Chem 2
Ch 19: Ionic Equilibria In Aqueous Solutions
BUFFERED SOLUTIONS: A solution that resists a change in its pH when either OH- or H+
ions are added.
 Ex: Blood, absorbs acid and bases without change in pH.
 The components of a buffer are the conjugate acid-base pair of a weak acid.
 WA and its salt-HF and NaF
 WB and its salt- NH3 and NH4Cl.
Common ion effect: The shift in equilibrium position that occurs because of the addition
of an ion already involved in equilibrium reaction is called common ion effect.
Consider a weak acid HF and its salt NaF.
NaF (s) _______ Na+ (aq)+ F- (aq) , Major species: Na+ F- HF H2O
HF (aq)↔ H+ (aq) + F- (aq)
F- is the common ion.
F- from the added NaF moves the equ position to the left according to LC principle.
If we dissolve HF in a NaF solution, the F-ion and H+ ion enter the solution. The F- ion
already present combines with the H+, which lowers the H+.
Buffered Solution Characteristics
 Buffers contain relatively large amounts of weak acid and corresponding base.
 Added H+ is consumed by A Added OH is consumed by HA
 The pH is determined by the ratio of the concentrations of the weak acid and its
conjugate base.
P.1. A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and 0.50 M sodium
acetate.
a) Calculate pH of this solution.
b) Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L
of buffered solution described in the above problem.Compare this pH change with the
one that occurs when 0.010 mol solid NaOH is added to 1.0 L of water.
c) Calculate the change in pH that occurs when 0.010 mol solid HCl is added to 1.0 L of
buffered solution.
Henderson-Hasselbalch Equation
Useful for calculating pH when the [A]/[HA] ratios are known.
pH=pKa + log ( [base]/[acid])
P. 2. Calculate the pH of a solution 0.75 M lactic acid (Ka= 1.4 x10-4 ) and 0.25 M
sodium lactate.
P.3. A buffer solution contains 0.25 M NH3 (Kb= 1.8 x 10-5) and 0.40 M NH4Cl.
Calculate pH of this solution.
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Buffering Capacity: It represents the amount of H+ or OH the buffer can absorb without
a significant change in pH.
 More concentrated the components of a buffer , the greater the buffer capacity.
 pH of a buffered solution is determined by the ratio [A-]/[HA] .
 If an acid or a base is added , the concentration ratio changes less when the buffer
component concentrations are similar than they are different.
 Buffer has highest capacity when component concentrations are equal.
 pH=pKa highest buffer capacity.
 The Buffer Range is the pH range over which the buffer works effectively.
 The further the buffering component concentration ratio is from 1 less effective is
the buffering action.
Preparing a Buffer:
1. Choose the conjugate acid-base pair. Ratio of concentrations should be
close to1. pH=pKa
2. Calculate ratio of buffer component concentrations. Use the Henderson
Hasselbalch equation.
3. Determine the buffer concentration. Higher the buffer concentrations
greater the buffer capacity.
4. Mix the solutions and adjust the pH by adding strong acid or strong base
with the help of a pH probe.
P.4. A chemist needs a solution buffered at pH= 4.30 and can choose from the following
acids and their sodium salts:
1. chloroacetic acid Ka= 1.35 x 10-3
2. propanoic acid Ka= 1.3 x 10-5
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3. benzoic acid Ka= 6.4 x 10
4. hypochlorous acid. Ka= 3.5 x 10-8
Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30, which
system will work the best?
ACID BASE TITRATION CURVES
A plot of pH of the solution being analyzed as a function of the amount of titrant added.
Acid-Base Indicators:
 It marks the end point of a titration by changing color.
Ex: phenolphthalein is colorless in its HIn form and pink in In- form, or basic
form.
 Indicator changes color over a range of about 2 pH units.
STRONG ACID- STRONG BASE TITRATION CURVES:
 pH is low initially, as base is added the pH increases slowly.
 The pH rises steeply when the moles of OH- nearly equals the moles of H3O+
 The additional drop of base neutralizes the tiny excess acid and introduces a tiny
excess of base.
 Then pH increases smoothly as more base is added.
Equivalence (stoichiometric) point: Enough titrant has been added to react exactly
with the solution being analyzed.
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Equivalence point is nearly the vertical portion of the curve.
The point at which the number of moles of added OH- =the number of moles of
H3O+
For a SA vs SB titration pH =7(neutral ions)
End point: It occurs when indicator changes color.The indicator chosen should be
close to the equivalence point.
Calculating pH:
1. Initial pH is the pH of the acid.
2. Before the equ. Point find initial moles present, moles reacted, change in volume
and then the pH.
3. At the equ point, pH =7
4. After the equ point find the excess of OH- present and then calculate pH.
P.5. 50.0 mL of 0.200 m HNO3 is titrated with 0.100M NaOH. Calculate the pH of the
solution at selected points during the course of titration, where 0 mL, 20.0 mL, 100.0 mL,
150.0 mL of 0.100 M NaOH has been added.
WEAK ACID-STRONG BASE TITRATIONS:
 Initial pH is high as weak acid dissociates slightly.
 Buffer Region: A gradual rising portion of the curve appearing before the
equ.point.
 At midpoint of the buffer region pH=pKa.
 pH at equ point is > 7.
