ACID BASE TITRATIONS

ACID BASE TITRATIONS

Chapter 15

Titration Curves

• A plot of pH versus amount of acid or base added.

• At the EQUIVALENCE POINT on a titration curve, the amount of acid = the amount of base.

• The END POINT of a titration is determined by a color change of an indicator.

• Ideally, the end point and equivalence point will be within 1 drop of each other.

INDICATORS• Select an indicator based on the pH range of the equivalence point. pKa of

indicator within +/- 1 pH unit of the equivalence point.• Indicators are organic dyes whose colors depend on the [H3O+] or pH of a solution.

Most are produced synthetically. – Ex.

• Phenolphthalein• Universal indicator – mix of organic acids that indicate over different ranges

Many are vegetable dyes.– Ex.

• Litmus• Purple cabbage indicator

INDICATORS

• Generally they are WEAK ORGANIC ACIDS– Symbolized HIn– Dissociation reaction:

HIn + H2O H30+ + In-

Ex. Bromthymol blue• HIn – yellow• In- - blue

HIn + H20 In- + H30+

• Ka = [H30+] [In-1] [HIn]

When the ratio goes to 1/10, a color change will occur.

• Adding an acid shifts the equilibrium left.• Adding a base shifts the equilibrium right.

Indicator sample problem• An indicator, HIn, has a Ka = 1.0 x 10-7 .

Determine the pH at which a color change will occur given the following scenarios:

• Acid titration

• Base titration

INDICATOR EXAMPLETitration of an acid

• The solution is initially acidic, [HIn] is dominant.

• The color change occurs when [In]/[HIn] = 1/10

Ka = 1.0 x 10-7 = [H+] (1/10) [H+] = 10 (1.0 x 10-7) = 1.0 x 10-

6

pH = 6.0

Titration of a base

• The solution is initially basic, [In] is dominant.

• The color change occurs when [In]/[HIn] = 10/1

Ka = 1.0 x 10-7 = [H+] (10/1) [H+] = (1.0 x 10-7)/10 = 1.0 x 10-

8

pH = 8.0

STRONG ACID/ STRONG BASE

• Consider the titration of 50.0 ml of 0.2 M nitric acid with 0.1 M NaOH

Calculate the pH @

• 0.0 ml base added



Major species in soln?

SA/SB




Major species in soln?

STRONG BASE/ STRONG ACID

• The pH Curve for the titration of 100.0 mL of 0.50 M NaOH with 1.0 M HCI

WEAK ACID/ STRONG BASE

• The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH

WEAK ACID/ STRONG BASE• Before any BASE is added, the pH depends only on the

weak acid.• After some base is added, but before the EQUIVALENCE

POINT, a series of weak acid/ salt buffer solutions determine the pH.

• At the EQUIVALNECE POINT, hydrolysis of the anion of the weak acid determines the pH.

• Beyond the equivalence point, EXCESS STRONG BASE determines the pH.

WA/SB Sample problem:• 30.0 ml of 0.10 M NaOH is added to 50.0 ml of 0.10 M

HF. What is the pH after all 30.0 ml are added? Ka = 7.2 x 10-4

1. Major species in soln?2. Rxn?3. HF initial? (Use mmol)4. OH- added?5. HF consumed?6. F- formed?

WA/SB After equilibrium1. Rxn?2. Ka expression3. Calculate concentrations using mmoles and

volumes M = mmol/ml4. ICE5. [H+] and pH

WEAK BASE/ STRONG ACID

• Calculate the pH at each of the following points in the titration of 50.00 ml of a 0.01000M sodium phenolate (NaOC6H5) solution with 1.000 M HCl soltuion. Ka for HOC6H5 = 1.05 x 10-10.

• Initial• Midpoint• Equivalence point

WB/SA

• Initial – weak base Kb • pH = pKa at the midpoint, so pOH = pKb since [BH+]/[B] = 1

WB/SA• At equivalence.• How many moles of HCl were needed to neutralize?MaVa = MbVb HCl + OC6H5-

• New volume?• Weak acid dissociation reaction.

HOC6H5- + H20 • Ka expression.• [H+]• pH

POLYPROTIC ACIDS

• Consider 20.00ml or 0.100 M polyprotic acid H2A titrated with 0.100 M NaOH.

• Ka1 = 1x10-3

• Ka2 = 1x10-7

• 2 equivalence points are expected.

