Chapter 15 Acids/Bases with common ion –This is when you have a mixture like HF and NaF. –The result is that you have both the acid and it’s conjugate.

Chapter 15

• Acids/Bases with common ion– This is when you have a mixture like HF and

NaF. – The result is that you have both the acid and

it’s conjugate base so you have to keep track of both initial concentrations.

– When doing a calculation make sure you include both initial concentrations

• Calculate the pH for a aqueous solution containing 1.0 M HF and 1.0 M NaF Ka = 7.2 x 10-4

– HF F- + H+

– Chem. [init.] [equil.]– HF 1.0 1.0 – x = 1.0– F- 1.0 1.0 + x = 1.0– H+ 0 + x

– 7.2 x 10-4 = (1.0)x / 1.0 x = 7.2 x 10-4

– pH = - log (7.2 x 10-4) = 3.14

• Buffered Solutions– Resists a change in pH by reacting with any

H+ or OH- added to it.– Done by having a weak acid and its conjugate

base or a weak base and its conjugate acid in the same solution (common ion)

– Ex. HNO2 / NaNO2 (NO2-)– Adding acid reacts with the NO2- to make

HNO2, adding base reacts with the HNO2 to make NO2-

• Henderson – Hasselbalch equation– From the definition of Ka we can create this

equation by taking the - log and simplifying

– pH = pKa + log ([A-] / [HA]) A- is the base HA is the acid

– This equation is an easy way to calculate the pH if you know the equilibrium concentrations of the acid and its conjugate base

• Buffering capacity

• Calculate the pH for a solution of 0.75M lactic acid (HC3H5O3) Ka = 1.4 x 10-4 and 0.25M sodium lactate (conj. Base).– Since the Ka and the conc. are not very close

we can assume very little acid changes to base or base changes to acid

– So the equilibrium conc. are equal to the initial conc. and we can use the Hend-Hassel eq.

– pH = pKa + log ([A-] / [HA])– pH = -log(1.4 x 10-4) + log((0.25)/(0.75)) – pH = 3.38

• Titrations– Systematic mixing of an acid with a base or

vice versa to neutralize the solution.– Moles H+ = moles OH- at the equiv. point– #H+(VA)(MA) = #OH-(VB)(MB)

• pH curve– The graph of the pH for the mixing of the

acid/base combo.

– Forms a s shape with usually a fairly steep center section

• Titration Examples– We will use millimoles instead of mole since

we usually do titrations using ml for volume– M = mmol/ml so ml x M = mmol

• Strong Acid/Strong Base– Before any base is added the pH is

completely due to the strong acid concentration

– After adding the base we have to convert the indicated amount of acid and then calculate the pH

• Calculate the pH for the titration of 50 ml of 0.2M HCl with 0.1M NaOH.– At 0 ml NaOH– pH = - log(0.2) = 0.699

– At 20 ml we have to react the base with the acid

– H+ OH-– init. 10 mmol 2 mmol– Change - 2 mmol -2 mmol– Final 8 mmol 0 mmol– pH = - log (8mmol/70ml) = 0.942

– At 100 ml we have

H+ OH-– init. 10 mmol 10 mmol– Change - 10 mmol -10 mmol– Final 0 mmol 0 mmol– pH = pH of water = 7

– At 200 ml we have

H+ OH-– init. 10 mmol 20 mmol– Change - 10 mmol -10 mmol– Final 0 mmol 10 mmol

– pH = 14 – pOH – pOH = - log ( 10mmol/250ml ) = 1.4– pH = 14 – 1.4 = 12.6

• Weak Acid/Strong Base– Similar process but we end up with a weak

acid equilibrium problem until we pass the equiv. pt.

