Topic: Titration
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Topic: Titration Do Now: . Acid-Base Titration. A procedure used in order to determine the molarity of an acid or base K nown volume of a solution with a known concentration (standard solution) and k nown volume of unknown concentration and a M A V A = M B V B acid-base indicator needed.

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Topic titration do now

Topic: Titration

Do Now:


Acid base titration

Acid-Base Titration

  • A procedure used in order to determine the molarity of an acid or base

  • Known volume of a solution with a known concentration (standard solution) and known volume of unknown concentration and a

  • MAVA = MBVB

  • acid-base indicator needed


Titration

Titration

  • Standard solution slowly added to unknown solution

  • As solutions mix:

    • neutralization reaction occurs

  • Eventually:

    • enough standard solution is added to neutralize the unknown solution Equivalence point


Equivalence point

Equivalence point

total # moles H+1 ions donated by acid equals

total # moles H+1 accepted by base

total moles H+1 = total moles OH-1


Titration1

Titration

  • End-point = point at which indicator changes color

    • if indicator chosen correctly:

      • end-point very close to equivalence point


Titration of a strong acid with a strong base

14-

Equivalence Pt 

0 ml

20 ml

Titration of a strong acid with a strong base

Phenolphthalein

Color change:

8.2 to 10 

pH

7-

0-

Volume of 0.100 M NaOH added (ml)

40ml


M h 1 v h 1 m oh 1 v oh 1

MH+1 VH+1= MOH-1 VOH-1

  • MH+1= molarity of H+1

  • MOH-1= molarity of OH-1

  • VH+1= volume of H+1

  • VOH-1= volume of OH-1


Titration problem 1

Titration Problem #1

  • In a titration of 40.0 mL of a nitric acid solution, the end point is reached when 35.0mL of 0.100M NaOH is added

    Calculate the concentration of the nitric acid solution

HNO3 + NaOH H2O + NaNO3


Variables

Variables

  • # of H’s = 1

  • Ma= ?

  • Va = 40.0 mL

  • # of OH’s = 1

  • Mb = 0.100 M

  • Vb = 35.0 mL


Topic titration do now

(1)(X) (40.0 mL) = (0.100 M )(35.0mL)(1)

X = 0.875 M HNO3


Titration problem 2

Titration Problem #2

  • What is the concentration of a hydrochloric acid solution if 50.0 mL of a 0.250M KOH solution is needed to neutralize 20.0mL of the HCl solution of unknown concentration?

KOH + HCl H2O + KCl


Variables1

Variables

  • # of H’s = 1

  • Ma= X

  • Va = 20.0 mL

  • # of OH’s = 1

  • Mb = 0.250 M

  • Vb = 50.0 mL


Topic titration do now

(1)(X)(20.0 mL) = (0.250 M) (50.0 mL)(1)

X = 0.625 M HCl


Titration problem 3

Titration Problem #3

  • What is the concentration of a sulfuric acid solution if 50.0mL of a 0.25 M KOH solution is needed to neutralize 20.0mL of the H2SO4 solution of unknown concentration?

H2SO4 + 2 KOH  2 H2O + K2SO4


Variables2

Variables

  • # of H’s = 2

  • Ma= X

  • Va = 20.0mL

  • # of OH’s = 1

  • Mb = 0.25M

  • Vb = 50.0mL


Topic titration do now

(2)(X)(20.0ml) = (0.25M)(50.0ml)(1)

X = 0.3125 M H2SO4(sulfuric acid)


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