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TITRATIONS

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14 diapositives · 280 Views · Aug 23, 2014

This resource explains titrations and neutralization reactions, fundamental concepts in chemistry. A neutralization reaction occurs between an acid and a base, producing water and a salt, exemplified by H₂SO₄ reacting with KOH. The guide also details the use of indicators like litmus, phenolphthalein, and bromothymol blue to identify pH changes during titrations. The equivalence point, where the amounts of acid and base are stoichiometrically equivalent, is highlighted alongside calculations for determining unknown concentrations and volumes using the molarity equation.

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  1. TITRATIONS

  2. Neutralization • Reaction between an acid and a base that produces water and a salt • (Type of reaction: Double displacement) Example: • H2SO4 (aq) + 2 KOH (aq)  K2SO4 (aq) + 2 HOH (l) • Acid Base Salt Water • Remember: HOH = H2O

  3. Acid/base Indicators • An indicator is a substance that changes color when the pH changes. • Common indicators: • Litmus – changes from red to blue • pH Range: 5.0-8.0 • Phenolphthalein – changes from clear (acidic) to pink (basic) • pH Range: 8.2-10.0 • Bromothymolblue – yellow (acidic) to blue (basic) • pH range: 6.0-7.6

  4. Video of the day: Cabbage pH

  5. Universal indicator

  6. Indicators

  7. Titration • Titration is controlled addition of a solution with a known concentration to a solution of unknown concentration. • Titrations enable us to find an unknown concentration by finding two things that are stoichiometrically equivalent

  8. Using a buret to do titrations Base or acid is slowly added until the solution changes color. The color change tells you that all of the base or acid has reacted so you can stop adding more.

  9. Equivalence point • Equivalence point: The point in a titration where the acid and base are stoichiometrically equivalent • There is no excess base or excess acid. All of the base and acid has reacted. • End point: The point in a titration when the color of the solution changes permanently.

  10. Titrationcurve An indicator is used to find the endpoint . You choose and indicator so that the endpoint is in the rapid rise region of the curve. Therefore the endpoint = equivalence point

  11. Finding an unknown Molarity or volume • Use (#H+)MaVa=MbVb(#OH-) • H+ = Number of hydrogen ions (H+) from acid (see next slide for an example) • Ma = molarity (M) of acid • Va= volume of acid used in titration • Mb = molarity (M) of base • Vb= volume of base used in titration • (#OH-) = Number of hydroxide (OH-) ions from base (see next slide for an example)

  12. Finding an unknown Molarity or volume • Use (#H+)MaVa=MbVb(#OH-) • Remember at equivalence point, moles of acid = moles of base • You must remember to use a mole ratio when the acid and base don’t react in a 1:1 mol ratio • Example: H3PO4 reacts with Ba(OH)2 • Mol H+ = 3 • Mol OH- = 2 • (3)MaVa=MbVb(2)

  13. Example – Per. 3 • When 10.0 mL of a 0.125 M solution of hydrochloric acid, HCl, is titrated with a 0.100 M solution of potassium hydroxide, KOH, what is the volume of base needed to neutralize the acid? • HCl+ KOH KCl + HOH • (#H+)MaVa= MbVb(#OH-) • (1)(0.125 M)(10.0 mL) = (0.100 M)Vb(1) • Mb = 0.125 M x 10.0 mL = 12.5 mL KOH 0.100 M

  14. Example – Per. 4 • When 10.0 mL of a 0.125 M solution of hydrochloric acid, HCl, is titrated with12.5 mL of potassium hydroxide, KOH, what is the concentration of base needed to neutralize the acid? • HCl + KOH KCl + HOH • (#H+)MaVa= MbVb(#OH-) • (1)(0.125 M)(10.0 mL) = Mb (12.5 mL) (1) • Mb = 0.125 M x 10.0 mL = 0.1 M KOH 12.5 mL