# Chapter 16: Acid-Base Equilibria

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1 Chapter 16: Acid-Base Equilibria In the 1 st half of this chapter we will focus on the equilibria that exist in aqueous solutions containing: weak acids polyprotic acids weak bases salts use equilibrium tables to determine: equilibrium composition of solutions ph % ionization Ka or Kb In the 2nd half of the chapter, our focus will shift to understanding solutions in which there is some combination of acidic and basic species: buffer solutions titration experiments We will need to consider that neutralization reactions can/will occur, as well as the equilibria that exist. use multiple steps: determine the ph of solutions after neutralization reactions are complete construct and interpret titration curves Steps for solving weak acid equilibrium problems: 1. Identify all major species in solution. 2. Identify all potential H + transfer reactions that could contribute to the [H3O + ]total in the sol n. 3. By considering K values, determine the dominant source of H3O + in the solution. 4. Set up the equilibrium calculation based on equilibrium identified in step Solve! x = [H3O + ] solve for ph, % ionization, or Ka

2 Determine the ph of 0.10 M HCN (aq). For HCN, Ka = 4.9 x Identify all major species in solution. this is an aqueous solution of a weak acid, so the major species are: HCN & H2O 2. Identify all potential H + transfer reactions that could contribute to the [H3O + ]total in the sol n. there are 2 possible sources of H3O + in this sol n: HCN (aq) + H2O (aq) CN (aq) + H3O + (aq); Ka = 4.9 x H2O (l) H3O + (aq) + OH (aq); KW = 1 x Determine the ph of 0.10 M HCN (aq). For HCN, Ka = 4.9 x By considering K values, determine the dominant source of H3O + in the solution. there are 2 possible sources of H3O + in this sol n: HCN (aq) + H2O (aq) CN (aq) + H3O + (aq); Ka = 4.9 x H2O (l) H3O + (aq) + OH (aq); KW = 1 x Ka > KW, so the acid ionization of HCN will be the dominant source of H3O + in his solution. OR [H3O + ]total = [H3O + ]HCN-ionization + [H3O + ]H2O-ionization but we will assume that [H3O + ]H2O-ionization is negligibly small so [H3O + ]total [H3O + ]HCN-ionization Determine the ph of 0.10 M HCN (aq). For HCN, Ka = 4.9 x Set up the equilibrium calculation based on equilibrium identified in step 3. HCN (aq) + H2O (l) CN (aq) + H3O + (aq) initial [ ] 0.10 M [ ] x x + x equil [ ] (0.10 x)m --- x M x M [H3O Ka = + ][CN ] x ; 4.9 x = 2 [HCN] 0.10 x Determine the ph of 0.10 M HCN (aq). For HCN, Ka = 4.9 x Solve! x = [H3O + ] [H3O Ka = + ][CN ] x ; 4.9 x = 2 [HCN] 0.10 x use a simplifying approximation Because Ka is very small, the reaction does not proceed very far forward (toward products) before reaching equilibrium. We will assume that x in the denominator will be negligibly small relative to [HCN] x 0.10 OR

3 Determine the ph of 0.10 M HCN (aq). For HCN, Ka = 4.9 x solution: x 2 x x = 0.10 x 0.10 so at equilibrium: x = 7.0 x 10 6 [H3O + ] = [CN ] = 7.0 x 10 6 M [HCN] = 0.10 M ph = log (7.0 x 10 6 ) = 5.15 Determine the ph of 0.10 M HCN (aq). For HCN, Ka = 4.9 x check the validity of the approximation: if x x 100 < 5%, the approximation is valid [HA]0 for our 7.0 x 10 6 x 100 = 0.007% 0.10 the approximation is valid: 0.10 x 0.10 Determine the ph and % ionization of M CH3COOH (aq). For acetic acid, Ka = 1.8 x Percent Ionization What percentage of a weak acid originally present is in its ionized form at equilibrium? [HA]ionized percent ionization = x 100 [HA]0 percent ionization is another way that we can assess the acidity of a solution and strength of an acid greater % ionization! higher [ion] higher [ion]! higher [H3O + ] higher [H3O + ]! lower ph! more acidic solution

4 Percent Ionization and Acid Concentration for a given acid, HA, % ionization will increase as [HA] decreases dilution effect - changing concentrations of species in solution results in Q < K reaction proceed forward to re-establish equilibrium new equilibrium [H3O + ] and [A ] are higher relative to new [HA]0 % ionization is greater Some Comparisons compare 2 solutions of different concentration of acetic acid (HC2H3O2, Ka = 1.8 x 10 5 ) compare 2 solutions of different concentration of hydrocyanic acid (HCN, Ka = 4.9 x ) compare acetic acid and hydrocyanic acid solutions of the same concentration M HC2H3O M HC2H3O M HCN 0.10 M HCN [H3O + ], M 4.2 x x x x 10 6 ph % ionization 4.2% 2.7% 0.014% % Be careful when comparing solutions and making qualitative statements about them. You can compare 2 different acids at the same concentration: acid with larger Ka is the stronger acid acid solution with larger Ka will have: higher [H3O + ] lower ph higher % dissociation Determination of Ka from Experimental Data: Given [HA]0 and ph The ph of M HF (aq) is Determine the Ka for HF. You can compare the same acid at 2 different concentrations: Ka is the same, so acid strength is the same solution with higher concentration will have: higher [H3O + ] lower ph lower % dissocation

