71 An aqueous solution is 0.25 M in formic acid, HCHO2 and 0.18 M in sodium formate, NaCHO2. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. neutralization or buffer? buffer NaCHO2(s) Na+(aq) + CHO2-(aq) 0.18 M 0.18 M HCHO2 (aq) + H2O(l) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [HCHO2] 0.25 -x 0.25 - x H3O+(aq) + CHO2- (aq) [H3O+] 0 +x x [CHO2-] 0.18 +x 0.18 +x buffer, similar conc can neglect x Note: 0.25 M HCHO2 pH = 2.18 0.25 M HCHO2/0.18 M NaCHO2 pH = 3.63 72 A solution is prepared to be 0.15 M HC2H3O2 and 0.20 M NaC2H3O2. What is the pH of this solution at 25oC? Ka HC2H3O2 @ 25oC = 1.8 x 10-5 neutralization or buffer? buffer NaC2H3O2 (s) Na+(aq) + C2H3O2 -(aq) 0.20 M 0.20 M HC2H3O2 (aq) + H2O(l) H3O+(aq) + C2H3O2 - (aq) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [HC2H3O2 ] 0.15 -x 0.15 - x [H3O+] 0 +x x [C2H3O2-] 0.20 +x 0.20 +x Note: 0.15 M HC2H3O2 pH = 2.78 0.15 M HC2H3O2 /0.18 M NaC2H3O2 pH = 4.87 73 Calculate the pH of 75 mL of a buffer solution (0.15 M HC2H3O2 / 0.20 M NaC2H3O2 pH = 4.87) to which 9.5 mL of 0.10 M HCl is added. Ka HC2H3O2 @ 25oC = 1.8 x 10-5 Step 1: Calculate the initial mmols of each component Step 2: Neutralization reaction NaC2H3O2(aq) + HCl(aq ) HC2H3O2(aq) + NaCl (aq) mmolo Dmmol mmolneu 15.00 -0.95 14.05 0.95 -0.95 0.00 basic salt 11.25 +0.95 12.20 0 +0.95 0.95 weak acid neutral salt Step 3: New concentrations for species affecting pH 74 Step 3: New concentrations for species affecting pH Step 4: Hydrolysis NaC2H3O2 (s) Na+(aq) + C2H3O2 -(aq) 0.166 M 0.166 M HC2H3O2 (aq) + H2O(l) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [HC2H3O2] 0.144 -x 0.144 - x H3O+(aq) + C2H3O2 - (aq) [H3O+] 0 +x x [C2H3O2-] 0.166 +x 0.166 +x 75 What was the change in pH to the buffer by additional HCl? Calculate the pH if we added the same 9.5 mL of 0.10 M HCl to 75 mL of pure H2O. HCl (aq) H+(aq) + Cl -(aq) 0.0112 M 0.0112 M HW 36 code: buffers buffer holds pH relatively constant when small amounts of acid or base or added 76 Buffers – Two important characteristics of a buffer are its buffer capacity and its pH. – Buffer capacity depends on the amount of acid and conjugate base present in the solution; typically, 1 pH unit. – The next topic illustrates how to calculate the pH of a buffer by special eq but suggest you do it just like we did to keep with thought process and not memorize a equation. 77 The Henderson-Hasselbalch Equation • How do you prepare a buffer of given pH? – A buffer must be prepared from selecting a conjugate acid-base pair in which the Ka of the acid is approximately equal to the desired H3O+ concentration. – Note: you can add extra conjugate base, higher pH, or less base, lower pH, to get to exact pH wanted. 78 The Henderson-Hasselbalch Equation – Consider a buffer of a weak acid HA and its conjugate base A-. HA(aq ) H 2O(l ) H 3O (aq ) A (aq ) – The acid ionization constant is: [H 3O ][ A ] Ka [HA ] – By rearranging, you get an equation for the H3O+ concentration. [HA ] [H 3O ] K a [A ] 79 The Henderson-Hasselbalch Equation – Taking the negative logarithm of both sides of the equation we obtain: [ HA] - log[H 3O ] log( K a ) [A ] [ HA] - log[H 3O ] log( K a ) ( log ) [A ] – The previous equation can be rewritten [A ] pH pK a log [HA ] 80 The Henderson-Hasselbalch Equation – More generally, you can write [base, A ] pH pK a log [acid , HA] Previous example: An aqueous solution is 0.25 M in formic acid, HCHO2 and 0.18 M in sodium formate, NaCHO2 , had pH 3.63. The Ka for formic acid is 1.7 x 10-4. – This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation (for buffers only). Also way to work previous buffer problem. Note pH = pKa when conc base=acid 81 • Let’s prepare a buffer with pH of 4.90 – So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid-base