Chemistry questions?

Question

A sample of vinegar is titrated with a sodium hydroxide solution to find the molarity of acetic acid. If 18.82mL of a 0.430M NaOH solution is required to titrate a 10.00mL vinegar solution, what is the molarity of acetic acid in the vinegar?

WHOAAA... I'm a little confused... Please EXPLAIN so I can understand in the future. Thanks

Answer 1

Simply use this Equation:

MaVa=MbVb

Ma = Molarity of the acid = ?

Va = Volume of the acid = 10.00mL

Mb = Molarity of the base = 18.82mL

Vb = Volume of the base = 0.430M

Since we are solving for Ma we can rearrange the eqn to:

Ma = (Mb)(Vb) / Va

Ma = (0.430M)(18.82mL) / 10.00mL

Thats it!  Simple eh?

The only catch is you can only use MaVa=MbVb when the H+ and OH- are in 1:1 ratio such as in this case.  

HC2H3O2 + NaOH --->  H2O + NaC2H3O2

Answer 2

CH3COOH + NaOH --------> CH3COONa + H2O

1 mole of acetic acid react with 1 mole of NaOH

18.82 ml of 0.43M NaOH = (18.82/1000)L * 0.43 mole/L= 0.0081 mole of NaOH = 0.0081 mole of acetic acid

10 ml of vinegar have 0.0081 mole of acetic acid.

molarity = 0.0081 mole / (10/1000)L = 0.81 M

Answer 3

I won't get too detailed, but the point of a titration is to add a known volume of NaOH (in this case) to reach the equivalence point (color change from indicator), which is the point in which the number of protons from acetic acid is balanced by the number of hydroxides from NaOH.  Calculate the number of moles of NaOH you added and go from there.