a 30 mot sampl of vinegar is treated with .4190M NaOH. if titration require a 27.83...
a 30 mot sampl of vinegar is treated with .4190M NaOH. if titration require a 27.83 no of NaOH what is the concentration of acetic acid
Homework Answers
To find the concentration on molarity of acetic acid
Given that
30 mL sample of vinegar
27.83 ml of a 0.4190 M NaOH
Sodium hydroxide (NaOH) reacts with acetic acid to give sodium acetate and water,
NaOH + CH3COOH -----> CH3COONa + H2O
first lets find the moles of NaOH
27.83 mL = 0.02783 L
moles = molarity x volume
= 0.4190 mol/L x 0.02783 L
= 0.0116 moles
Find the moles of CH3COOH
there is NaOH and CH3COOH is 1:1 mole ratio
so moles of CH3COOH = 0.0116 moles
find the concentration on (molarity) of CH3COOH
Molarity = moles of solute / liter of solution
30 ml = 0.03 L
= 0.0116 moles / 0.03 L
= 0.386 M
hence the concentration of acetic acid in the vinegar = 0.386 M
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For vinegar titration in this experiment, would the following errors cause the % acetic acid in vinegar determined to be...
For vinegar titration in this experiment, would the
following errors cause the % acetic acid in vinegar determined to
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information to answer. Assume that the mistake given is the only
mistake made in the experiment. EXPLAIN.
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For vinegar titration in this experiment, would the
following errors cause the % acetic acid in vinegar determined to
be (A) too large, (B) too small, (C) no effect, or (D) not enough
information to answer. Assume that the mistake given is the only
mistake made in the experiment. EXPLAIN.
1.) The burst is rinsed with distilled water, but not NaOH solution
before the vinegar sample is titrated.
2.) After standardization, the NaOH solution is not stoppered and
obsorbs CO2...
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