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Corr 2018 SRC Analytical Chemistry Module 5 JG

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9/7/2018

Distribution plots – Alpha values

Module 5 – Acid-Base Titration


𝐻+ 3 + 𝐾𝑎1 𝐻+ 2 + 𝐾𝑎1 𝐾𝑎2 𝐻+ + 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3
𝐶𝐻3 𝑃𝑂4 = 𝐻3 𝑃𝑂4
𝐻+ 3

Jonyl L. Garcia, RCh., M.Sc. 𝐻2 𝑃𝑂4− = 𝐻3 𝑃𝑂4 𝐾𝑎1 / 𝐻+


𝐻𝑃𝑂42− = 𝐻3 𝑃𝑂4 𝐾𝑎1 𝐾𝑎2 / 𝐻+ 2

𝑃𝑂43− = 𝐻3 𝑃𝑂4 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 / 𝐻+ 3


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Triprotic acid with Strong Base Triprotic acid with Strong Base
𝑎𝑐𝑖𝑑
Region Constituents Equations / Remarks
𝑉𝐻3 𝑃𝑂4 𝑥𝑀𝐻3 𝑃𝑂4 − 𝑉𝑁𝑎𝑂𝐻 − 𝑉𝐸𝑃1 𝑥𝑀𝑁𝑎𝑂𝐻𝑥𝑆𝐹
=
A. Initial point H3PO4 𝐻+ = 𝐾𝑎1 ∙ 𝐶𝐻3 𝑃𝑂4 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙.
D. Second Buffer
𝑉𝐻3 𝑃𝑂4 𝑥𝑀𝐻3 𝑃𝑂4 − 𝑉𝑁𝑎𝑂𝐻𝑥𝑀𝑁𝑎𝑂𝐻𝑥𝑆𝐹 H2PO4- + HPO42- 𝑉𝑁𝑎𝑂𝐻 − 𝑉𝐸𝑃1 𝑥𝑀𝑁𝑎𝑂𝐻𝑥𝑆𝐹
𝑎𝑐𝑖𝑑 = Region 𝑠𝑎𝑙𝑡 =
𝑉𝐻3 𝑃𝑂4 + 𝑉𝑁𝑎𝑂𝐻 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙.

𝑉𝑁𝑎𝑂𝐻 𝑥𝑀𝑁𝑎𝑂𝐻𝑥𝑓 𝑠𝑎𝑙𝑡


B. First Buffer Region H3PO4 + H2PO4- 𝑠𝑎𝑙𝑡 = 𝑝𝐻 = 𝑝𝐾𝑎2 + 𝑙𝑜𝑔
𝑉𝐻3 𝑃𝑂4 + 𝑉𝑁𝑎𝑂𝐻 𝑎𝑐𝑖𝑑
E. Second HPO4 + H2O
2-
𝐻+ = 𝐾𝑎2 𝐾𝑎3
𝑠𝑎𝑙𝑡 Equivalence Point (Na2HPO4)
𝑝𝐻 = 𝑝𝐾𝑎1 + 𝑙𝑜𝑔
𝑎𝑐𝑖𝑑
C. First Equivalence H2PO4- + H2O
𝐻+ = 𝐾𝑎1 𝐾𝑎2
Point (NaH2PO4)

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9/7/2018

Triprotic acid with Strong Base


𝑎𝑐𝑖𝑑
𝑉𝐻3 𝑃𝑂4 𝑥𝑀𝐻3 𝑃𝑂4 − 𝑉𝑁𝑎𝑂𝐻 − 𝑉𝐸𝑃2 𝑥𝑀𝑁𝑎𝑂𝐻𝑥𝑆𝐹
=
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙.
F. Third Buffer
HPO42- + PO43- 𝑉𝑁𝑎𝑂𝐻 − 𝑉𝐸𝑃2 𝑥𝑀𝑁𝑎𝑂𝐻𝑥𝑆𝐹
Region 𝑠𝑎𝑙𝑡 =
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙.

