1 applications of aqueous equilibria chapter 15 ap chemistry seneca valley shs

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Applications of Aqueous Applications of Aqueous EquilibriaEquilibria

Chapter 15Chapter 15

AP ChemistryAP Chemistry

Seneca Valley SHSSeneca Valley SHS

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The Common Ion EffectThe Common Ion Effect• The solubility of a partially soluble salt is decreased

when a common ion is added.• Consider the equilibrium established when acetic

acid, HC2H3O2, is added to water.

• At equilibrium H+ and C2H3O2- are constantly moving

into and out of solution, but the concentrations of ions is constant and equal.

• If a common ion is added, e.g. C2H3O2- from

NaC2H3O2 (which is a strong electrolyte) then [C2H3O2

-] increases and the system is no longer at equilibrium.

• So, [H+] must decrease.

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The Common Ion EffectThe Common Ion Effect

When a solution contains a salt having an ion common with one in equilibrium, the position of the equilibrium is driven away from the side containing that ion.

Compare the pH values for sample problems 5 and 7 to see this!

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Buffered SolutionsBuffered SolutionsComposition and Action of Buffered SolutionsComposition and Action of Buffered Solutions• A buffer resists a change in pH when a small amount

of OH- or H+ is added.• A buffer consists of a mixture of a weak acid (HX)

and its conjugate base (X-):

• The Ka expression isHX(aq) H+(aq) + X-(aq)

.]X[

[HX]]H[

.[HX]

]X][H[

-

-

a

a

K

K

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Buffered SolutionsBuffered SolutionsComposition and Action of Buffered SolutionsComposition and Action of Buffered Solutions• When OH- is added to the buffer, the OH- reacts with

HX to produce X- and water. But, the [HX]/[X-] ratio remains more or less constant, so the pH is not significantly changed.

• When H+ is added to the buffer, X- is consumed to produce HX. Once again, the [HX]/[X-] ratio is more or less constant, so the pH does not change significantly.

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Buffered SolutionsBuffered SolutionsKey Points on Buffered SolutionsKey Points on Buffered Solutions

1. They are weak acids or bases containing a common ion.

2. After addition of strong acid or base, deal with stoichiometry first, then equilibrium.

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Buffered SolutionsBuffered SolutionsAddition of Strong Acids or Bases to BuffersAddition of Strong Acids or Bases to Buffers• We break the calculation into two parts:

stoichiometric and equilibrium.• The amount of strong acid or base added results in a

neutralization reaction:

X- + H3O+ HX + H2O

HX + OH- X- + H2O.

• By knowing how must H3O+ or OH- was added (stoichiometry) we know how much HX or X- is formed.

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Buffered SolutionsBuffered SolutionsAddition of Strong Acids or Bases to BuffersAddition of Strong Acids or Bases to Buffers

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Buffered SolutionsBuffered SolutionsAddition of Strong Acids or Bases to BuffersAddition of Strong Acids or Bases to Buffers• With the concentrations of HX and X- (note the

change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation

acid conjugatebase conjugate

logppH

[HX]]X[

logppH-

a

a

K

K

Henderson-Hasselbalch Equation: For a particular buffering system, all solutions that have the same ratio [A-]/[HA] will have the same pH.

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Buffered SolutionsBuffered SolutionsBuffer Capacity and pH – Henderson-Buffer Capacity and pH – Henderson-

Hasselbach EquationHasselbach Equation• If Ka is small (i.e., if the equilibrium concentration of

undissociated acid is close to the initial concentration), then

.[HX]

]X[logppH

.]X[

[HX]loglog]Hlog[

]X[

[HX]]H[

-

-

-

a

a

a

K

K

K

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Buffered SolutionsBuffered SolutionsBuffer Capacity and pHBuffer Capacity and pH• Buffer capacity is the amount of acid or base

neutralized by the buffer before there is a significant change in pH.

• Buffer capacity depends on the composition of the buffer.

• The greater the amounts of conjugate acid-base pair, the greater the buffer capacity.

• The pH of the buffer depends on Ka.

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Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations• Consider adding a strong base (e.g. NaOH) to a

solution of a strong acid (e.g. HCl).– Before any base is added, the pH is controlled by the strong

acid solution. Therefore, pH < 7.

– When base is added, before the equivalence point, the pH is controlled by the amount of strong acid left in excess. Therefore, pH < 7.

– At equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid originally present. Therefore, the pH is determined by the salt solution. Therefore, pH = 7.

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Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations• Problems only involve a stoichiometry calculation.• We know the pH at equivalent point is 7.00. • To detect the equivalent point, we use an indicator

that changes color somewhere near 7.00.• The equivalence point in a titration is the point at

which the acid and base are present in stoichiometric quantities.

• The end point in a titration is the observed point.• The difference between equivalence point and end

point is called the titration error.

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Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations

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Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• Consider the titration of acetic acid, HC2H3O2 and

NaOH.• Before any base is added, the solution contains only

weak acid. Therefore, pH is given by the equilibrium calculation.

• As strong base is added, the strong base consumes a stoichiometric quantity of weak acid:

HC2H3O2(aq) + NaOH(aq) C2H3O2-(aq) + H2O(l)

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Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations

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Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• There is an excess of acetic acid before the

equivalence point. • Therefore, we have a mixture of weak acid and its

conjugate base.– The pH is given by the buffer calculation.