5 Determination of Ka from Experimental Data: Given [HA]0 and % Ionization A M solution of HNO2 (aq) is 3.65% dissociated at equilibrium. Determine Ka for nitrous acid, and the ph of the solution. Polyprotic Acids acid with more than one acidic proton polyprotic acids dissociate in a step-wise manner each step corresponds to the dissociation of one H+ each step has a unique Ka value consider oxalic acid, H2C2O4: 1 st dissociation step: H2C2O4 (aq) + H2O (l) HC2O4 (aq) + H3O + (aq); Ka1 = 5.9 x nd dissociation step: HC2O4 (aq) + H2O (l) C2O4 2 (aq) + H3O + (aq); Ka2 = 6.4 x 10 5 Ka1 > Ka2 this is always true for polyprotic acids the acids become weaker with each successive dissociation step H2C2O4 is a stronger acid than HC2O4 Why? Polyprotic Acids Consider a M solution of carbonic acid. Determine the ph of this solution as well as the equilibrium concentrations of: [H2CO3], [HCO3 ], [H3O + ], [CO3 2 ], and [OH ]. For H2CO3, Ka1 = 4.3 x 10 7 and Ka2 = 5.6 x st ionization equation: H2CO3 (aq) + H2O (l) HCO3 (aq) + H3O + (aq) 2nd ionization equation: HCO3 (aq) + H2O (l) CO3 2 (aq) + H3O + (aq)

6 solution: because Ka1 > Ka2 & Ka1 >> Kw, the primary source of H3O + in the solution will be the 1 st ionization step for H2CO3: H2CO3 (aq) + H2O (l) HCO3 (aq) + H3O + (aq) initial [ ] M [ ] x x + x equil [ ] (0.040 x)m --- x M x M use Ka1; solve for x: x = 1.3 x 10 4 so: [H2CO3] M [HCO3 ] = [H3O + ] = 1.3 x 10 4 M ph = 3.89 solution: to determine [CO3 2 ] we will have to consider the 2 nd ionization step: HCO3 (aq) + H2O (l) CO3 2 (aq) + H3O + (aq) initial [ ] 1.3 x 10 4 M x 10 4 M [ ] x x + x equil [ ] (1.3 x 10 4 x)m --- x M (1.3 x x)m use Ka2; solve for x: x = 5.6 x so: [CO3 2 ] = 5.6 x M [OH ]? [OH ] = KW/[H3O + ] = 7.7 x M Steps for solving weak base equilibrium problems: 1. Identify all major species in solution. 2. Identify all potential H + transfer reactions that could contribute to the [OH - ]total in the sol n. 3. By considering K values, determine the dominant source of OH - in the solution. 4. Set up the equilibrium calculation based on equilibrium identified in step Solve! x = [OH - ] solve for ph, % ionization, or Kb Calculate ph and % dissociation of 0.40 M NH3 (aq). For NH3, Kb = 1.8 x NH3 (aq) + H2O (l) NH4 + (aq) + OH (aq) initial [ ] 0.40 M [ ] x x + x equil [ ] (0.40 x)m --- x M x M

7 Some Comparisons compare 2 solutions of different concentration of ammonia (NH3, Kb = 1.8 x 10 5 ) compare 2 solutions of different concentration of pyridine (C5H5N, Kb = 1.4 x 10 9 ) compare ammonia and pyridine solutions of the same concentration 0.15 M NH M NH M C5H5N 0.80 M C5H5N [OH ], M x x 10 5 ph % ionization 1.1% 0.68% % % Be careful when comparing solutions and making qualitative statements about them. You can compare 2 different bases at the same concentration: base with larger Kb is the stronger base base solution with larger Kb will have: higher [OH ] higher ph higher % dissociation You can compare the same base at 2 different concentrations: Kb is the same, so base strength is the same solution with higher concentration will have: higher [OH ] higher ph lower % dissocation Codeine (C18H21NO3) is a naturally occurring amine. The ph of a M solution of codeine is determined to be Determine the base ionization constant for codeine, and the % ionization of the solution. Relationship Between Ka & Kb for Conjugate Acid/Base Pair consider the conjugate acid/base pair of NH3 & NH4 + so: NH3 (aq) + H2O (l) NH4 + (aq) + OH (aq) [NH4 + ][OH ] Kb = [NH3] NH4 + (aq) + H2O (l) NH3 (aq) + H3O + (aq) [NH3][H3O Ka = + ] [NH4 + ] [NH3][H3O + ] [NH4 + ][OH ] Ka x Kb = x [NH4 + ] [NH3] Ka x Kb = [H3O + ][OH ] Ka x Kb = KW