pair with a pKa close to the desired pH or 10-4.90 to find Ka close to 1.3 x 10-5 (desired hydronium conc. equivalent to pH desired) HC2H3O2 Ka = 1.8 x 10-5 H2CO3 Ka = 4.3 x 10-7 HNO2 Ka = 4.5 x 10-4 want Ka close to [H3O+] – The Ka for acetic acid is 1.8 x 10-5, and its pH/pKa is 4.74 (calculated using 1.8 x 10-5 as [H3O+]) if equal conc of acid/salt are used. – You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid], meaning more basic salt. 82 What if I wanted to use 0.100 M HC2H3O2 (Ka = 1.8 x 10-5), what concentration of NaC2H3O2 would I need to obtain a buffer with pH =4.90. HC2H3O2 (aq) + H2O(l) H3O+(aq) + C2H3O2 - (aq) [base, A ] pH pK a log [acid , HA] pKa = -log 1.8 x 10-5 = 4.74 [C H O ] 4.90 4.74 log 2 3 2 0.100 [C2 H 3O2 ] 4.90 4.74 0.16 log 0.100 [ C H O 0.16 2 3 2 ] 10 0.100 100.16 0.100 [C2 H 3O2 ] 1.45 0.100 0.14 M [C2 H 3O2 ] [ NaC 2 H 3O2 ] HW 37 code: mixture 83 33.5 Acid-Base Titrations or Mixtures • An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). H 2CO3 NaOH H 2O NaHCO3 HCl + NaOH NaCl + H2O titrant NaHCO 3 NaOH H 2O Na 2CO3 ___________________________ H 2CO3 2 NaOH 2 H 2O Na 2CO3 – Such curves are used to gain insight into the titration process. – You can use titration curves to choose an appropriate indicator that will show when the titration is complete. 84 Figure: Curve for the titration of a strong acid by a strong base. end pt end pt Ebbing, D. D.; Gammon, S. D. General 85 Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. • Figure shows a curve for the titration of HCl with NaOH. – Note that the pH changes slowly until the titration approaches the equivalence point. – The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added; contrary to end point (observable point). – At this equivalence point of SA and SB, the pH of the solution is 7.0 because it contains a neutral salt, NaCl, that does not hydrolyze. NaOH + HCl NaCl + H2O – To detect the end point, you need an acid-base indicator that changes color beyond equivalence point (titration error). – Phenolphthalein can be used because it changes color in the pH range 8.2-10. 86 Previous example equivalent pt was at 7 but not all neutralization reactions eq pts are at 7. During a titration or just mixing an acid and base together to make a solution, the resulting solution will contain a salt and water Acid + Base salt + water One of the chemical properties of acids and bases is that they neutralize one another; however, that doesn’t mean that the product will be neutral. The misconception is that if the acid and base are in stoich proportions that the resulting solution is neutral. This is not true. The salt formed may be a acidic, basic, or neutral salt and will dictate the pH of the solution. Common sense can help you predict the pH of a mixture. SA + SB neutral salt only true neutralization pH =7 WA + SB basic salt original acid and base neutralize but product has basic properties and basic pH SA + WB acidic salt original acid and base neutralize but product has acidic properties and acidic pH 87 When working acid-base titration problems or mixtures treat them all the same but must be able to examine the components remaining after neutralization to determine what affects pH or not (strong, weak, salts). They are all done the same. The problem will dictate where you go. We can derive titration curve by doing a calculation at different points and draw curve or measure with pH meter to measure and draw curve. 