– Calculate the pH for the titration of 50.0 ml of 0.10M acetic acid with 0.10M NaOH. Ka = 1.8 x 10-5

– The equiv. pt is at 50.0 ml of NaOH so before that we have an equil. problem to solve

– At 10.0 ml of 0.1 M NaOH

– HC2H3O2 OH- C2H3O2-

– init. 5 mmol 1 mmol 0 mmol– Change - 1 mmol -1 mmol 1 mmol– Final 4 mmol 0 mmol 1 mmol

– Chem [Init.] [Equil]– HC2H3O2 4/60 4/60 – x = 4/60– C2H3O2- 1/60 1/60 + x = 1/60– H+ 0 + x

– Chem [Init.] [Equil]– HC2H3O2 4/60 4/60 – x = 4/60– C2H3O2- 1/60 1/60 + x = 1/60– H+ 0 + x

– pH = pKa + log [A-]/[HA]– pH = -log(1.8 x 10-5) + log(1/60 / 4/60)– pH = 4.14

– At 25.0 ml of 0.1M NaOH

– HC2H3O2 OH- C2H3O2-

– init. 5 mmol 2.5 mmol 0 mmol– Change - 2.5 mmol -2.5 mmol 2.5 mmol– Final 2.5 mmol 0 mmol 2.5 mmol

– Chem [Init.] [Equil]– HC2H3O2 2.5/75 2.5/75 – x = 2.5/75– C2H3O2- 2.5/75 2.5/75 + x = 2.5/75– H+ 0 + x

– Chem [Init.] [Equil]– HC2H3O2 2.5/75 2.5/75 – x = 2.5/75– C2H3O2- 2.5/75 2.5/75 + x = 2.5/75– H+ 0 + x

– pH = pKa + log [A-]/[HA]– pH = -log(1.8 x 10-5) + log(2.5/75 / 2.5/75)– pH = - log(1.8 x 10-5) + 0 = 4.74

– So pH = pKa at the half-way pt.

– At 50.0 ml of 0.1M NaOH (eq. pt.)

– HC2H3O2 OH-C2H3O2-

– init. 5 mmol 5 mmol 0 mmol– Change - 5 mmol -5 mmol 5 mmol– Final 0 mmol 0 mmol 5 mmol– since all of the acid has been converted to the

conjugate base we have to do a Kb problem– C2H3O2- + H2O HC2H3O2 + OH-

– Chem [Init.] [Equil]– C2H3O2- 5/100 5/100 – x = 5/100– HC2H3O2 0 + x– OH- 0 + x

– Kb = Kw/Ka = 1.0 x 10-14/1.8 x 10-5

– Kb = 5.6 x 10-10

– 5.6 x 10-10 = x2/(5/100) x = 5.3 x 10-6

– pOH = -log(5.3 x 10-6) = 5.28– pH = 14 – 5.28 = 8.72

– Past the eq. pt. we have the weak base as well as the excess strong base so we can focus on the excess strong base to determine pH

– 60 ml 0.1MNaOH– HC2H3O2 OH-

C2H3O2-

– init. 5 mmol 6 mmol 0 mmol

– Change - 5 mmol -5 mmol 5 mmol– Final 0 mmol 1 mmol 5 mmol

– HC2H3O2 OH-C2H3O2-

– init. 5 mmol 6 mmol 0 mmol

– Change - 5 mmol -5 mmol 5 mmol– Final 0 mmol 1 mmol 5 mmol

– So we focus on the new [OH-]– [OH-] = 1 mmol / 110 ml = 9.1 x 10-3

– pOH = - log(9.1 x 10-3) = 2.04– pH = 14 – 2.04 = 11.96

• Weak Acid / Strong Base overview– Acid only Ka problem– Before eq. pt. Ka problem– ½ way pt. pH = pKa

– Eq. pt. Kb problem– Past eq. pt. strong base only

• Calculating Ka

– We can use a titration set of data and the pH to determine an unknown Ka value.

– Just determine the final concentrations of the weak acid and conjugate base then insert into the equation

– pH = pKa + log [A-]/[HA]

• Weak Base/Strong Acid– Before any acid Kb problem– Before eq. pt. Kb problem– ½ way pt. pOH = pKb– At eq. pt. Ka problem– Past eq. pt. focus on strong acid conc.

