Archive for March 31st, 2008

Mass of Acetic Acid in Vinegar Using Titration Method

Chemistry Tutor Site Question

The balance reaction of acetic acid that reacts with sodium hydroxide is,

HC2H3O2 (aq) + NaOH (aq) ——–> H2O (l) + NaC2H3O2 (aq)

If 2.50 mL of vinegar needs 34.9 ml of 0.0960 M NaOH to reach the equivalence point in titration, how many grams of acetic acid are in a 1.00-qt sample of this vinegar?

Chemistry Tutor Site Answer

This is an example of titration weak acid with strong base. Acetic acid is a weak acid and sodium hydroxide is a strong base. At the end point the mol equivalent of acid will be equal to the mol equivalent of base. Use this formula when you solve problem that involve titration calculation:

N x V acid = N x V base
n x M x V (acid) = n x M x V ( base)

Where n is the total amount of ion H+ in acid or OH- in base, N is Normality, M is Molarity, and V is Volume. Thus for the question above we have:

1 x M x 2.50 mL = 1 x 0.0960 M x 34.9 mL
M = 1.340 M
This means that in 1 L of vinegar contains 1.340 mol of acetic acid. Multiply with its molar mass thus we can get the mass of acetic acid in the 1 L of vinegar.

mass of acetic acids = mol x molar mass
mass of acetic acids = 1.340 mol x 60 =
mass of acetic acids = 80.4 g
because 1 quart, liquid (U.S.) is equal to 0.946 liter, thus the mass of acetic acid per 1.00 qt ia
= 80.4 g /L x 0.946 L/qt
= 76.06 g/qt
Thus there are 76.06 g of acetic acid per 1 qt of vinegar

Add comment March 31st, 2008

Volume of Solution in the End Point of H2SO4 and NaOH Titration

Chemistry Tutor Site

How much of total volume in the conical flask at the end point of titration, when 0.1 M of sulphuric acid (H2SO4) is used to titrate 20 mL of 0.2 M of sodium hydroxide (NaOH)?

Chemistry Tutor Site

This is an acid-base titration that involve sulphuric acid (H2SO4) and sodium hydroxide (NaOH). H2SO4 is a strong acid and NaOH is a strong base, in the end point the pH of the solution will be 7 or neutral. The balance reaction is:

H2SO4(aq)+2NaOH(aq) –> 2H2O (l) +Na2SO4(aq)

Always use this formula when you calculate titration problem, in the end point the mole equivalent titrant is equal to the mol equivalent of titer, thus we can write:

N1V1=N2V2
n1.M1.V1 (acid) = n2.M2.V2 (base)

Where are n1 or n2 are the amount of H+ in acid or OH- in the base. In this case n1 is 2 because H2SO4 has 2 ion H+, and n2 is 1 because NaOH has one OH-, thus

n1.M1.V1 (acid) = n2.M2.V2 (base)
2×0.1xV1 = 1 x 0.2 x 20
V1 = 20 mL

So the total volume in the end point is 20 mL + 20 mL = 40 mL

Add comment March 31st, 2008


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