Chp2_titrationstoich

# Chp2_titrationstoich - mL mmol x mL mmol 100 00 10 10 53 2...

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Titration Example From problem 67, page 58. See statement of the problem there. Generally, the key to these problems is to work oneself backwards to the units one requires. Here, we need mg Ca 2+ / mL of blood; the problem provides 10.00mL blood sample, so we only need mg Ca 2+ and are given: (1) a balanced stoichiometric reaction and (2) information on the volume and concentration of a KMnO 4 solution. So, our logic path is: KMnO 4 > C 2 O 4 2– > CaC 2 O 4 > Ca 2+ mmol MnO 4 : mmol mL mmol mL 2 3 10 01 . 1 10 86 . 8 14 . 1 - - × = × × mmol C 2 O 4 2– : mmol MnO mmol O C mmol 2 4 2 4 2 2 10 53 . 2 2 5 10 01 . 1 - - - - × = × × mmol CaC 2 O 4 : mmol O C mmol O CaC mmol mmol 2 2 4 2 4 2 2 10 53 . 2 1 1 10 53 . 2 - - - × = × × This is the number of mmol of calcium oxylate precipitated from the 10.00 mL aliquot. Since that aliquot was taken from a solution of 100 mL total volume, the amount (of mols) calcium oxylate in the original solution may be determined:
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Unformatted text preview: mL mmol x mL mmol . 100 00 . 10 10 53 . 2 2 = ×-mmol x 1 10 53 . 2-× = ‘x’ represents the number of mmol calcium oxylate precipitated from the original blood sample. Since there is a one-to-one mole ratio of Ca 2+ to calcium oxylate, this is also the number of mmol of Ca 2+ ion present in the original 10.00 mL blood sample. Thus, to determine the mass of Ca 2+ we simply convert using the atomic mass of calcium (the two electrons lost making a negligible contribution): + +-= × × 2 2 1 1 . 10 1 . 40 10 53 . 2 Ca mg Ca mmol Ca mg Ca mmol And the final concentration: blood mL Ca mg blood mL Ca mg / 01 . 1 . 10 1 . 10 2 2 + + =...
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