POLY pH calculations @

• 0 ml base added• H2A dissociation

• H2A H+ + HA-

• 10.0 ml base added• H2A/ HA- bufferH2A + OH- HA- + H20

POLY pH calculations @

• 20.0 ml base added – first equivalence point

• 30.0 ml base added – ½ way between 1st and 2nd equivalence point.

H2A + OH- HA- + OH- A2-

POLY pH calculations @• 40.0 ml base added – 2nd

equivalence pointH2A + OH- HA- + OH- A2-

A2- + H20 HA- + OH-

• 50.0 ml NaOH added – excess OH

H2A + OH- HA- + OH- A2-

SOLUBILITY EQUILIBRIUM

Ksp

Ksp SOLUBILITY PRODUCT EQUILIBRIA

• Problems dealing with solubility of PARTIALLY soluble ionic compounds (in other words, salts that barely dissociate/dissolve in water)

• General Form, called the solubility product:MX(s) n M+

(aq) + p X – (aq)

• Ksp = [M+]n[X -]p

• Ex Ba(OH)2 (s)

Solubility

• Is NOT the solubility product.• Uses an equilibrium problem to determine

how much can dissolve at a certain temperature.

• Is related stoichiometrically to the initial formula. Ex. Ba(OH)2 yields twice as many OH-

1 ions.

Calculate Ksp

• Given the solubility of FeC2O4 at eq. = 65.9mg/L

• Given the solubility of Li2CO3 is 5.48 g/L.

Calculate solubility

• Of SrSO4 with a Ksp of 3.2 x 10-7 in M and g/L

• Of Ag2CrO4 with Ksp of 9.0 x 10-12 in M and g/L

Relative solubilities

• Ksp can be used to compare the solubility of solids that break apart into the same number of ions.

• The bigger the Ksp, the more soluble.• An ICE table is necessary if different

numbers of ions are produced.

Common Ion Effect

• If we try to dissolve the solid in solution with either the cation or anion present, less will dissolve.

• Calculate the solubility of strontium sulfate in a 0.100M solution of Na2SO4

pH and solubility• OH- can be a common ion.• More soluble in acid, since OH- will be removed

from the reaction.• For other anions, if they come from a weak acid

they are more soluble in acid than in water.Ex. CaC2O4 Ca+2 + C2O4 -2

H+ + C2O4-2 HC2O4

-

Reduces the C2O4-2 in acidic solution

Precipitation

• Ion product, Q = [M+]n[X -]p

• If Q > Ksp, the precipitate forms• Q< Ksp, no precipitate• Q = Ksp at equilibrium

Sample PPT problem• A solution of 750.0 ml of 4.00 x 10-3 M cerium

(III) nitrate is added to 300.0 ml of 2.00 x 10-2 M potassium iodate. Will cerium (III) iodate, Ksp = 1.9 x 10-10 ppt. and if so what is the concentration of ions in solution?

Selective precipitation• Used to separate mixtures of metal ions

in solution.• Add anions that will only ppt. certain

metals at a time.• Used to purify mixtures.• Often use H2S because in acidic solution,

Hg+2, Cd+2, Bi+3, Cu+2, and Sn+4 will ppt.

Selective precipitation

• In basic solution, adding OH- solution, S-2 will increase so more soluble sulfides will ppt.

• Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3

Selective precipitation

• Follow the steps with first insoluble chlorides (Ag, Pb, Ba)

• Then sulfides in acid• Then sulfides in base• Then insoluble carbonates (Ca, Ba, Mg)• Alkali metals and NH4

+ remain in solution– Flame test, NH4+ yields an ammonia smell when

heated.

Complex ion equilibria

• A charged ion surrounded by LIGANDS.• Ligands are LEWIS BASES using their lone pair

to stabilize the charged metal ions.• Common ligands are NH3, H2O, CN-, Cl-

• COORDINATION NUMBER is the number of attached ligands – usually twice the cations charge.– Ex. Cu(NH3)4

+2 has a coordination # = 4

LIGAND attachment

• The addition of each ligand has its own equilibrium.

• Usually the ligand is in large EXCESS.• The complex ion will be the biggest ion in

solution.• Complex formation helps dissolve otherwise

insoluble compounds.

Complex ion equilibrium

• Calculate the concentrations of Ag+ and Ag(CN-)2

-1 in a solution prepared by mixing 100.0 ml of 5.0 x 10-3 M AgNO3 with 100.0 ml of 2.00 M KCN.

Ag+ + 2 CN- Ag(CN)2-1 K1 = 1.3 x10-21

Documents

ACID BASE TITRATIONS