• First the amount of C2H3O2- generated is calculated, as well as the

amount of HC2H3O2 consumed. (Stoichiometry.)

• Then the pH is calculated using equilibrium conditions. (Henderson-Hasselbalch.)

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Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• At the equivalence point, all the acetic acid has been

consumed and all the NaOH has been consumed. However, C2H3O2

- has been generated.

– Therefore, the pH is given by the C2H3O2- solution.

– This means pH > 7.• More importantly, pH 7 for a weak acid-strong base titration.

• After the equivalence point, the pH is given by the strong base in excess.

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Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• The inflection point is not as steep for a weak acid-

strong base titration.• The shape of the two curves after equivalence point is

the same because pH is determined by the strong base in excess.

• Two features of titration curves are affected by the strength of the acid:– the amount of the initial rise in pH, and

– the length of the inflection point at equivalence.

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Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations

• The weaker the acid, the smaller the equivalence point inflection.

• For very weak acids, it is impossible to detect the equivalence point.

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Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• Titration of weak bases with strong acids have similar

features to weak acid-strong base titrations.

Weak Acid-Strong Base Titrations: A SummaryWeak Acid-Strong Base Titrations: A SummaryStep 1 - A stoichiometry problem - reaction is assumed

to run to completion - then determine remaining species.Step 2 - An equilibrium problem - determine position of

weak acid equilibrium and calculate pH.

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Acid-Base TitrationsAcid-Base Titrations Titrations of Polyprotic AcidsTitrations of Polyprotic Acids• In polyprotic acids, each ionizable

proton dissociates in steps.• Therefore, in a titration there are

n equivalence points corresponding to each ionizable proton.

• In the titration of Na2CO3 with HCl there are two equivalence points:– one for the formation of HCO3

-

– one for the formation of H2CO3.

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Indicator Color ChangeIndicator Color Change• Another way determine the equivalence point of an

acid-base titration is through the use of an acid-base indicator.

• Careful selection of the indicator will ensure that the end point is close to the equivalence point.

• Most common acid-base indicators are complex molecules that are themselves weak acids (HIn). They exhibit one color when the proton is attached and a different color when the proton is absent.

• Page 749 contains a list of acid-base indicators.• Henderson-Hasselbach equation is very useful in

determining the pH at which an indicator changes color.

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Solubility EquilibriaSolubility Equilibria Solubility-Product Constant, Solubility-Product Constant, KKspsp

• Consider

• for which

• Ksp is the solubility product. (BaSO4 is ignored because it is a pure solid so its concentration is constant.)

BaSO4(s) Ba2+(aq) + SO42-(aq)

]SO][Ba[ -24

2spK

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Solubility EquilibriaSolubility Equilibria Solubility-Product Constant, Solubility-Product Constant, KKspsp

• In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.

• Solubility is the amount (grams) of substance that dissolves to form a saturated solution.

• Molar solubility is the number of moles of solute dissolving to form a liter of saturated solution.

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Solubility EquilibriaSolubility Equilibria Solubility and Solubility and KKspsp

To convert solubility to Ksp

• solubility needs to be converted into molar solubility (via molar mass);

• molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium calculation),

• Ksp is the product of equilibrium concentration of ions.

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Exercise 15.12 Page 752

Copper (I) bromide has a measured solubility of 2.0x10-4 mol/L at 25C. Calculate its Ksp value.

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Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0x10-15

mol/L at 25C.

Solubility EquilibriaSolubility Equilibria

Solubility and Solubility and KKspsp


The Ksp value for copper (II) iodate is 1.4 x 10-7 at 25º C. Calculate its solubility.

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A potassium chromate solution being added to aqueous silver nitrate, forming silver chromate.

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Factors That Affect SolubilityFactors That Affect Solubility Common-Ion EffectCommon-Ion Effect• Solubility is decreased when a common ion is added.• This is an application of Le Châtelier’s principle:

• as F- (from NaF, say) is added, the equilibrium shifts away from the increase.

• Therefore, CaF2(s) is formed and precipitation occurs.

• As NaF is added to the system, the solubility of CaF2 decreases.

CaF2(s) Ca2+(aq) + 2F-(aq)

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Factors That Affect SolubilityFactors That Affect Solubility Common-Ion EffectCommon-Ion Effect

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Factors That Affect SolubilityFactors That Affect Solubility Common-Ion EffectCommon-Ion Effect


Calculate the solubility of solid CaF2 (Ksp = 4.0x10-11) in a 0.025 M NaF solution.

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Factors That Affect SolubilityFactors That Affect Solubility Solubility and pHSolubility and pH• Again we apply Le Châtelier’s principle:

– If the F- is removed, then the equilibrium shifts towards the decrease and CaF2 dissolves.

– F- can be removed by adding a strong acid:

– As pH decreases, [H+] increases and solubility increases.

• The effect of pH on solubility is dramatic.

CaF2(s) Ca2+(aq) + 2F-(aq)

F-(aq) + H+(aq) HF(aq)

1 applications of aqueous equilibria chapter 15 ap chemistry seneca valley shs

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