Acid and Base Reaction Question?

Question

A dilute solution of acetic acid is called vinegar. Acetic aced reacts with sodium hydroxide according to the equation

A 5.00 mL sample of vinegar needed 42.5 mL of 0.115M NaOH to reach the equivalence point

a. Calculate the molarity of the vinegar

b. Calculate the percent (m/v) of vinegar in the sample

c. What type of Chemical Reaction does this represent?

Answer 1

a. First find moles of NaOH. .0425 x .115M = 4.9 x10^ -5

The moles of the NaOH is equal to the value of vinegar.

Divide by liters of vinegar. .005 L.

4.9x106 -5 / .005 = .98 M

b. I'm not quite sure about b... Sorry... =(

c. This is an Acid Base Neutralization.

Answer 2

1. Write a balanced equation

This is a simple neutralization (a type of double displacement, metathesis reaction) and the ratio is 1 mole to 1 mole of the

reactants

CH3COOH + NaOH - CH3COONa + H2O

first find the moles of NaOH you have:

0.0425L X 0.115 mol / 1L = 0.0048875 mol NaOH

The ratio is 1:1 so you need 0.0048875 mol acetic acid

a. You now the moles and volume now use the formula, molairty(M) = moles/L = 0.0048875 mol / 0.005L = 0.9775 M

There fore the molarity of the vinegar is  0.9775 M

b. The mixed percent is mass(g)/volume(L) X100.

To get the mass we now 1 mole of acetic acid is 60.058, its "molar mass"

we have 0.0048875 mol, so multiply my the molar mass you get 0.2935334 grams. To find the percent, take the mass of the viniger and divide it by the volume of the solution in liters 0.2935334 grams / 0.005L = 58.7

c. This represents a special type of double displacement reaction called a "NEUTRALIZATION" reaction in which acid and base react to form salt an water.

Hear you were actually doing a titration to determine the pH of an unknown sample of acetic acid

GOOD LUCK!