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?

Answer 1

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