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Topic: Titration Do Now: PowerPoint Presentation

Topic: Titration Do Now:

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Topic: Titration Do Now:

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Topic: Titration

Do Now:

- 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

- 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

total # moles H+1 ions donated by acid equals

total # moles H+1 accepted by base

total moles H+1 = total moles OH-1

- End-point = point at which indicator changes color
- if indicator chosen correctly:
- end-point very close to equivalence point

- if indicator chosen correctly:

14-

Equivalence Pt

0 ml

20 ml

Phenolphthalein

Color change:

8.2 to 10

pH

7-

0-

Volume of 0.100 M NaOH added (ml)

40ml

- MH+1= molarity of H+1
- MOH-1= molarity of OH-1
- VH+1= volume of H+1
- VOH-1= volume of OH-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

- # of H’s = 1
- Ma= ?
- Va = 40.0 mL
- # of OH’s = 1
- Mb = 0.100 M
- Vb = 35.0 mL

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

X = 0.875 M HNO3

- 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

- # of H’s = 1
- Ma= X
- Va = 20.0 mL
- # of OH’s = 1
- Mb = 0.250 M
- Vb = 50.0 mL

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

X = 0.625 M HCl

- 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

- # of H’s = 2
- Ma= X
- Va = 20.0mL
- # of OH’s = 1
- Mb = 0.25M
- Vb = 50.0mL

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

X = 0.3125 M H2SO4(sulfuric acid)