Use the figure below to answer the following questions. Show all work for full c

Question

Use the figure below to answer the following questions. Show all work for full credit. Write the neutralization reaction for this experiment. Calculate the concentration of acetic acid in flask A. Calculate the concentration of acetic acid in flask B. Calculate the concentration of acetic acid in flask C. Determine which concentration/flask should be used to calculate the percentage of acetic acid in our vinegar. Calculate the percentage of acetic acid. Assume the density of vinegar is 1.00g/mL.

Explanation / Answer

a)

CH3COOH(aq) + NaOH(aq) --> H2O(l) + NaCH3COO(aq)

b)

the concnetration of A

vinegar is usuallaly about 5% in volume/volume

M = 0.833 M

c)

If you add an extra 10 ml

VT = 10 + 10 = 20 ml

the concnetration dilutes to half

M2 = 1/2*M1 = 1/2*0.833 = 0.4165 M

d)

in flask C

after 0.4 M and V = 30 ml of base

mmol of base = 30*0.4 = 12 mmol of base

reaction:

CH3COOH(aq) + NaOH(aq) --> H2O(l) + NaCH3COO(aq)

ratio is 1;1:

0.833*0.1 = 0.0833 mol of acid - 0.012 mol of base = 0.0713 mol of acid left

VT = 10+10+30 = 50

M3 = (0.0713 )/(50/1000) = = 1.426 M

e)

You should use the first concnetration and the first volume

find moles = 0.0833 mol of acid

then; as stated before, the V/V% = 5 % of acetic acid in vinegar

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