8 Salt Solutions and ph Considerations salt - an ionic compound soluble salts dissolve in water to produce solutions that may be acidic, basic, or neutral consider the cation and anion separately assess the potential acidic or basic nature of each recall the inverse nature between strengths in a conjugate acid/base pair: the stronger an acid or base, the weaker its conjugate the weaker an acid or base, the stronger its conjugate Steps for solving salt solution ph problems: 1. Identify all major species in solution. 2. Identify all potential H + transfer reactions that could contribute to the [H3O + ]total OR [OH - ]total in the sol n. 3. By considering K values, determine the dominant source of H3O + OR OH - in the solution. 4. Set up the equilibrium calculation based on equilibrium identified in step Solve! -- x = [H3O + ] OR [OH - ] -- solve for ph a closer look at a weak acid/conjugate base pair: HA & A - HA (aq) + H2O (l)! A (aq) + H3O + (aq) the stronger the acid (HA), the weaker its conjugate base (A - ): strong monoprotic acids produce conjugate bases that are not effective bases in solution - they are neutral anions neutral anions are: Cl, Br, I, NO3, ClO4 the weaker the acid (HA), the stronger its conjugate base (A - ): weak acids produce conjugate bases that can function as bases in solution - they are basic anions base ionization equilibrium for A - : A - (aq) + H2O (l)! HA (aq) + OH - (aq); Kb examples of basic anions include: F -, CO3 2-, ClO2 -, SO3 2-, PO4 3-, NO2 -, BrO4 - a closer look at a weak base/conjugate acid pair: B & BH + B (aq) + H2O (l)! BH + (aq) + OH - (aq) the stronger the base (B), the weaker its cation: strong bases produce cations that are not effective acids in solution - they are neutral cations neutral cations are: Li +, Na +, K +, Rb +, Cs +, Ca 2+, Sr 2+, Ba 2+ the weaker the base (B), the stronger its conjugate acid (BH + ): weak bases produce conjugate acids that can function as acids in solution - they are acidic cations acid ionization equilibrium for BH + : BH + (aq) + H2O (l)! B (aq) + H3O + (aq); Ka examples of acidic cations include: NH4 +, CH3NH3 +, C2H5NH3 +, C5H5NH +, C6H5NH3 +

9 recall the relationship between Ka and Kb for a conjugate acid/base pair: Ka x Kb = Kw OR pka + pkb = pkw salts that produce neutral solutions: ph = 7.00 at 25 o C neutral cation with a neutral anion examples: KNO3, NaCl, Ca(ClO4)2, SrBr2, CsI and: at 25 o C, pkw = -log(1 x ) = so: pka + pkb = ph < 7.00 at 25 o C salts that produce acidic solutions: type 1: acidic cation with neutral anion examples: NH4Cl, CH3NH3NO3, C5H5NHBr example problem: calculate the ph of 0.10 M NH4Cl (aq) NH4 + (aq) + H2O (l)! NH3 (aq) + H3O + (aq); Ka = Kw/Kb for NH3 = 5.6 x type 2: neutral cation with anion from a polyprotic acid examples: KHSO4, NaHCO3, Ca(HC2O4)2 example problem: calculate the ph of.20 M Ca(HC2O4)2 (aq) HC2O4 - (aq) + H2O (l)! C2O4 2- (aq) + H3O + (aq); Ka = Ka2 for H2C2O4 = 5.1 x 10-5 type 3: hydrated metal cation of high positive chargedensity with neutral anion recall: hydrated metal ion can behave as a Bronsted-Lowry acid: [Al(H2O)6] 3+ (aq) + H2O(l)! Al[(H2O)5(OH)] 2+ (aq) + H3O + (aq) examples: MgCl2, AlBr3, Zn(NO3)2, Cr(NO3)3 example problem: calculate the ph of M AlCl3 (aq); for [Al(H2O)6] 3+, Ka = 1.4 x 10-5

10 salts that produce basic solutions: ph > 7.00 at 25 o C neutral cation with basic anion examples: KF, CaSO4, Na2CO3, K3PO4, Na2C2O4, NaNO2 example problem: calculate the ph of 0.20 M NaNO2 (aq) NO2 - (aq) + H2O (l)! HNO2 (aq) + OH - (aq); Kb = Kw/Ka for HNO2 = 2.2 x The Common Ion Effect What happens to the ph of a weak acid (HA) or weak base (B) solution when a salt containing its conjugate is added? resulting solution contains a conjugate acid base pair calculate and compare [H3O + ] or [OH - ], ph, % ionization demonstration of LeChatelier s Principle 0.10 mol HC2H3O2 and 0.10 mol NaC2H3O2 are combined in a solution with a total volume of 1.0 L. Determine the [H3O + ], ph, and % ionization in this solution. For HC2H3O2, Ka = 1.8 x the equilibrium that controls the ph of this sol n: HC2H3O2(aq) + H2O (l) C2H3O2 (aq) + H3O + (aq) initial [ ] 0.10 M M 0 [ ] x x + x for comparison: 0.10 M HC2H3O2 (aq) 0.10 M HC2H3O2 (aq) M NaC2H3O2 (aq) [H3O + ] 1.3 x 10-3 M 1.8 x 10-5 M ph % ionization 1.3% 0.018% equil [ ] (0.10 x)m --- (0.10+x)M x M

11 for comparison: 0.15 M NH3 (aq) 0.15 M NH3 (aq) M NH4Cl (aq) [OH ] 1.6 x 10 3 M 6.0 x 10-6 M ph common ion solutions Buffer Solutions sol ns that contain a conjugate acid/base pair HA & A - or B & BH + solution resists change in ph when small amounts of strong acid (H + ) or strong base (OH - ) are added % ionization 1.1% % Consider, again, 0.10 M HC2H3O2 & 0.10 M NaC2H3O2 common ion solution; starting ph = How does the ph change after the addition of mol HCl to 1.00 L of this solution? 1 st : addition of strong acid (H + ) results in a neutralization reaction (note: Cl - is a spectator ion; this is the net ionic equation): C2H3O2 - (aq) + H + (aq) " HC2H3O2 (aq) 2 nd : after the strong acid is consumed, equilibrium is established that determines the ph of the solution: HC2H3O2 (aq)! C2H3O2 - (aq) + H + (aq) resulting solution ph = 4.66; ph = -.08 How does the ph change after the addition of mol HCl to 1.00 L of a solution composed of 0.10 M HC2H3O2 & 0.10 M NaC2H3O2 (initial ph = 4.74). 1 st : addition of strong acid (H + ) results in a neutralization reaction (note: Cl - is a spectator ion; this is the net ionic equation): the added strong acid (H + ) will react with the base (C2H3O2 - ) in the buffer solution C2H3O2 (aq) + H + (aq) HC2H3O2 (aq) before rxn 0.10 mol mol 0.10 mol mol mol mol after rxn mol mol

