c161l09_09_21b

# c161l09_09_21b - Calculations with titrations Calculations...

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Calculations with titrations Calculations with titrations W Step 1: Balanced chemical reaction W Step 2: Convert volume of titrant into number of moles W Step 3: Find number of moles of analyte by stoichiometry W Step 4: Find concentration of analyte

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Example Example W Determine the concentration of oxalic acid (H 2 C 2 O 4 ) if 25.00 mL of oxalic acid is titrated against 0.500M NaOH and it took 38.0 mL to reach the equivalence point. Step 1: H 2 C 2 O 4 (aq)+2NaOH(aq) Na 2 C 2 O 4 (aq)+ H 2 O(l) Step 2: 38.0mL×0.500mol/L = 19.0 mmol NaOH used Step 3: 19.0 mmol NaOH × 1 mol H 2 C 2 O 4 /2mol NaOH = 9.5 mmol H 2 C 2 O 4 Step 4: 9.5 mmol H 2 C 2 O 4 / 25.00 mL = 0.380 mol/L
Dilution Dilution W Frequently we make one solution from another, more concentrated one. W We need to know how many moles are in the new solution n = MV W The concentration of the new solution is found from the number of moles divided by the new volume M=n/V W M 1 V 1 =M 2 V 2

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Acidic and Basic Periodic Acidic and Basic Periodic Properties Properties 18 1 2 13 14 15 16 17 Li Be B C N F
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## This note was uploaded on 11/12/2011 for the course CHEM 160:161 taught by Professor Siegel during the Fall '09 term at Rutgers.

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c161l09_09_21b - Calculations with titrations Calculations...

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