In this Acid-Base Titration, find the molarity of NaOH from the given values. Th

Question

In this Acid-Base Titration, find the molarity of NaOH from the given values. Then find the molarity of Acetic Acid from the given values. Show all calculations.

1. mass of KC8H5O4 : 0.497g

volume of (0.3M) NaOH solution added : 4.15mL

Molarity of NaOH : ____________

2. volume of vinegar : 10.00mL

volume of NaOH added : 11.55mL

Molarity of Acetic Acid : ___________

Mass percent of acetic acid (C2H4O2) in sample (assuming density of vinegar is 1.0g/mL) : ___________

If your standardized NaOH was used to titrate 20.00 mL of H2SO4, what concentration of H2SO4 would you determine if 24.66 mL of the NaOH solution was required by the titration? Give the balanced equation for this reaction.

Explanation / Answer

1)

Molarity of NaOH , as stated before is 0.3 M i.e. 0.3 mol of NaOH per liter

Q2

For the acetic acid:

mol of base = mol of acid

mol of base = MV = 0.3*11.55 = 3.465 mmol of OH-

so mol of H+ = 3.465 mmol

[Acid] = mmol/V = 3.465/10 = 0.3465 M

So

% mass = mass of aci d/ total Volume

mass of acid = mol*MW = (3.465 *10^-3)(60) = 0.2079 g of acid

% = 0.2079 /10 *100 = 2.079% is acetic acid

Q3.

mol of NaOH = MV = 0.3*24.66 = 7.398 mmol of NaOH

so

mmol of H+ = 7.398 mmol

2 mol of H+ per mol of H2SO$

so

H2SO4 = 1/2*7.398 = 3.699 mmol of H2SO4

M = mmol/mL = 3.699 / (20) = 0.18495M

H2SO4 + 2NaOH = Na2SO4 + 2H2O

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