12 How does the ph change after the addition of mol HCl to 1.00 L of a solution composed of 0.10 M HC2H3O2 & 0.10 M NaC2H3O2 (initial ph = 4.74). 2 nd : after the strong acid is consumed, equilibrium is established that determines the ph of the solution: HC2H3O2 (aq) + H2O (l)! C2H3O2 - (aq) + H3O + (aq) initial [HC2H3O2] and [C2H3O2 ] in equilibrium problem determined by consideration of what is in solution after the neutralization reaction is complete HC2H3O2 (aq) + H2O (l) C2H3O2 (aq) + H3O + (aq) initial [ ] 0.11 M M 0 [ ] x x + x equil [ ] (0.11 x)m --- (0.090+x)M x M the result of the neutralization reaction: C2H3O2 (aq) + H + (aq) HC2H3O2 (aq) C2H3O2 is consumed [C2H3O2 ] decreases strong acid, H +, is the limiting reactant H + is completely consumed ph of solution decreases slightly solution becomes slightly more acidic HC2H3O2 is formed [HC2H3O2] increases Consider, again, 0.10 M HC2H3O2 & 0.10 M NaC2H3O2 common ion solution; ph = How does the ph change after the addition of mol NaOH to 1.00 L of this solution? 1 st : addition of strong base (OH - ) results in a neutralization reaction (note: Na + is a spectator ion; this is the net ionic equation): HC2H3O2 (aq) + OH - (aq) " C2H3O2 - (aq) + H2O (l) 2 nd : after the strong base is consumed, equilibrium is established that determines the ph of the solution: HC2H3O2 (aq)! C2H3O2 - (aq) + H + (aq) resulting solution ph = 4.82; ph = +.08 How does the ph change after the addition of mol NaOH to 1.00 L of a solution composed of 0.10 M HC2H3O2 & 0.10 M NaC2H3O2 (initial ph = 4.74). 1 st : addition of strong base (OH ) results in a neutralization reaction (note: Na + is a spectator ion; this is the net ionic equation): the added strong base (OH ) will react with the acid (HC2H3O2) in the buffer solution HC2H3O2 (aq) + OH (aq) C2H3O2 (aq) + H2O (l) before rxn 0.10 mol mol 0.10 mol mol mol mol --- after rxn mol mol ---

13 How does the ph change after the addition of mol NaOH to 1.00 L of a solution composed of 0.10 M HC2H3O2 & 0.10 M NaC2H3O2 (initial ph = 4.74). 2 nd : after the strong base is consumed, equilibrium is established that determines the ph of the solution: HC2H3O2 (aq) + H2O (l)! C2H3O2 - (aq) + H3O + (aq) initial [HC2H3O2] and [C2H3O2 ] in equilibrium problem determined by consideration of what is in solution after the neutralization reaction is complete HC2H3O2 (aq) + H2O (l) C2H3O2 (aq) + H3O + (aq) initial [ ] M M 0 [ ] x x + x equil [ ] (0.090 x)m --- (0.11+x)M x M the result of the neutralization reaction: HC2H3O2 (aq) + OH (aq) C2H3O2 (aq) + H2O (l) HC2H3O2 is consumed [HC2H3O2] decreases C2H3O2 is formed [C2H3O2 ] increases strong base, OH, is the limiting reactant OH is completely consumed ph of solution increases slightly solution becomes slightly more basic Change of ph of a Buffer Solution With H + or OH Added ph is controlled by [H + ] How Does a Buffer Work? in an HA/A buffer solution: HA (aq) H + (aq) + A (aq) [H Ka = + ][A ] [HA] OR [H + ] = Ka [HA] [A ] to keep [H + ] (and therefore ph) relatively constant, [HA]/[A ] must also remain relatively constant

14 in buffer solution with HA & A : addition of H + addition of OH neutralization: neutralization: H + + A HA OH + HA A + H2O [HA] increases slightly [HA] decreases slightly [A ] decreases slightly [A ] increases slightly [HA]/[A ] increases [HA]/[A ] decreases [H + ] increases [H + ] decreases ph decreases ph increases [HA] [H + ] = Ka [A ] Buffer Capacity and Buffer Failure buffer capacity is the amount of strong acid or strong base that can be added to a buffer solution before it fails a buffer fails when enough strong acid or strong base is added to cause a ph > 1 unit ph of buffer solution depends on the [HA]/[A ] ratio capacity of a buffer solution depends on magnitude of [HA] and [A ] Buffer Capacity compare 2 HF + NaF buffer solutions: 0.25 M HF & 0.50 M NaF [HF]/[F ] = 0.50 ph = 3.76 after add n of mol H + to 100 ml solution [HF]/[F ] = 0.56 ph = 3.71 ph = 0.05 better buffer capacity M HF & 0.10 M NaF [HF]/[F ] = 0.50 ph = 3.76 after add n of mol H + to 100 ml solution [HF]/[F ] = 0.88 ph = 3.51 ph = 0.25 Henderson-Hasselbalch Equation [HA] [H + ] = Ka [A ] take log of both sides of equation [A ph = pka + log ] [HA] OR [base] ph = pka + log [acid] buffer solutions work best when [base]/[acid] is close to 1 when [base] = [acid]: [base]/[acid] = 1 ph = pka

