Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

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Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte

Transcript of Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

Page 1: Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

Procedure for calculating pH of a titration

Volume of titrant needed is based on amount of analyte

Page 2: Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal volume = 200 mL

Kb =[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05 – x 0.05 x 1.05 x 10-6 = [OH-]

pOH = 5.98

pH = 14 – pOH = 8.02

Page 3: Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

A sample of 0.1276 g of an unknown monoprotic acid was dissolved in 25.0 ml of water and titrated with 0.0633 M NaOH solution. The volume of the NaOH required to get to the equivalence point was 18.4 ml. What is the molar mass of the acid?

Moles Base = (0.0633 moles/L)(0.0184 L) = 0.00116 moles

Moles Base = Moles Acid @ Equiv. Point

MW Acid = 0.1276 g / 0.00116 moles = 110. g/mole

After 10.0 ml NaOH solution was added, the pH was 5.87. What is the Ka of the unknown acid?

pKa = pH - log [A-][HA]

10.0 ml NaOH => (0.0633 moles/L)(0.0100 L) = 0.000633 moles = moles A-

Page 4: Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

Moles HA = (total moles HA) - (moles A- )

= 0.00116 moles - 0.000633 moles

= 0.000527 moles

pKa = pH - log [A-][HA]

= 5.87 - log

[A-][HA]

= moles / Vmoles / V

Volumes are the same since they are in the same

solution

0.000633

0.000527= 5.87 - 0.079

pKa = 5.79

Ka = 10-5.79 = 1.62 x 10-6

Page 5: Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

Acid-Base Indicators

HIn (aq) H+ (aq) + In- (aq)

10[HIn]

[In-]Color of acid (HIn) predominates

10[HIn]

[In-]Color of conjugate base (In-) predominates

16.5

Page 6: Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

The titration curve of a strong acid with a strong base.

16.5

Page 7: Procedure for calculating pH of a titration Volume of titrant needed is based on amount of analyte.

Which indicator(s) would you use for a titration of of HNO2 versus KOH ?

Weak acid titrated with strong base.

At equivalence point, will have conjugate base of weak acid.

At equivalence point, pH > 7

Use cresol red or phenolphthalein

Low pKa acids (stronger weak acids) change color

@ low pH

High pKa acids (weaker weak acids) change color

@ high pH

Use small amount of indicator so experiment is

not altered.