Quantitative Chemical Analysis 7e - UCSBdevries/chem150/lecture notes/ch 08-12… · 25.00 mL of...
Transcript of Quantitative Chemical Analysis 7e - UCSBdevries/chem150/lecture notes/ch 08-12… · 25.00 mL of...
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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