Titration Calculation (1 Viewer)

[xwrathbringerx]

The acidity of a particular white wine was determined by titrating 25 mL of the wine with 0.0511 mol/L NAOH solution; 8.7 mL was required. Calculate the molarity of hydrogen ions in the wine. Assume that the hydrogen ions come entirely from diprotic tartaric acid and calculate the concentration of tartaric acid in the wine in grams per 100 mL.

I got 26.7 g/ 100 mL as my answer BUT the textbook says it's meant to be 0.133 g/100 mL.

Could someone please show me how to do this problem properly?

 

[js992]

Tartaric Acid is C4H6O6

8.7ml of 0.0511mol/L NaOH = 4.4567x10^-4 moles.

Since NaOH is monoprotic (1 mole of OH per mole of NaOH)
and tartaric acid is diprotic,
1 mole of tartaric acid neutralises 2 moles of NaOH

As you used 4.4567x10^-4 moles of NaOH , that means in the wine, there was only half as many moles of tartaric acid

==> 2.22285 x 10^-4 moles of C4H6O6

Times by the molar mass of C4H6O6 (150.088)
gives you 0.03336231108 grams of tartaric acid.
This is the amount in 25 mL, times by 4 to get /100mL

= 0.133449g/100mL