Determine the amount of tartaric acid in a sample of wine... 25.0 ml of wine is titrated with 0.0122 M NaOH. The tiration requires 12.46 ml of NaOH. What is the molarity of tartaric acid in the wine?

Answer 1

molar mass = mass / moles Molarity = mole of solute/ L of solution Therefore, mole of solute = 0.1008 M * 0.0150 L = 0.001512 mole molar mass of unknown acid= 0.2053g / 0.001512 mol =136 g/mol

Answer 2

Well, i done titration at school doing this... lol.

It come out at something like; 0.9879M to 0.1098M

But done with 20MLs of White Wine, and 0.2M NaOH.

Answer to u is,

C=(0.0122x12.46)/25

C=0.00608048 M

This seems Really low, Are u sure u have the right Amounts? Or i have done mine wrong.

CNaOH*VNaOH = C<wine>?*V<wine>

:. (CNaOH*VNAOH)/V<wine>

yes?

C=Concertration (M)

V= Volume

Or is my Method Correct?

If amounts of NaOH shouldn't effect the end result as it all goes into [CNaOH*VNaOH = C<wine>?*V<wine>] If it is less strength, u will need less wine for it to change yes?

Answer 3

moles NaOH used = 0.02196 L x 0.456 mol/L = 0.01001 moles
moles acid present = 0.01001 moles b/c NaOH + HA ==> NaA + H2O with a 1:1 ratio of NaOH to HA
molarity of acid = 0.01001 moles/0.01694 L = 0.5911 M = 0.591 M (to 3 significant figures)

Answer 4

126g/mol

Answer 5

0.12