15 Consider the following acids and their pka values: H2PO4 pka = 7.21 HF pka = 3.14 NH4 + pka = 9.25 HC7H5O2 pka = 4.19 What would be the best acid/conjugate base to prepare a buffer solution with ph = 7.00? What would be the best acid/conjugate base to prepare a buffer solution with ph = 9.00? Since buffer solutions work best when [base]/[acid] is close to 1, the best acid to pick for a buffer is one with pka close to the desired ph. ideally: pka of acid = sol n ph + 1 What is the ph of a buffer solution prepared by mixing 100 ml of M NaF with 200 ml of M HF? For HF, pka = What concentration of NaC7H5O2 is required to prepare a buffer solution with ph = 4.10 with M HC7H5O2 (aq)? For HC7H5O2, pka = Acid-Base Titrations: A Quick Review titration is an analytical technique in which one reactant is added to another in a very controlled way stop when the reaction is just complete at this point there is no limiting reactant, and no excess reactants reactants are completely converted to products stoichiometrically correct mol ratio of reactants this point is called the: stoichiometric point equivalence point end point usually some visible indication that you are at the stoichiometric point (use of indicators) Acid-Base Titrations we will look in detail at the following titrations: for each we will: strong acid + strong base weak acid + strong base strong acid + weak base calculate the ph at points before, at, and beyond the stoichiometric point discuss characteristics of each type of titration create and interpret titration curves end with a discussion of acid-base indicators

16 basic strategy: Strong Acid + Strong Base Titrations 1. write the net ionic equation for the neutralization reaction that will occur 2. using V and M, calculate mol of each reactant present 3. set up a reaction table; identify the limiting and excess reactant 4. determine [H + ] or [OH ] after neutralization reaction is complete remember to use the total solution volume! 5. calculate ph of solution Strong Acid + Strong Base Titrations 40.0 ml of M HCl (aq) is titrated with M NaOH (aq). Determine the following: initial ph of the solution ph after the addition of 30.0 ml of NaOH ph at the stoichiometric point ph after the addition of 60.0 ml NaOH 40.0 ml of M HCl (aq) is titrated with M NaOH (aq). neutralization reaction (Na + and Cl spectator ions): H + (aq) + OH (aq) H2O (l) calculate mol H + (aq): mol H + = ( L)(0.110 mol/l) = mol H + calculate volume NaOH (aq) required to reach the stoichiometric point: mol H + x 1 mol OH x 1 L sol n = L 1 mol H mol OH or 46.3 ml 40.0 ml of M HCl (aq) is titrated with M NaOH (aq). Determine the initial ph; ph of the solution before adding any NaOH (aq). another way to say this... What is the ph of M HCl (aq)? [H + ] = M ph = log (0.110) = 0.96

17 40.0 ml of M HCl (aq) is titrated with M NaOH (aq). Determine the ph after the addition of 30.0 ml NaOH (aq). mol OH added = ( L)(0.095 mol/l) = mol OH set up a reaction table to identify limiting and excess reactant: H + (aq) + OH (aq) H2O (l) before rxn: mol mol --- change: mol mol --- after rxn: mol ml of M HCl (aq) is titrated with M NaOH (aq). Determine [excess reactant] after neutralization reaction is complete: [H + ] = mol H + after rxn total sol n volume [H + ] = mol H + ( ) L [H + ] = M ph = ml of M HCl (aq) is titrated with M NaOH (aq). Determine the ph at the stoichiometric point ml NaOH (aq) added to reach stoichiometric pt at stoichiometric point: mol OH added = mol H + present = mol set up reaction table: H + (aq) + OH (aq) H2O (l) before rxn: mol mol --- change: mol mol --- after rxn: after reaction the solution is neutral; ph = ml of M HCl (aq) is titrated with M NaOH (aq). Determine the ph after the addition of 60.0 ml NaOH (aq). mol OH added = ( L)(0.095 mol/l) = mol OH set up a reaction table to identify limiting and excess reactant: H + (aq) + OH (aq) H2O (l) before rxn: mol mol --- change: mol mol --- after rxn: mol ---

18 40.0 ml of M HCl (aq) is titrated with M NaOH (aq). Determine [excess reactant] after neutralization reaction is complete: [OH ] = mol OH total sol n volume [OH ] = mol ( ) L Some Titration Data/Titration Curves Strong Acid/Strong Base Titrations: Consider the titration of 40.0 ml of M HCl (aq) with M NaOH (aq). ml NaOH added total sol n vol. [H 3O + ] after rxn ph before M 0.96 stoichiometric M 1.07 point: M 1.15 [OH ] = M poh = 1.89; ph = M M x 10!4 M 3.46 at stoich. point: x 10!7 M 7.00 [OH ] after rxn beyond M stoichiometric M point: M M M Some Titration Data/Titration Curves Strong Acid/Strong Base Titrations: Consider the titration of 40.0 ml of M HCl (aq) with M NaOH (aq). ml NaOH added total sol n vol. [H 3 O + ] after rxn ph before M 0.96 stoichiometric M 1.07 point: M M M x 10!4 M 3.46 at stoich. point: x 10!7 M 7.00 [OH ] after rxn beyond M stoichiometric M point: M M M Titration of 40.0 ml of M HCl (aq) with M NaOH (aq) Titration Curve for Strong Acid + Strong Base ph notes: start at low (acidic) ph Titration of 40.0 ml of M HCl (aq) with M NaOH (aq) stoichiometric pt volume NaOH added, ml ph < 7 before stoichiometric point (OH limiting reactant) ph = 7 at stoichiometric point (neutral solution) ph > 7 beyond stoichiometric point (H + limiting reactant) end at high (basic) ph strategy: 1. write 8 the net ionic equation for the neutralization ph 4 2 present Weak Acid + Strong Base Titrations reaction that will occur 6 strong base is completely ionized weak acid is only partially ionized 2. using V and M, calculate mol of each reactant 0 3. set up 0 a reaction 20 table; identify 40 the limiting 60 and 80 volume NaOH added, ml excess reactant