 Beyond equ point pH increases slowly as excess OH- is added.
Calculating pH;
1. When only HA present, Use ICE table calculate H+ and pH like a weak acid type
calculation.
2. As we add base, find change in moles and solve for H+
3. At equ. Point pH >7, Find Kb from Ka and then calculate OH- and pH.
4. Beyond the equ. point , calculate excess of OH- present and calculate pH.
P.6. 50.0 mL of 0.1 M acetic acid (Ka=1.8 x 10-5) with 0.10 M NaOH. Calculate the pH
at various points representing volumes of 0mL, 10.0 mL , 25.0 mL, 50.0 mL , 60.0 mL
of added NaOH.
WEAK BASE –STRONG ACID TITRATION CURVES:
 Same shape curve as WA vs SB , but inverted.
 Initially pH above 7 as it is a weak base.
 pH decreases in buffer region.
 At midpoint pH=pKa
 After buffer region curve drops vertically to equ.point .
 pH at equ.point is below 7
 Beyond equ point , pH decreases slowly as more acid is added.
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P.7. 100.0 mL of 0.050 M NH3 is titrated with 0.10 M HCl. Calculate the pH at various
points: 1) before adding HCl 2) Before equ.point 3) At equ point (setup)
P.8 If 2.00 mmol of solid acid in 100.0 mL water is titrated with 0.0500 M NaOH. After
20.0 mL NaOH has been added the pH is 6.00. What is the Ka value for acid?
TITRATION CURVES OF POLYPROTIC ACIDS:
 Number of curves =Number of H+ ions.
EQUILIBRIA OF SLIGHTLY SOLUBLE IONIC COMPOUNDS:
The Ionic product Qsp and The Solubility product Ksp:
 An ionic substance dissociates completely in water to form hydrate cations and
anions.
 Both the forward and reverse reactions take place.
 A stage is reached when no more solid dissolves. This is called as dynamic
equilibrium.
 CaF2 (s) ↔Ca2+ (aq) + 2F- (aq)
 Ksp= [Ca2+ ] [F-]2
 Ksp is called as the solubility product constant or solubility product for the
equilibrium expression.
 For the hypothetical compound, MpXq
 At equilibrium Qsp = [Mn+]p [Xz-]q= Ksp
 Higher the Ksp greater the solubility for formulas containing same total number
of ions.
 If salts being compared produce different number of ions.
Compare their solubilities, which is opposite of Ksp
P.9. Write the ion-product expression for each of the following:
1) magnesium carbonate 2) Iron(II) hydroxide 3) Calcium phosphate 4) Silver sulfide
P.10. CuBr has a measured solubility of 2.0 x 10-4 mol/L at 25 C. Calculate Ksp.
P.11. Calculate Ksp value for Bi2S3 which has a solubility of 1.0 x 10-15 mol/L at 25 C.
P.12. Calculate the solubility of copper (II) iodate at 25 C, if Ksp = 1.4 x 10-7
Common ion Effect:
 Presence of a common ion decreases solubility of a slightly soluble ionic
compound.
P.13. Calculate the solubility of CaF2 (Ksp=4.0 x 10-11) in a 0.025 M NaF solution.
pH and solubility:
If the compound contains the anion of a weak acid, addition of H3O+ increases its
solubility.
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P.14. Write balanced equations to explain whether addition of H3O+ from a strong acid
affects the solubility of these ionic compounds:
1) Lead (II) bromide 2) copper (II) hydroxide 3) iron (II) sulfide.
Predicting formation of precipitate:
If Qsp > Ksp ppt occurs
If Qsp < Ksp solution is unsaturated, no ppt occurs.
If Qsp = Ksp solution is saturated and no change occurs.
P.15. A common laboratory method for preparing a precipitate is to mix solutions of the
component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed
with 0.200L of 0.060M NaF?
EQUILIBRIA INVOLVING COMPLEX IONS:
 Complex ion: A central metal ion covalently bonded to two or more anions, or
molecules called ligands.
 Ligand: Lewis base that contains a lone pair of electron that can be donated to an
empty orbital on the metal ion to form a covalent bond.
Ex: H2O, NH3, CN-,Cl Coordination number: Number of ligands attached to a metal ion.ex; 6,4,2
 Formation constants/Stability Constants : Metal ions add ligands one at a time in
steps.
 A ligand increases the solubility of a slightly soluble ionic compound if it forms a
complex ion with the cation.
P.16. An industrial chemist converts Zn(H2O)42+ to the more stable Zn(NH3)42+ by
mixing 50.0L of 0.0020M Zn (H2O)42+ and 25.0L of 0.15M NH3. What is the final
[Zn (H2O)4 2+]? Kf of Zn(NH3)42+ is 7.8x108.
P.17.In black-and-white film developing, excess AgBr is removed from the film negative
by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of
the complex ion Ag(S2O3)2 3 .Calculate the solubility of AgBr in (a) H2O; (b) 1.0M
hypo. Kf of Ag(S2O3)2 3-is 4.7x1013 and Ksp AgBr is 5.0x10-13
P.18.A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that
would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp
of Cu(OH)2 is 2.2x10-20
Qualitative analysis:
Separating and identifying ions in complex mixture.
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