Topic: TitrationDo Now:

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

• 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

total # moles H+1 ions donated by acid equals

total # moles H+1 accepted by base

total moles H+1 = total moles OH-1

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

Volume of 0.100 M NaOH added (ml)

pH

0-

14-

7- Equivalence Pt

Phenolphthalein Color change: 8.2 to 10

0 ml

40ml

20 ml

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• 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

• # 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

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

Variables

• # 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

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

Variables

• # 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)