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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