19 Weak Acid + Strong Base Titrations 4. identify the species present when the neutralization reaction is complete, and determine which of them are key to determining the ph of the solution before the stoichiometric point, the neutralization reaction results in a HA/A buffer solution at the stoichiometric point, the base ionization of A will determine the ph beyond the stoichiometric point excess OH determines the ph Weak Acid + Strong Base Titrations 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH (aq). Determine the following: initial ph of the solution ph after the addition of 20.0 ml of NaOH ph at the stoichiometric point ph after the addition of 50.0 ml NaOH 5. using post-neutralization concentrations of species, set up the appropriate calculation and determine ph of solution 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. neutralization reaction (Na + spectator ion): HC2H3O2 (aq) + OH (aq) C2H3O2 (aq) + H2O (l) calculate mol HC2H3O2 (aq): mol HC2H3O2 = ( L)(0.125 mol/l) = mol HC2H3O2 calculate volume NaOH (aq) required to reach the stoichiometric point: mol HC2H3O2 x 1 mol OH x 1 L sol n = L 1 mol HC2H3O2.100 mol OH or 37.5 ml 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. Determine the initial ph; ph of the solution before adding any NaOH (aq). another way to say this... What is the ph of M HC2H3O2 (aq)? weak acid calculation; Ka for HC2H3O2 = 1.8 x 10 5 x = [H + ] = M ph = log (0.0015) = 2.82

20 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. Determine the ph after the addition of 20.0 ml NaOH (aq). mol OH added = ( L)(0.100 mol/l) = mol OH set up a reaction table to identify limiting and excess reactant: HC2H3O2 + OH C2H3O2 + H2O before rxn: mol mol change: mol mol mol --- after rxn: mol mol ml of M HC2H3O2 (aq) is titrated with M NaOH. Determine [HC2H3O2] and [C2H3O2 ] after neutralization reaction is complete: [HC2H3O2] = mol [C2H3O2 ] = mol ( )L ( )L [HC2H3O2] = M this is a buffer solution [C2H3O2 ] = M determine [H + ] and ph by using an equilibrium table OR the Henderson-Hasselbalch equation 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. using an equilibrium table (Ka = 1.8 x 10 5 ): HC2H3O2 C2H3O2 + H + initial [ ]:.0350 M.0400 M 0 change [ ]: x +x +x equil [ ]: (.0350 x) M ( x) M x M x = [H + ] = 1.6 x 10 5 M; ph = 4.80 using Henderson-Hasselbalch (pka = 4.74): ph = log ph = ml of M HC2H3O2 (aq) is titrated with M NaOH. Determine the ph at the stoichiometric point ml NaOH (aq) added to reach stoichiometric pt at stoichiometric point: mol OH added = mol HC2H3O2 present = mol set up reaction table: HC2H3O2 + OH C2H3O2 + H2O before rxn: mol mol change: mol mol mol --- after rxn: mol ---

21 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. after the neutralization reaction, HC2H3O2 and OH are completely consumed base ionization of C2H3O2 controls the ph of sol n: C2H3O2 (aq) + H2O (l) HC2H3O2 (aq) + OH (aq) Kb = KW (Ka for HC2H3O2) = 5.6 x determine [C2H3O2 ] after neutralization reaction: [C2H3O mol ] = = M.0675 L 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. the equilibrium table and calculation: C2H3O2 + H2O HC2H3O2 + OH initial [ ]:.0556 M [ ]: x x + x equil [ ]: (.0556 x)m --- x M x M solve for x: Kb = 5.6 x x = x x = [OH ] = 5.6 x 10 6 M poh = 5.25 ph = 8.75 for a weak acid titrated with a strong base, the ph > 7 at the stoichiometric point 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. Determine the ph after the addition of 50.0 ml NaOH (aq). mol OH added = ( L)(0.100 mol/l) = mol OH set up a reaction table to identify limiting and excess reactant: HC2H3O2 + OH C2H3O2 + H2O before rxn: mol mol change: mol mol mol --- after rxn: mol mol ml of M HC2H3O2 (aq) is titrated with M NaOH. determine [excess reactant] after neutralization reaction is complete: [OH mol OH ] = total sol n volume [OH mol ] = ( ) L [OH ] = M poh = 1.81; ph = 12.19

