III. Titrations

III. TitrationsIII. Titrations• In an acid-base titration, a basic (or

acidic) solution of unknown [ ] is reacted with an acidic (or basic) solution of known [ ].

• An indicator is a substance used to visualize the endpoint of the titration.

• At the equivalence point, the number of moles of acid equals the number of moles of base.

III. Visual of a TitrationIII. Visual of a Titration

III. Titration/pH CurvesIII. Titration/pH Curves

• A titration or pH curve is a plot of how the pH changes as the titrant is added.

• It is possible to calculate the pH at any point during a titration.

• Multiple pH’s can be calculated, and the results plotted to create the theoretical titration curve.

III. Find the Equivalence Point!III. Find the Equivalence Point!• The keys to these types of problems

are writing the titration equation and finding the equivalence point of the titration.

• The calculation then depends on what region of the titration curve you are in:

1) Before titration begins2) Pre-equivalence3) Equivalence point4) Post-equivalence

III. Illustrative ProblemIII. Illustrative Problem

• Sketch the pH curve for the titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

1) Write the titration equation. HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

2) Calculate the equivalence point. What volume of NaOH is needed to completely react with HCl?

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

3) Calculate initial pH before titration. Since HCl is strong, 0.100 M HCl has

[H3O+] = 0.100 M, and pH = 1.000.4) Calculate the pH of some points in the

pre-equivalence region. As NaOH is added, the neutralization

reaction OH-(aq) + H3O+

(aq) H2O(l) takes place.

We calculate the pH after addition of 5.00 mL of NaOH.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

• To calculate the pH after 5.00 mL, we need to calculate initial moles of acid and the number of moles of base.

• We put these moles into a reaction chart.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution H3O+ + OH- 2H2O

Initial 0.00250 mol 0.000500 mol 0 mol

Change -0.000500 mol -0.000500 mol +0.001000 mol

Final 0.00200 mol 0 mol 0.001000 mol

The H3O+ leftover is in a larger volume so we calculate its concentration.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution• We do the same thing for some other

points: 10.0 mL, 15.0 mL, and 20.0 mL.• Results summarized below.

Volume (ml) Mol H3O+ [H3O+] (M) pH

5.00 0.00200 0.06667 1.176

10.0 0.00150 0.04286 1.368

15.0 0.00100 0.02500 1.602

20.0 0.00050 0.01111 1.954

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

5) Calculate the pH at the equivalence point.

For a strong-strong titration, pH always equals 7.00 at the equivalence point!

6) Calculate the pH of some points in the post-equivalence region.

In this region, the pH depends on the excess OH- added.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

• To find the excess added, calculate how many mL past the equivalence point have been added, convert to moles, and divide by total volume. For 30.0 mL:

pH can then be found from pOH.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution• Again, calculate for additional points like

35.0, 40.0, and 50.0 mL. Results summarize below.

Volume (mL) [OH-] (M) pOH pH30.0 0.009091 2.0414 11.959

35.0 0.01667 1.7781 12.222

40.0 0.02307 1.6370 12.363

50.0 0.03333 1.4771 12.523

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

• Now we plot the data points and sketch the pH titration curve!

III. Sample ProblemIII. Sample Problem

• A 0.0500 L sample of 0.0200 M KOH is being titrated with 0.0400 M HI. What is the pH after 10.0 mL, 25.0 mL, and 30.0 mL of the titrant have been added?

III. Weak Acid/Base TitrationsIII. Weak Acid/Base Titrations• The situation becomes a little more

complicated when a weak acid/base is titrated with a strong base/acid.

• Again, the keys are to identify the titration reaction and the equivalence point.

• The method of calculating the pH will then depend on the region of the titration curve.

III. Four Different RegionsIII. Four Different Regions• For a weak acid titrated with a strong base,

there are 4 regions as well:1) Before titration: only HA in solution, so it’s a weak

acid problem!2) Pre-equivalence: a mixture of HA and A-, so it’s a

buffer!3) Equivalence point: only A- in solution, so it’s a

weak base problem!4) Post-equivalence: adding excess OH-, so it’s a

dilution problem!• For a weak base titrated with a strong acid,

everything is just rewritten w/ conjugates!

