Experiment 29: Choice I



Experiment 29: Choice I

Acid-Base Titration

Goals: (1) To standardize a newly prepared solution of NaOH. Standardization is the process of determining the exact concentration of the solution. This is the case where concentrations are reported in at least 3 significant figures.

2) To determine the concentration of commercial vinegar.

Calculations: (1) Standardization

Equation 1: (MV)acid = (MV)base

Since: Mole of acid = mole of base, therefore:

Equation 2: moles of acid = Molarity of NaOH x Vol. of NaOH in L

Example: If 0.600 grams of KHP were used, then moles of KHP would be: mole KHP = 0.600 grams = 0.00294 mole

204.2 gms/mole

Then: 0.0029 mole = (MV)base

So: Molarity of NaOH = 0.00294 mole / volume of NaOH used in L

(2) Analysis of Unknown Acid: Commercial Vinegar

Given: CH3COOH + NaOH ( CH3COO- Na+ + H2O

Acetic Acid

To calculate for % Mass of Acetic Acid in Vinegar, do the following:

Since ratio of acid to base is 1:1, then

(MV)acid = (MV)base therefore,

A. To find moles of acetic acid in vinegar:

molesAcetic Acid = Mean Molarity of vinegar x Vol. of vinegar used in L

B. To Convert Molarity of Vinegar to grams

If: Molarity = [grams / molar mass] / Liter of solution

Then:

x grams Acetic Acid = MolarityAcetic Acid x Liter of soln. x Molar MassCH3COOH

Note: Molar Mass of CH3COOH (Acetic Acid) = 60.05 grams /mole

C. To find mass of vinegar

Given that the density of vinegar is 1.01 gm/mL

If Density = mass/volume, then

Mass of 5.00 grams of vinegar = Density x 5.00 mL = 5.05 grams

D. Calculate % Mass of Acetic Acid in vinegar

% by Mass of CH3COOH = Mass of CH3COOH x 100

mass of vinegar

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