𝑠𝑎𝑙𝑡
𝑝𝐻 = 𝑝𝐾𝑎2 + 𝑙𝑜𝑔
𝑎𝑐𝑖𝑑
𝑉𝑁𝑎𝑂𝐻 − 𝑉𝐸𝑃3 𝑥𝑀𝑁𝑎𝑂𝐻 𝑥𝑆𝐹
𝐶𝑆𝑎𝑙𝑡 =
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙.
G. Third Equivalence PO43- + H2O
𝑂𝐻− = 𝐾𝑏1 ∙ 𝐶𝑆𝑎𝑙𝑡
Point (Na3PO4)
𝐾𝑤
𝐾𝑏1 =
𝐾𝑎3

H. After All Acid is 𝑉𝑁𝑎𝑂𝐻 − 𝑉𝐸𝑃3 𝑥𝑀𝑁𝑎𝑂𝐻


Excess NaOH 𝑂𝐻− =
Consumed 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙. Figure 2-21cd p39

Figure 2-21ab p39 Figure 2-21ef p39

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Effect of Reactant Concentration Effect of Reactant Concentration

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Effect of Reaction Completeness Polyprotic acids

A – H3PO4
B – H2C2O4
C – H2SO4

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Identifying Titration Curves


HA w/ NaOH HA 1 + HA 2 w/ NaOH Na 2 CO3 w/ HCl

Ka HA1 >> KaHA2


HA 1 > HA 2

HCl w/ NaOH NaHCO3 + Na 2 CO3 w/ HCl BOH w/ HCl

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Acid – Base (Neutralization) – Mixed Alkali


 OH-. CO3-2, HCO3-,
OH- + CO32-, CO32- + HCO3-
 Makes use of two
indicators: alkaline range
(e.g. phenolpthalein
[PHTH]) and acidic range 𝒙 𝒎𝒐𝒍𝑨 𝟏
%𝑨𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 = 𝑽𝑯𝑪𝒍 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 × 𝑴𝑯𝑪𝒍 × × 𝑴𝑾𝑨 × × 𝟏𝟎𝟎
𝒚 𝒎𝒐𝒍𝑯𝑪𝒍 𝒎𝒔𝒂𝒎𝒑𝒍𝒆
(e.g. bromocresol green
[BCG], methyl red [MR],
methyl orange [MO]) 𝒙 𝒎𝒐𝒍𝑨 𝟏
𝒑𝒑𝒎𝑨 = 𝑽𝑯𝑪𝒍 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 × 𝑴𝑯𝑪𝒍 × × 𝑴𝑾𝑨 ×
𝒚 𝒎𝒐𝒍𝑯𝑪𝒍 𝑽𝒔𝒂𝒎𝒑𝒍𝒆

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Sample Problem - 1 Sample Problem - 2


A 1.200-g sample containing NaOH (40 g/mol), NaHCO3 A solution contains NaHCO3, Na2CO3, and NaOH, either
(84.01 g/mol), Na2CO3 (105.99 g/mol), alone or in any alone or in permissible combination. Titration of a 50.00-
possible combinations plus some inert impurities is mL portion to a phenolphthalein end point requires 22.10
dissolved and titrated cold with 0.5000 M HCl with phth mL of 0.100 M HCl. A second aliquot requires 48.40 mL
as indicator, the soln turns colorless after addition of of the HCl when titrated to a bromocresol green end
30.00mL of acid. MO is then added and 5.00mL more of point. Determine the composition and the molar solute
the acid was required to reach its end point color. What concentrations of the original solution,
is/are the constituent/s of the sample and their
percentages?

NaOH: 41.67% Na2CO3: 0.0442 M


Na2CO3: 22.08% NaHCO3: 8.400 x10 -3 M
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Sample Problem - 3
 A series of solutions containing NaOH (FM 39.997),
Na3AsO4 (FM 207.89) and Na2HAsO4 (FM 185.91),
alone or in compatible combination, was titrated
with 0.08601 M HCl. For a 25.00-mL aliquot of the
sample, 16.00 mL of the acid was required to reach the
phenolphthalein end point. Another 25.00-mL aliquot of
the sample required 37.37 mL of the acid to reach the
end point with methyl orange. Find the composition as
well as mg/mL of each component in the sample.

Na3AsO4: 11.44 mg/mL


Na2HAsO4: 3.435 mg/mL
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