III. Illustrative ProblemIII. Illustrative Problem

• A 50.00 mL sample of 0.02000 M CH3COOH is being titrated with 0.1000 M NaOH. Calculate the pH before the titration begins, after 3.00 mL of the titrant have been added, at the equivalence point, and after 10.20 mL of the titrant have been added. Note that Ka = 1.75 x 10-5 for acetic acid.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

1) First, we need to write the titration eqn. CH3COOH(aq) + OH-

(aq) CH3COO-(aq) + H2O(l)

2) Next, we calculate the equivalence point.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

3) Before titration begins, it’s just a weak acid problem.

CH3COOH + H2O H3O+ + CH3COO-

Initial 0.02000 M --- 0 0Change -x --- +x +xEquil. 0.02000-x --- x x

Solving this (with simplification), we get [H3O+] = 5.916 x 10-4, so pH = 3.228.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

4) After 3.00 mL of 0.1000 M NaOH have been added, we will have a mixture of CH3COOH and CH3COO- in solution.

Since it’s a buffer, we can use the Henderson-Hasselbalch eqn.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution• In the H-H equation, we need a ratio of

CH3COO- to CH3COOH.• After adding 3.00 mL, we are 3.00/10.00

to equivalence. We can use a relative concentration chart.

CH3COOH + OH- CH3COO- + H2O

Relative Initial 1 0 0 ---Change -3.00/10.00 +3.00/10.00 +3.00/10.00 ---Relative Final 7.00/10.00 0 3.00/10.00 ---

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

• Now we just plug the relative final row into the H-H equation.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution5) At the equivalence point, all CH3COOH has

been converted to CH3COO-. Initial moles of CH3COOH = moles of CH3COO- at

the equivalence point, but the volume has increased.

Must calculate [CH3COO-] at equiv. pt.

III. Illustrative Problem SolutionIII. Illustrative Problem Solution• Now we solve a weak base problem.

CH3COO- + H2O OH- + CH3COOH

Initial 0.016667 M --- 0 0

Change -x --- +x +xEquil. 0.016667-x --- x x

Using Kb = 5.714 x 10-10 and the simplification, x = 3.089 x 10-6. Thus, pOH = 5.5102 and pH = 8.490.Note that pH does not equal 7.00!!

III. Illustrative Problem SolutionIII. Illustrative Problem Solution

6) At 10.20 mL of added titrant, we are 0.20 mL past equivalence, and the pH depends only on excess OH-. Thus:

Of course, this means that pOH = 3.479 andpH = 10.52.

III. Sample ProblemIII. Sample Problem

• A 25.00 mL sample of 0.08364 M pyridine is being titrated with 0.1067 M HCl. What’s the pH after 4.63 mL of the HCl has been added? Note that pyridine has a Kb of 1.69 x 10-9.

III. Sample Weak/Strong CurvesIII. Sample Weak/Strong Curves• Important aspects about pH curves for

weak/strong titrations: At equivalence, pH does not equal 7.00. At ½ equivalence, pH = pKa.

III. Polyprotic Acid TitrationIII. Polyprotic Acid Titration

• If the Ka’s are different enough, you will see multiple equivalence points.

• Since protons come off one at a time, 1st equiv. pt. refers to the 1st proton, 2nd to the 2nd, etc.

III. Detecting the Equiv. Pt.III. Detecting the Equiv. Pt.

• During a titration, the equivalence point can be detected with a pH meter or an indicator.

• The point where the indicator changes color is called the endpoint.

• An indicator is itself a weak acid that has a different color than its conjugate base.

III. PhenolphthaleinIII. Phenolphthalein

III. IndicatorsIII. Indicators

• The indicator has its own equilibrium: HIn(aq) + H2O(l) In-

(aq) + H3O+(aq)

• The color of an indicator depends on the relative [ ]’s of its protonated and deprotonated forms. If pH > pKa of HIn, color will be In-. If pH = pKa of HIn, color will be in between. If pH < pKa of HIn, color will be HIn.

