acid base titrations n.
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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.

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titration curves
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.
  • 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

Generally they are WEAK ORGANIC ACIDS

    • Symbolized HIn
    • Dissociation reaction:

HIn+ H2O  H30+ + In-

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

HIn+ H20  In- + H30+

  • Ka = [H30+] [In-1]


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
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 example

Titration of an acid

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

  • 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

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 @
Calculate the pH @
  • 0.0 ml base added
  • 10.0 ml base added
  • 20.0 ml base added

Major species in soln?

sa sb
  • 50.0 ml base added
  • 100.0 ml base added
  • 200.0 ml base added

Major species in soln?

strong base strong acid
  • The pH Curve for the titration of 100.0 mL of 0.50 MNaOH 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 MNaOH
weak acid strong base1
  • 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
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
  • Major species in soln?
  • Rxn?
  • HF initial? (Use mmol)
  • OH- added?
  • HF consumed?
  • F- formed?
wa sb after equilibrium
WA/SB After equilibrium
  • Rxn?
  • Ka expression
  • Calculate concentrations using mmoles and volumes M = mmol/ml
  • ICE
  • [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 HClsoltuion. 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 sa1
  • At equivalence.
  • How many moles of HCl were needed to neutralize?

MaVa = MbVbHCl + 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 @
POLY pH calculations @
  • 0 ml base added
  • H2A dissociation
  • H2A  H+ + HA-
  • 10.0 ml base added
  • H2A/ HA- buffer

H2A + OH- HA- + H20

poly ph calculations @1
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 @2
POLY pH calculations @
  • 40.0 ml base added – 2nd equivalence point

H2A + OH- HA- + OH-  A2-

A2- + H20  HA- + OH-

  • 50.0 ml NaOH added – excess OH

H2A + OH- HA- + OH-  A2-

k sp 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
  • ExBa(OH)2 (s)
  • 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
Calculate Ksp
  • Given the solubility of FeC2O4at eq. = 65.9mg/L
  • Given the solubility of Li2CO3 is 5.48 g/L.
calculate solubility
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
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
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
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

  • Ion product, Q = [M+]n[X -]p
  • If Q > Ksp, the precipitate forms
  • Q< Ksp, no precipitate
  • Q = Ksp at equilibrium
sample ppt problem
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
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 precipitation1
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 precipitation2
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
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
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
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