1- A 30.00 mL wine sample is titrated with a 1.10×10 -1 M solution of sodium hyd
ID: 1052131 • Letter: 1
Question
1- A 30.00 mL wine sample is titrated with a 1.10×10-1M solution of sodium hydroxide. The equivalence point is determined to be 34.705 mL. What is the acidity of the wine expressed as (g tartaric acid / L; MW tartaric acid = 150.085 g/mole)?
2- The acidity of a wine sample is 1.15×101, expressed as (g citric acid / L wine; MW citric acid = 192.124 g/mole) . The sample is titrated with a 0.100 M solution of sodium hydroxide. The equivalence point is determined to be 23.391 mL. What is the volume of the wine sample analyzed in mL?
Explanation / Answer
1.
volume of wine sample = 30 ml
no of mole of NaOH = M * V
= 1.1*10^-1*34.705
= 3.82 mmole
1 mole wine (tartaric acid ) = 2 mole NaOH
so that,
no of mole of wine = 3.82/2 = 1.91 mole
concentration of wine = 1.91/30 = 0.064 M
strength of tartaricacid = M * 150.085
= 0.064 * 150.085
= 9.6 g/l
2.
no of mole of NaOH = M * V
= 0.1*23.391
= 2.34 mole
acidity of a wine sample = 11.5 g/L
COncentration of wine = acidity/Mwt
= 11.5/192.124
= 0.06 M
volume of sample analyzed = ?
citric acid is triprotic acid
1 mole citric acid = 3 mole NaOH
M1V1/n1 = M2V2/n2
(0.06*V1/1) = (0.1*23.391/3)
volume of sample analyzed ( V1 ) = 13 ml
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