III. Selecting an IndicatorIII. Selecting an Indicator

IV. SolubilityIV. Solubility

• In 1st semester G-chem, you memorized solubility rules and regarded compounds as either soluble or insoluble.

• Reality is not as clear cut – there are degrees of solubility.

• We examine solubility again from an equilibrium point of view.

IV. Solubility EquilibriumIV. Solubility Equilibrium

• If we apply the equilibrium concept to the dissolution of CaF2(s), we get: CaF2(s) Ca2+

(aq) + 2F-(aq)

• The equilibrium expression is then: Ksp = [Ca2+][F-]2

• Ksp is the solubility product constant, and just like any other K, it tells you how far the reaction goes towards products.

IV. Some KIV. Some Kspsp Values Values

IV. Calculating SolubilityIV. Calculating Solubility

• Recall that solubility is defined as the amount of a compound that dissolves in a certain amount of liquid (g/100 g water is common).

• The molar solubility is obviously the number of moles of a compound that dissolves in a liter of liquid.

• Molar solubilities can easily be calculated using Ksp values.

IV. KIV. Kspsp Equilibrium Problems Equilibrium Problems

• Calculating molar solubility is essentially just another type of equilibrium problem.

• You still set up an equilibrium chart and solve for an unknown. Pay attention to stoichiometry!

Fe(OH)3(s) Fe3+(aq) + 3OH-

(aq)

Initial --- 0 0Change --- S 3S

Equil. --- S 3S

IV. Sample ProblemIV. Sample Problem

• Which is more soluble: calcium carbonate (Ksp = 4.96 x 10-9) or magnesium fluoride (Ksp = 5.16 x 10-11)?

IV. The Common Ion EffectIV. The Common Ion Effect

• The solubility of Fe(OH)2 is lower when the pH is high. Why? Fe(OH)2(s) Fe2+

(aq) + 2OH-(aq)

Le Châtelier’s Principle!• common ion effect: the solubility of an

ionic compound is lowered in a solution containing a common ion than in pure water.

IV. Sample ProblemIV. Sample Problem

• Calculate the molar solubility of lead(II) chloride (Ksp = 1.2 x 10-5) in pure water and in a solution of 0.060 M NaCl.

IV. pH and SolubilityIV. pH and Solubility

• As seen with Fe(OH)2, pH can have an influence on solubility.

• In acidic solutions, need to consider if H3O+ will react with cation or anion.

• In basic solutions, need to consider if OH- will react with cation or anion.

IV. Sample ProblemsIV. Sample Problems

a) Which compound, FeCO3 or PbBr2, is more soluble in acid than in base? Why?

b) Will copper(I) cyanide be more soluble in acid or base? Why?

c) In which type of solution is AgCl most soluble: acidic, basic, or neutral?

IV. PrecipitationIV. Precipitation• Ksp values can be used to predict when

precipitation will occur.• Again, we use a Q calculation.

If Q < Ksp, solution is unsaturated. Solution dissolve additional solid.

If Q = Ksp, solution is saturated. No more solid will dissolve.

If Q > Ksp, solution is supersaturated, and precipitation is expected.

IV. Sample ProblemsIV. Sample Problems

a) Will a precipitate form if 100.0 mL 0.0010 M Pb(NO3)2 is mixed with 100.0 mL 0.0020 M MgSO4?

b) The concentration of Ag+ in a certain solution is 0.025 M. What concentration of SO4

2- is needed to precipitate out the Ag+? Note that Ksp = 1.2 x 10-5

for silver(I) sulfate.

V. Complex IonsV. Complex Ions• In aqueous solution, transition metal

cations are usually hydrated. e.g. Ag+

(aq) is really Ag(H2O)2+

(aq). The Lewis acid Ag+ reacts with the Lewis

base H2O.• Ag(H2O)2

+(aq) is a complex ion.

A complex ion has a central metal bound to one or more ligands.

A ligand is a neutral molecule or an ion that acts as a Lewis base with the central metal.

