acid base titrations
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/1.jpg)
ACID BASE TITRATIONS
Chapter 15
![Page 2: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/2.jpg)
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.
![Page 3: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/3.jpg)
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
![Page 4: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/4.jpg)
INDICATORS
![Page 5: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/5.jpg)
• Generally they are WEAK ORGANIC ACIDS– Symbolized HIn– Dissociation reaction:
HIn + H2O H30+ + In-
![Page 6: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/6.jpg)
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.
![Page 7: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/7.jpg)
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
![Page 8: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/8.jpg)
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
![Page 9: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/9.jpg)
STRONG ACID/ STRONG BASE
• Consider the titration of 50.0 ml of 0.2 M nitric acid with 0.1 M NaOH
![Page 10: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/10.jpg)
Calculate the pH @
• 0.0 ml base added
• 10.0 ml base added
• 20.0 ml base added
Major species in soln?
![Page 11: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/11.jpg)
SA/SB
• 50.0 ml base added
• 100.0 ml base added
• 200.0 ml base added
Major species in soln?
![Page 12: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/12.jpg)
STRONG BASE/ STRONG ACID
• The pH Curve for the titration of 100.0 mL of 0.50 M NaOH with 1.0 M HCI
![Page 13: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/13.jpg)
WEAK ACID/ STRONG BASE
• The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH
![Page 14: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/14.jpg)
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.
![Page 15: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/15.jpg)
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?
![Page 16: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/16.jpg)
WA/SB After equilibrium1. Rxn?2. Ka expression3. Calculate concentrations using mmoles and
volumes M = mmol/ml4. ICE5. [H+] and pH
![Page 17: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/17.jpg)
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
![Page 18: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/18.jpg)
WB/SA
• Initial – weak base Kb • pH = pKa at the midpoint, so pOH = pKb since [BH+]/[B] = 1
![Page 19: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/19.jpg)
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
![Page 20: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/20.jpg)
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.
![Page 21: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/21.jpg)
POLY pH calculations @
• 0 ml base added• H2A dissociation
• H2A H+ + HA-
• 10.0 ml base added• H2A/ HA- bufferH2A + OH- HA- + H20
![Page 22: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/22.jpg)
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-
![Page 23: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/23.jpg)
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-
![Page 24: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/24.jpg)
SOLUBILITY EQUILIBRIUM
Ksp
![Page 25: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/25.jpg)
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)
![Page 26: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/26.jpg)
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.
![Page 27: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/27.jpg)
Calculate Ksp
• Given the solubility of FeC2O4 at eq. = 65.9mg/L
• Given the solubility of Li2CO3 is 5.48 g/L.
![Page 28: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/28.jpg)
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
![Page 29: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/29.jpg)
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.
![Page 30: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/30.jpg)
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
![Page 31: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/31.jpg)
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
![Page 32: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/32.jpg)
Precipitation
• Ion product, Q = [M+]n[X -]p
• If Q > Ksp, the precipitate forms• Q< Ksp, no precipitate• Q = Ksp at equilibrium
![Page 33: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/33.jpg)
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?
![Page 34: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/34.jpg)
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.
![Page 35: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/35.jpg)
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
![Page 36: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/36.jpg)
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.
![Page 37: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/37.jpg)
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
![Page 38: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/38.jpg)
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.
![Page 39: ACID BASE TITRATIONS](https://reader036.vdocuments.net/reader036/viewer/2022082215/568165c0550346895dd8c4f0/html5/thumbnails/39.jpg)
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