• Acid/Base Indicators– Weak acids/bases that change color at

specific pH points

– Usually use the ratio [I-]/[HI] = 1/10 to determine when an indicator changes color

– Phenolphthalein is most common indicator and is clear in acid solution and turns pink in base solution

• Solubility Equilibria– The equilibrium for a partially soluble ionic

substance in water– Solubility is the amount of solid that

dissociates– PbCl2 (s) Pb2+ + 2Cl-

• Ksp– Equilibrium constant for solubility– Ksp = [Pb2+][Cl-]2

• Calculate the Ksp for Bi2S3 that has a solubility of 1.0 x 10-15 M.– Bi2S3 (S) 2Bi3+ + 3S2-

– Ksp = [Bi3+]2[S2-]3

– 1.0 x 10-15 M Bi2S3 = 2.0 x 10-15 M Bi3+

– 1.0 x 10-15 M Bi2S3 = 3.0 x 10-15 M S2-

– Ksp = (2.0 x 10-15 )2(3.0 x 10-15 )3

– Ksp = 1.1 x 10-73

• Calculate the solubility for Cu(IO3)2 with a Ksp = 1.4 x 10-7

– Cu(IO3)2 Cu2+ + 2IO3-

– Ksp = [Cu2+][IO3-]2

– 1.4 x 10-7 = (x)(2x)2 = 4x3

– x = 3.3 x 10-3 M

• Common Ion Effect– Presence of one of the ions that are formed

by the ionic substance

– AgCl with Cl- (from NaCl) this would impact the solubility of the AgCl since some Cl- is already present

• Calculate the solubility of CaF2 in a 0.025 M NaF solution if the Ksp = 4.0 x 10-11

– Ksp = [Ca2+][F-]2

– Chem [init.] [equil.]– Ca2+ 0 + x– F- 0.025 0.025 + 2x = 0.025

– 4.0 x 10-11 = (x)(0.025)2 – x = 6.4 x 10-8 M– Solubility is 6.4 x 10-8 M

• pH effect– A change in pH can effect some salts

– Sr(OH)2 Sr2+ + 2OH-

– Adding OH- would shift the reaction left– Adding H+ would shift the reaction right

because it removes some of the OH-

• Precipitation and Qualitative Analysis– We can determine if a precipitate forms by

comparing a set of concentrations to the Ksp by calculating Q

– Remember Q is the same formula as Ksp

– Will Ce(IO3)3, Ksp = 1.9 x 10-10, precipitate from a solution made from 750 ml of 0.004 M Ce3+ and 300 ml of 0.02 M IO3-?

– Q = [Ce3+]o[IO3-]3o

– [Ce3+]o = (750ml)(0.004M)/1050ml

= 2.86 x 10-3M– [IO3-]o = (300ml)(0.02M)/1050ml

= 5.71 x 10-3M

Q = [Ce3+]o[IO3-]3o = (2.86 x 10-3)(5.71 x 10-3)3

Q = 5.32 x 10-10 Q>Ksp so it precipitates

• Calculate the concentrations of Mg2+ and F- in a mixture of 150 ml of 0.01 M Mg2+ and 250 ml of 0.1 M F-.

• Ksp for MgF2 is 6.4 x 10-9

– [Mg2+]o = (150ml)(0.01M)/400ml = 0.00375 M– [F-]o = (250ml)(0.1M)/400ml = 0.0625 M

– Q = (0.00375)(0.0625)2 = 1.46 x 10-5

– Since Q > Ksp a precipitate forms and we have to do an equilibrium problem

– Chem. init. Change– Mg2+ (150ml)0.01M

1.5 mmol - 1.5

2F- (250ml)0.1M 25 – 2(1.5)

25 mmol 22 mmol

We have excess F-

[F-] = 22 mmol/400ml = 0.055 M

– Chem. [init.] [equil.]– Mg2+ 0 + x– F- 0.055 0.055 + 2x = 0.055

– Ksp = 6.4 x 10-9 = [Mg2+][F-]2

– 6.4 x 10-9 = (x)(0.055)2

– x = 2.1 x 10-6 M

– [Mg2+] = 2.1 x 10-6 M– [F-] = 0.055 M

• We can use Ksp values to determine which precipitate forms first since the smaller the Ksp the less ions that will exist in solution.

– By adding NaI to a mixture of Pb2+ and Cu+ which will precipitate first.

– PbI2 Ksp = 1.4 x 10-8

– CuI Ksp = 5.3 x 10-12

– Since CuI is much smaller we can predict that it precipitates first