Analytical Chemistry Guided Inquiry Activity Predicting Acid/Base Titration Curv

Question

Analytical Chemistry Guided Inquiry Activity Predicting Acid/Base Titration Curves Titration 2: 0.1069 g Na CO3 are dissolved in a 100 mL volumetric flask which is filled to the mark with DI. The full 100 mL are transferred to a 250 mL Erlenmeyer, and then titrated with 0.0532 M HCI. a) Titrand species: b) Titrant species: c) Molecular equation to describe reaction of titrand and titrant species. d) Total ionic equation to describe reaction of titrand and titrant species. e) Net ionic equation to describe reaction of titrand and titrant species. f) Titrand milli-moles l) Titrant milli- equivalents at equivalence m) Titrant milli- moles at equivalence g) Titrand milli- equivalents h) Titrant volume added at (final) equivalence pt. (mL) i) Total volume at (final) equivalence (mL) j) [Na') at equivalence 86 F4 F3 FS F6 FB

Explanation / Answer

Titration Na2CO3 with HCl

molar concentration of Na2CO3 solution = 0.1069 g/106 g/mol x 0.1 L = 0.0101 M

a) Titrand species = CO3^2- from Na2CO3

b) Titrant species = H+ from HCl

c) molecular equation,

Na2CO3- + 2HCl <==> H2CO3 + 2NaCl

d) ionic equation

2Na+ + CO3^2- + 2H+ + 2Cl- <==> H2CO3 + 2Na+ + 2Cl-

e) net-ionic equation,

CO3^2- + 2H+ <==> H2CO3

f) Titrand millimoles = 0.0101 M x 100 ml = 1.01 mmol

g) Titrand milliequivalents = 1 meq

h) Titrant volume at equivalence point = 2 x 1.01 mmol/0.0532 M = 37.97 ml

i) Total volume at equivalence point = 137.97 ml

j) [Na+] at equivalence = 2 x 1.01 mmol/137.97 ml = 0.015 M

l) Titrant milliequivalents = 2 meq

m) Titrant millimoles at equivalence point = 2 x 1.01 mmol = 2.02 mmoles

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