V. Formation ConstantsV. Formation Constants

• Stronger Lewis bases will replace weaker ones in a complex ion. e.g. Ag(H2O)2

+(aq) + 2NH3(aq) Ag(NH3)2

+

(aq) + 2H2O(l)

For simplicity, it’s common to write Ag+

(aq) + 2NH3(aq) Ag(NH3)2+

(aq) Since this is an equilibrium, we can write

an equilibrium expression for it.

V. Formation ConstantsV. Formation Constants

• Kf is called a formation constant.• Unlike other equilibrium constants we’ve

seen, Kf’s are large, indicating favorable formation of the complex ion.

Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq)

V. Sample Formation ConstantsV. Sample Formation Constants

V. Calculations w/ KV. Calculations w/ Kff’s’s

• Since Kf’s are so large, calculations with them are slightly different.

• We assume the equilibrium lies essentially all the way to the right.

• This changes how we set up our equilibrium chart.

V. Illustrative ProblemV. Illustrative Problem

• Calculate the concentration of Ag+ ion in solution when 0.085 g silver(I) nitrate is added to a 250.0 mL solution that is 0.20 M in KCN.

V. Illustrative Problem SolutionV. Illustrative Problem Solution

1) First, we must identify the complex ion.

In solution, we will have Ag+, NO3-, and

K+, and CN-. The complex ion must be made from Ag+

and CN-. Looking at table of Kf’s, we find that

Ag(CN)2- has Kf = 1 x 1021.

V. Illustrative Problem SolutionV. Illustrative Problem Solution

2) Next, we need concentrations. Already know that [CN-] = 0.20 M. We calculate the [Ag+].

V. Illustrative Problem SolutionV. Illustrative Problem Solution

3) Now we set up our equilibrium chart. Since Kf is so big, we assume the

reaction essentially goes to completion.

Ag+(aq) + 2CN-

(aq) Ag(CN)2-(aq)

Initial 0.00200 M 0.20 M 0

Change ≈ -0.00200 ≈ -0.00400 ≈ +0.00200

Equil. x 0.196 0.00200

V. Illustrative Problem SolutionV. Illustrative Problem Solution4) Finally, we solve for x.

Thus, [Ag+] = 5 x 10-23. It is very small, so ourapproximation is valid. Note that book would use 0.20 for [CN-] in the calculation.

V. Sample ProblemV. Sample Problem

• A 125.0-mL sample of a solution that is 0.0117 M in NiCl2 is mixed with a 175.0-mL sample of a solution that is 0.250 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+

(aq) remains?

V. Complex Ions & SolubilityV. Complex Ions & Solubility

• Formation of complex ions enhances the solubility of some normally insoluble ionic compounds.

• Typically, Lewis bases will enhance solubility.• e.g. Adding NH3 to a solution containing

AgCl(s) will cause more AgCl(s) to dissolve.

AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 1.77 x 10-10

Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq) Kf = 1.7 x 107

V. Complex Ion FormationV. Complex Ion Formation

V. Metal HydroxidesV. Metal Hydroxides

• All metal hydroxides can act as bases. e.g. Fe(OH)3(s) + 3H3O+

(aq) Fe3+(aq) + 6H2O(l)

• Some metal hydroxides can act as acids and bases; they are amphoteric. In addition to the above, Al(OH)3(s) can also

absorb hydroxide. Al(OH)3(s) + OH-

(aq) Al(OH)4-(aq)

V. Aluminum HydroxideV. Aluminum Hydroxide

• In acidic solutions: Al(H2O)6

3+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)

• As OH- is added, solution becomes neutral. Al(H2O)5(OH)2+

(aq) + 2OH-(aq) Al(H2O)3(OH)3(s) +

2H2O(l)

• In basic solutions: Al(H2O)3(OH)3(s) + OH-

(aq) Al(H2O)2(OH)4-(aq)

• Thus, solubility is very pH dependent.

V. Aluminum HydroxideV. Aluminum Hydroxide

V. Amphoteric HydroxidesV. Amphoteric Hydroxides

• There are not that many metal hydroxides that are amphoteric.

• Only Al3+, Cr3+, Zn2+, Pb2+, and Sn2+.