22 point: M M Titration Curve for Weak Acid + Strong Base 14 Titration of M HC 2 H 3 O 2 (aq) with M NaOH (aq) Weak Acid/Strong Base Titrations: Consider the titration of 30.0 ml of M HC 2 H 3 O 2 (aq) with M NaOH (aq). ml NaOH added total sol n vol. [H 3 O + ] after rxn ph before x 10!3 M 2.82 stoichiometric x 10!4 M 3.93 point: x 10!5 M x 10!5 M x 10!5 M x 10!6 M x 10!7 M 6.61 [OH ] after rxn at stoich. point: x 10!6 M 8.75 beyond M stoichiometric M point: M M Titration of M HC 2 H 3 O 2 (aq) with M NaOH (aq) ph buffer zone 4 stoichiometric pt notes: volume NaOH added, ml start at low (acidic) ph ph < 7 before stoichiometric point (OH limiting reactant) ph > 7 at stoichiometric point (basic solution) ph > 7 beyond stoichiometric point (H + limiting reactant) end at high (basic) ph ml of M HC2H3O2 (aq) is titrated with M NaOH. 10 an important point in a weak acid + strong base titration is the half-way point ph 8 half-way to the stoichiometric point vol 6 required =!(vol to reach stoich.pt.) for this titration: 4 vol required =!(37.5 ml) 2 half-way to the stoichiometric point is after volume NaOH added, ml the addition of ml NaOH 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. Determine the ph after the addition of ml NaOH (aq). mol OH added = ( L)(0.100 mol/l) = mol OH set up a reaction table to identify limiting and excess reactant: HC2H3O2 + OH C2H3O2 + H2O before rxn: mol mol change: mol mol mol --- after rxn: mol mol ---

23 30.0 ml of M HC2H3O2 (aq) is titrated with M NaOH. determine [HC2H3O2] and [C2H3O2 ] after neutralization reaction is complete: mol [HC2H3O2] = [C2H3O mol ] = L L [HC2H3O2] = M [C2H3O2 ] = M this is a buffer solution with [acid] = [base] using Henderson-Hasselbalch (pka = 4.74): ph = log ph = pka = 4.74 at the half-way point in a weak + strong titration, the ph of the solution equals the pka of the acid strategy: Strong Acid + Weak Base Titrations 1. write the net ionic equation for the neutralization reaction that will occur strong acid is completely ionized weak base is only partially ionized 2. using V and M, calculate mol of each reactant present 3. set up a reaction table; identify the limiting and excess reactant Strong Acid + Weak Base Titrations 4. identify the species present when the neutralization reaction is complete, and determine which of them are key to determining the ph of the solution before the stoichiometric point, the neutralization reaction results in a B/BH + buffer solution at the stoichiometric point, the acid ionization of BH + will determine the ph beyond the stoichiometric point excess H + determines the ph Strong Acid + Weak Base Titrations 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3 (aq). Determine the following: initial ph of the solution ph after the addition of 10.0 ml of HNO3 ph at the stoichiometric point ph after the addition of 50.0 ml HNO3 5. using post-neutralization concentrations of species, set up the appropriate calculation and determine ph of solution

24 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. neutralization reaction (NO3 spectator ion): CH3NH2 (aq) + H + (aq) CH3NH3 + (aq) calculate mol CH3NH2 (aq): mol CH3NH2 = ( L)(1.20 mol/l) = mol CH3NH2 calculate volume HNO3 (aq) required to reach the stoichiometric point: mol CH3NH2 x 1 mol H + x 1 L sol n = L 1 mol CH3NH2.750 mol H + or 32.0 ml 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. Determine the initial ph; ph of the solution before adding any CH3NH2 (aq). another way to say this... What is the ph of 1.20 M CH3NH2 (aq)? weak base calculation; Kb for CH3NH2 = 3.7 x 10 4 x = [OH ] = M poh = log (0.021) = 1.68 ph = ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. Determine the ph after the addition of 10.0 ml HNO3 (aq). mol H + added = ( L)(0.750 mol/l) = mol H + set up a reaction table to identify limiting and excess reactant: CH3NH2 + H + CH3NH3 + before rxn:.0240 mol mol 0 change: mol mol mol after rxn:.0165 mol mol 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. determine [CH3NH2] and [CH3NH3 + ] after neutralization reaction is complete: [CH3NH2] =.0165 mol [CH3NH3 + ] = mol ( )L ( )L [CH3NH2] = M this is a buffer solution [CH3NH3 + ] = M determine [OH ] and ph by using an equilibrium table OR the Henderson-Hasselbalch equation

25 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. using an equilibrium table (Kb = 3.7 x 10 4 ): CH3NH2 + H2O CH3NH3 + + OH initial [ ]:.550 M M 0 [ ]: x x + x equil [ ]: (.550 x)m --- (.250+x) M x M x = [OH ] = 8.1 x 10 4 M; poh = 3.89 ph = using Henderson-Hasselbalch (pka of CH3NH3 + = 10.57): ph = log ph = ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. Determine the ph at the stoichiometric point ml HNO3 (aq) added to reach stoichiometric pt at stoichiometric point: mol H + added = mol CH3NH2 present = mol set up reaction table: CH3NH2 + H + CH3NH3 + before rxn:.0240 mol.0240 mol 0 change:.0240 mol.0240 mol mol after rxn: mol 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. after the neutralization reaction, CH3NH2 and H + are completely consumed acid ionization of CH3NH3 + controls the ph of sol n: CH3NH3 + (aq) CH3NH2 (aq) + H + (aq) Ka = KW (Kb for CH3NH2) = 2.7 x determine [CH3NH3 + ] after neutralization reaction:.0240 mol [CH3NH3 + ] = = M.0520 L 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. the equilibrium table and calculation: CH3NH3 + CH3NH2 + H + initial [ ]: M 0 0 [ ]: x + x + x equil [ ]: (0.462 x) M x M x M Ka = 2.7 x x = x solve for x: x = [H + ] = 3.5 x 10 6 M ph = 5.46 for a weak base titrated with a strong acid, the ph < 7 at the stoichiometric point

