Quantitative Chemical Analysis

Seventh Edition

Chapter 8-12Acid-Base Titrations

Copyright © 2007 by W. H. Freeman and Company

Daniel C. Harris

ACTIVITY

ACTIVITY

How to calculate activity?1: calculate Ionic strength2: Use Table 8-1

BUFFERS

BUFFERS

BUFFERS

= 8.3

H2CO3 ⇔ H+ + HCO3-

]CO[H][HCO log pK pH

32

-3

a +=

5-

3-

10 1.210 1.02 log 6.35 pH

×

×+=

So the lake is a buffer!

Consider the titration of 50.00 mL of 0.020 00 M KOH with 0.100 0 M HBr.

STEP 2: Ve

STEP 1: reaction

Strong baseWith Strong acid

STEP 3: before equivalence

When 3.00 mL of HBr have been added, the reaction is three-tenths complete because Ve = 10.00 mL.

Strong baseWith Strong acid

STEP 4: at equivalence

pH = 7

Strong baseWith Strong acid

A – 13B – 12C – 10D – 7E - 1

STEP 5: after equivalence

The concentration of excess H+ at, say, 10.50 mL is given by

Strong baseWith Strong acid

Strong baseWith Strong acid

Weak acidWith Strong base

EXAMPLE:titration of 50.00 mL of 0.020 00 M MES (pKa=6.27) with 0.100 0 M NaOH.

Weak acidWith Strong base

HA + OH-A- + H2OSTEP 1: reaction

STEP 2: Ve

mols base = mols acid

Weak acidWith Strong baseNB: before any base is added

Weak acidWith Strong baseSTEP 3: before equivalence

BUFFER!After adding 30% of Ve

After adding half of Ve

Weak acidWith Strong base

STEP 4: at equivalence

A < 7B = 7C > 7

Weak acidWith Strong baseSTEP 5: after equivalence

excess OH-

Weak acidWith Strong base

Calculated titration curvefor the reaction of50.00 mL of 0.020 00 M MESwith 0.100 0 M NaOH.Landmarks occur at half of the equivalence volume (pH = pKa) and at the equivalence point, \which is the steepest part of the curve.

Weak acidWith Strong base

Weak baseWith Strong acid

• 1.435 g sample of dry CaCO3 and CaCl2mixture was dissolved in 25.00 mL of 0.9892 M HCl solution.

• What was CaCl2 percentage in original sample, if 21.48 mL of 0.09312 M NaOH was used to titrate excess HCl?

• During titration 21.48×0.09312=2.000 mmole HCl was neutralized. • Initially there was 25.00×0.9892=24.73 mmole of HCl used, so during

CaCO3 dissolution 24.73-2.000=22.73 mmole of acid reacted. • As calcium carbonate reacts with hydrochloric acid 1:2 (2 moles of

acid per 1 mole of carbonate), original sample contained 22.73/2=11.37 mmole of CaCO3, or 1.137 g (assuming molar mass of CaCO3 is 100.0 g).

• So original sample contained 1.137/1.435×100%=79.27% CaCO3 and 100.0-72.27%=20.73% CaCl2.

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