For the analysis of vinegar part of the experiment one pair used 16.31 mL of the

Question

For the analysis of vinegar part of the experiment one pair used 16.31 mL of their 0.1016 M NaOH to reach the end point of their titration of 2.00 mL of vinegar. How many moles of NaOH were used? How many moles of acetic acid were present in the 2.00 mL sample of vinegar? How many grams of acetic acid (MM 60.05 g/mol) is this? What is the w/v% concentration of acetic acid in their sample? Again volumes should be to 2 decimal places. The concentration of the NaOH should have 4 decimal places. moles moles

Explanation / Answer

moles of NaOH= molarity* volume in L, 1000ml= 1L, hence moles of sodium hydroxide = 0.1016*16.3/1000=0.001656

the reaction between NaOH and acetic acid (CH3COOH) is CH3COOH+ NaOH-------->CH3COONa+ H2O

1 mole of acetic acid requires 1 mole of sodium hydroxide for complete neutralization

0.001656 moles of NaOH hence requires 0.001656 moles of acetic acid for neutralization.

mass of acetic acid = moles* molar mass =0.001656*60.05 gm =0.09945 gm

this many gram of acetic acid is present in 2 ml of solution.

hence W/V% of acetic acid = 100*0.009945/2 =4.9725%

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