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JulieT JulieT
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13 years ago
Lactic acid, HC3H5O3, is a monoprotic acid that forms when milk sours. An18.5 mL sample of a solution of lactic acid required 17.25 mL  of 0.155 M NaOH to reach an end point in a titration. How many grams of lactic acid were in the sample?
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#1
13 years ago
moles NaOH = 0.01725 L x 0.155 M=0.00267
because the lactic acid is a momoprotic acid
moles lactic acid = 0.00267
mass lactic acid = 0.00267 mol x 90.08 g/mol=0.241 g
wrote...
#2
13 years ago
CH3CH(OH)COOH(aq)+NaOH(aq)
->CH3CH(OH)COONa(aq)+H2O(l)

No of moles of NaOH
= [NaOH]x(volume of NaOH in L)
= 0.155x17.25x10^-3
= 2.67375x10^-3

The required ratio of the no of moles of CH3CH(OH)COOH to that of NaOH is 1:1.
Therefore, the no of moles of CH3CH(OH)COOH required
= no of moles of NaOH
= 2.67375x10^-3

Relative Molecular Mass(RMM) of CH3CH(OH)COOH= 90
1 mole of CH3CH(OH)COOH= 90g
2.67375x10^-3 moles of CH3CH(OH)COOH
= 90x2.67375x10^-3
= 0.241g(mass of lactic acid in the sample)

I hope this helps and feel free to send me an e-mail if you have any doubts!
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