volumetric calculations

14
Volumetric Volumetric Calculations Calculations How to analyze titration How to analyze titration data data and and perform dilution perform dilution calculations calculations

Upload: komala

Post on 10-Feb-2016

90 views

Category:

Documents


0 download

DESCRIPTION

Volumetric Calculations. How to analyze titration data and perform dilution calculations. Volumetric procedures. Involve the careful measurement of volumes of solutions - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Volumetric Calculations

Volumetric CalculationsVolumetric CalculationsHow to analyze titration dataHow to analyze titration data

andandperform dilution calculationsperform dilution calculations

Page 2: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Volumetric proceduresVolumetric procedures• Involve the careful measurement of Involve the careful measurement of

volumes of solutionsvolumes of solutions• Dissolve materials in water and run Dissolve materials in water and run

reaction in solution - titration in reaction in solution - titration in Erlenmeyer flask (designed for swirling to Erlenmeyer flask (designed for swirling to mix reactants)mix reactants)

• Buret is the measuring device - can read Buret is the measuring device - can read volumes to 0.01 mLvolumes to 0.01 mL

• Pipets may also be used, especially to Pipets may also be used, especially to transfer solutions - good to 0.01 mLtransfer solutions - good to 0.01 mL

Page 3: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Some terminologySome terminology• AnalyteAnalyte• TitrantTitrant• Standard Standard

solutionsolution• Equivalence Equivalence

pointpoint• End pointEnd point

Substance being analyzedSolution in the buret

Solution of known concentrationStoichiometric point

Page 4: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

At the equivalence pointAt the equivalence point• For a 1-to-1 reaction such as:For a 1-to-1 reaction such as:

HCl + NaOH HCl + NaOH NaCl + H NaCl + H22OO mole mole acidacid = mole = mole basebase

• moles = molarity x volume in litersmoles = molarity x volume in liters• SoSo

MMacidacid x V x Vacidacid = M = Mbasebase x V x Vbasebase

Page 5: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

What is the end point?What is the end point?• The end point is the point where an The end point is the point where an

indicator changes color.indicator changes color.• This is the This is the visual signvisual sign to stop or end to stop or end

the titration. If the indicator is the titration. If the indicator is selected correctly, the change is at selected correctly, the change is at or very near the equivalence point!or very near the equivalence point!

• Example is phenolphthaleinExample is phenolphthaleincolorless in acidcolorless in acidpinkpink in basein base

Page 6: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Example of calculationExample of calculation• A 25.07 mL sample of vinegar is A 25.07 mL sample of vinegar is

titrated with 37.31 mL of 0.5119 M titrated with 37.31 mL of 0.5119 M NaOH. What is the molarity of the NaOH. What is the molarity of the acetic acid in vinegar?acetic acid in vinegar?

Moles NaOH = 0.5119 moles/L x 37.31 mL x 1L/1000mL

Moles acetic acid = moles NaOH = 0.01910 moles

Maa = moles/volume = 0.01910 moles/0.02507 L = 0.7618 M

Page 7: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

• What is the percent acetic acid in the What is the percent acetic acid in the vinegar? Vinegar has a density near vinegar? Vinegar has a density near water so our 25.07 mL sample has a water so our 25.07 mL sample has a mass of 25.07 g.mass of 25.07 g.

Mass of acetic acid = moles x MM = 0.01910 moles x 60.00 g/mole= 1.146 g acetic acid

Percent = part x 100/total = 1.146 g acetic acid x 100/ 25.07 g sample = 4.47% acetic acid

Page 8: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

A Back Titration -A Back Titration -• An indirect analysis for antacids, since An indirect analysis for antacids, since

they are insolublethey are insoluble• Antacids, which are bases, are reacted Antacids, which are bases, are reacted

with excess acid and this excess acid is with excess acid and this excess acid is titrated with standard base.titrated with standard base.

Antacid + HCl (initial) neutralized antacid + HCl (leftover)

HCl (leftover) + NaOH NaCl + H2O

Page 9: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Back Titration Calculation-Back Titration Calculation-• moles HCl initially = Mmoles HCl initially = MHClHCl x V x VHClHCl

• moles HCl left-over = moles NaOHmoles HCl left-over = moles NaOH = M= MNaOHNaOH x V x VNaOHNaOH

• mole HCl reacted = mole HCl reacted = moles HCl initially - moles HCl left-overmoles HCl initially - moles HCl left-over

• moles antacids - depends on reaction moles antacids - depends on reaction stoichiometry such as:stoichiometry such as:CaCOCaCO33 + 2HCl + 2HCl CaCl CaCl22 + H + H22O + COO + CO22

Page 10: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Dilution -Dilution -• Process of adding solvent, Process of adding solvent,

usually water, to lower the usually water, to lower the concentration of a solutionconcentration of a solution

• Used to make working standards Used to make working standards for spectrophotometryfor spectrophotometry

• Used to bring a sample into Used to bring a sample into range for analysisrange for analysis

• Typically done using a pipet (TD) Typically done using a pipet (TD) snd volumetric flask (TC)snd volumetric flask (TC)

Solute is constant. Solvent is being added.

Page 11: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Dilution continuedDilution continued• Since the solute is constant, then Since the solute is constant, then

molesmolesinitialinitial = moles = molesfinalfinal

• So the dilution equation becomesSo the dilution equation becomesCCinitinit V Vinitinit = C = Cfinalfinal V Vfinalfinal

where Vwhere Vfinalfinal = V = Vinitinit + V + Vwaterwater

Page 12: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Dilution CalculationDilution Calculation• What initial volume of 0.50 M CuSOWhat initial volume of 0.50 M CuSO44

must be used to make 100 mL of must be used to make 100 mL of 0.025 M CuSO0.025 M CuSO44 solution? solution?

•CCinitinit V Vinitinit = C = Cfinalfinal V Vfinalfinal and rearrange and rearrange

•VVinit init = C= Cfinalfinal V Vfinalfinal / / CCinitinit=0.025 M x 100 mL/ 0.50 M = 5.0 mL=0.025 M x 100 mL/ 0.50 M = 5.0 mL

CCinitinitVVfinalfinal

CCfinalfinal

Page 13: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Dilution continuedDilution continued• A graduated cylinder is a volume A graduated cylinder is a volume

measuring device to contain (TC). measuring device to contain (TC). Could you do a dilution in it? Explain Could you do a dilution in it? Explain why or why not.why or why not.

• Dilution factor = VDilution factor = Vinitinit/V/Vfinalfinal

1 mL into 100 mL is a df of 1/1001 mL into 100 mL is a df of 1/100

Page 14: Volumetric Calculations

PGCC CHM 103 SinexPGCC CHM 103 Sinex

Dilution continuedDilution continued• Serial dilution – successive dilutions Serial dilution – successive dilutions

for large scale dilutions in stepsfor large scale dilutions in stepsA is diluted by 1/100 to B A is diluted by 1/100 to B B is diluted by 1/100 to C B is diluted by 1/100 to C C is 1/10,000 of AC is 1/10,000 of A