1.) Calculate mmols initially 2.) write the neutralization reaction and find mmols of components left after neutralization 3.) find new conc. of species that affect pH 4.) do hydrolysis of main species that affect pH (like earlier in chapter) 88 A Problem To Consider • Calculate the pH of a solution in which 50.00 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HCl. Step 1: Calculate the initial mmols of each component Step 2: neutralization reaction HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) mmolo 50.00 50.00 0 Dmmol -50.00 -50.00 +50.00 mmolneu 0.00 0.00 50.00 neutral salt Step 3: New concentrations for species affecting pH Sodium Chloride is a neutral salt; therefore, no further calculations are needed, pH = 7.00. Note: this only happened because we are mixing the stoichiometric amounts of a strong acid and a strong base. If we mix a weak and a strong species, the equivalence point will not be at 7 as we will see in later examples. 89 • Calculate the pH of a solution in which 49.99 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HCl. Step 1: Calculate the initial mmols of each component Step 2: neutralization reaction HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) mmolo 50.00 49.99 0 Dmmol -49.99 -49.99 +49.99 mmolneu 0.01 0.00 49.99 Strong acid neutral salt Step 3: New concentrations for species affecting pH Step 4: Hydrolysis HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq) 90 • Calculate the pH of a solution in which 50.01 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HCl. Step 1: Calculate the initial mmols of each component HW 38 code: strong Step 2: neutralization reaction HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) mmolo Dmmol mmolneu 50.00 -50.00 0.00 50.01 -50.00 0.01 Strong base 0 +50.00 50.00 neutral salt Step 3: New concentrations for species affecting pH Step 4: Hydrolysis NaOH(aq) Na+ (aq) + OH- (aq) 91 • Calculate the pH of a solution in which 20.00 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5 Step 1: Calculate the initial mmols of each component Step 2: neutralization reaction HC2H3O2 (aq) + NaOH (aq) NaC2H3O2 (aq) + mmolo Dmmol mmolneu 50.00 -20.00 30.00 Weak acid 20.00 -20.00 0.00 H2O (l) 0 +20.00 20.00 Basic salt BUFFER Step 3: New concentrations for species affecting pH 92 Step 4: Hydrolysis NaC2H3O2 (s) Na+ (aq) + C2H3O2- (aq) 0.2857 M HC2H3O2 (aq) + H2O (l) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [HC2H3O2] 0.4286 -x 0.4286 - x 0.2857 M H3O+ (aq) + C2H3O2- (aq) [H3O+] 0 +x x [C2H3O2-] 0.2857 +x 0.2857 +x 93 • Calculate the pH of a solution in which 50.00 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5 Step 1: Calculate the initial mmols of each component Step 2: neutralization reaction HC2H3O2 (aq) + NaOH (aq) NaC2H3O2 (aq) + mmolo Dmmol mmolneu 50.00 -50.00 0.00 50.00 -50.00 0.00 H2O (l) 0 +50.00 50.00 Basic salt Step 3: New concentrations for species affecting pH 94 Step 4: Hydrolysis NaC2H3O2 (s) Na+ (aq) + C2H3O2- (aq) 0.5000 M C2H3O2- (aq) + H2O (l) Initial, [ ]o Change, D[ ] Equilibrium, [ ]eq [C2H3O2-] 0.5000 -x 0.5000 - x 0.5000 M HC2H3O2(aq) + OH- (aq) [HC2H3O2] 0 +x x [OH-] 0 +x x Notice at eq pt and pH is not 7 because weak/strong 95 • Calculate the pH of a solution in which 51.00 mL of 1.00 M NaOH is added to 50.00 mL of 1.00 M HC2H3O2. Ka HC2H3O2= 1.8x10-5 Step 1: Calculate the initial mmols of each component Step 2: neutralization reaction HC2H3O2 (aq) + NaOH (aq) NaC2H3O2 (aq) + mmolo Dmmol mmolneu 50.0 -50.0 0.00 51.0 -50.0 1.0 Strong base H2O (l) 0 +50.0 50.0 Basic salt Step 3: New concentrations for species affecting pH [OH-] from basic salt is extremely small as compared to strong base ; therefore, neglect WA/conj basic salt – buffer conj acidic salt/WB - buffer SA/WA SB/WB SA/SA – add [H3O+] SB/SB - add [OH-] acid/base -neutralization rxn 96 Step 4: Hydrolysis NaOH (aq) Na+ (aq) + OH- (aq) Notice the pH - 9.22 and [OH-] - 1.7 x 10-5 M changed tremendously compared to the previous example. This indicates that neglecting the contribution of NaC2H3O2 would not affect the overall answer. HW 39 code: titration 97