Question
In: Chemistry
A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of...
A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of vinegar. If you titrate a 10.00 mL sample of undiluted vinegar, what volume (in mL) of 0.1014 M NaOH should you expect to need to use to reach the endpoint of the titration?
Solutions
Expert Solution
molarity = W*1000/G.M.Wt * volume in L
= 5.38*1000/60*100
= 0.896M
CH3COOH + NaOH------------> CH3COONa + H2O
1 mole 1 mole
CH3COOH NaOH
M1 = 0.896M M2 = 0.1014M
V1 = 10ml V2 =
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 0.896*10*1/0.1014 = 88.36ml of NaOH >>>>answer
queen_honey_blossom answered 1 year agoRelated Solutions
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I'm a little confused-if you solve it can you be sure to specify
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