26 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. Determine the ph after the addition of 50.0 ml HNO3 (aq). mol H + added = ( L)(0.750 mol/l) = mol H + set up a reaction table to identify limiting and excess reactant: CH3NH2 + H + CH3NH3 + before rxn:.0240 mol.0375 mol 0 change:.0240 mol.0240 mol mol after rxn: mol.0240 mol 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. determine [excess reactant] after neutralization reaction is complete: [H + mol H ] = + total sol n volume [H mol ] = ( ) L [H + ] = M ph = ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. Determine the ph after the add n of 16.0 ml HNO3 (aq). mol H + added = ( L)(0.750 mol/l) = mol H + set up a reaction table to identify limiting and excess reactant: CH3NH2 + H + CH3NH3 + before rxn:.0240 mol.0120 mol 0 change:.0120 mol.0120 mol mol after rxn:.0120 mol mol 20.0 ml of 1.20 M CH3NH2 (aq) is titrated with M HNO3. determine [CH3NH2] & [CH3NH3 + ] after neutralization reaction is complete: [CH3NH2] =.0120 mol [C2H3O2 ] =.0120 mol.0360 L.0360 L [HC2H3O2] = M [C2H3O2 ] = M this is a buffer solution with [acid] = [base] using Henderson-Hasselbalch (pka = 10.57): ph = log ph = pka = at the half-way point in a weak + strong titration, the ph of the solution equals the pka of the acid

27 M M Titration Curve for Strong Acid + Weak Base 14 Titration of 20.0 ml of 1.20 M CH 3 NH 2 (aq) with 0.75 M HNO 3 (aq) Weak Base/Strong Acid Titrations: Consider the titration of 20.0 ml of 1.20 M CH 3 NH 2 (aq) with 0.75 M HNO 3 (aq). ml HNO 3 added total sol n vol. [OH ] after rxn ph before M stoichiometric M point: x 10!4 M x 10!4 M x 10!4 M x 10!4 M x 10!5 M 9.40 [H 3 O + ] after rxn at stoich. point: x 10!6 M 5.46 beyond M 1.38 stoichiometric M 1.00 point: M M M Titration of 20.0 ml of 1.20 M CH 3 NH 2 (aq) with 0.75 M HNO 3 (aq) ph buffer zone stoichiometric pt 0 notes: volume nitric acid added, ml start at high (basic) ph ph > 7 before stoichiometric point (H + limiting reactant) ph < 7 at stoichiometric point (acidic solution) ph < 7 beyond stoichiometric point (OH limiting reactant) end at low (acidic) ph 12 Acid-Base Indicators an indicator is used to indicate the stoichiometric 10 point in a titration - typically by change in color 8 you want to choose an indicator that will change color very close to the stoichiometric point in your ph 6 titration 4 2 results in minimal experimental error ph of indicator color change should be close (+1) to ph at stoichiometric point acid-base 0 indicators tend to be large organic molecules that are weak acids: volume nitric acid added, ml HIn (aq) + H2O (l) In (aq) + H3O + (aq) acid form base form Acid-Base Indicators a few examples of indicators: phenolphthalein acid form: colorless base form: pink ph range of color change: 8 10 bromocresol green acid form: yellow base form: blue ph range of color change: methyl red acid form: orange base form: yellow ph range of color change:

28 Acid-Base Indicators the color of an indicator solution depends on the ph (and [H3O + ]) and the relative amounts of HIn and In present HIn (aq) + H2O (l) In (aq) + H3O + (aq) acid form base form [In KIn = ][H3O + ] [HIn] ; [H3O + ] = KIn [HIn] [In ] ph affects the equilibrium position, and therefore the ratio of acid form : base form of indicator in sol n 2 extremes: Acid-Base Indicators in acidic solution low ph high [H3O + ] equilibrium position far to the left [HIn] high [In ] low indicator is in its acid form and color in basic solution high ph low [H3O + ] equilibrium position far to the right [HIn] low [In ] high indicator is in its base form and color

29 Choosing an Indicator for a Titration In general, an indicator will change color at a solution ph = pkin + 1. Strong Acid + Strong Base Titration: ph at stoichiometric point = 7 You should choose an indicator with pkin value that is close to the ph of the solution at the stoichiometric point of your titration. Weak Acid + Strong Base Titration: ph at stoichiometric point > 7 Strong Acid + Weak Base Titration: ph at stoichiometric point < 7

30 Final Thoughts on Acid-Base Titrations neutralization reactions happen 1 st fast go to completion set up reaction table (with mol of species) for the neutralization reaction assess what is in solution when the neutralization reaction is complete calculate post-neutralization [ ] of important species [ ] = mol total sol n volume Final Thoughts on Acid-Base Titrations if H3O + or OH is present after the neutralization reaction, go straight to ph calculation at any point in a strong acid + strong base titration beyond the stoichiometric point in weak + strong titration if HA & A or B & BH + are present after the neutralization reaction: buffer solution use equilibrium calculation or Henderson-Hasselbalch equation before the stoichiometric point in weak + strong titration if A or BH + are present after the neutralization reaction: equilibrium calculation based on ionization: A (aq) + H2O (l) HA (aq) + OH (aq) BH + (aq) B (aq) + H + (aq) at the stoichiometric point in weak + strong titration Final Thoughts on Acid-Base Titrations ph titration curves: know the characteristic profiles for the 3 categories of titrations know the important points on the curve: stoichiometric point half-way point buffer zone calculations for titrations: volume of acid or base req d to reach stoichiometric pt initial ph of solution ph before, at, beyond & half-way to the stoichiometric pt

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