Molarity of standard NaOH solution: x/1 H 2/3 Unknown + 2/3 NaOH -> 2/3 H2O +Na

Question

Molarity of standard NaOH solution: x/1 H 2/3 Unknown + 2/3 NaOH -> 2/3 H2O +Na 2/3 Acetate

Unknown Acid code and number of reactive hydrogens: H+3

1. Find the volumes of acid solution (vinegar) and base solution used in each of your vinegar titrations.

2. Find the moles of base (NaOH) used in each of your vinegar titrations.

3. Using stoichiometry and the balanced equation, find the moles of acetic acid used in each of your vinegar titrations.

Vinegar: Titration 1 Titration 2 Tirtration 3 Initial V of Acid .01 mL 5.56 mL 11.32 mL Final V of Acid 5.56 mL 10.71 mL 16.02 mL Inital V of Base .01 mL 27.12 mL 3.01 mL Final V of Base 22.23 mL 44.71 mL 24.90 mL

Explanation / Answer

x/1 H 2/3 Unknown + 2/3 NaOH -> 2/3 H2O + Na 2/3 Acetate

1. In first titration -

Volume of acid = V(final) - V(Initial)

= 5.56 - 0.01

= 5.55 mL

= 0.00555 L

Volume of base = 22.23 - 0.01

= 22.22 mL

= 0.02222 L

In second titration -

Volume of acid = 10.71 - 5.56

= 5.15 mL

= 0.00515 L

Volume of base = 44.71 - 27.12

= 17.59 mL

= 0.01759 L

In third titration -

Volume of acid = 16.02 - 11.32

= 4.7 mL

= 0.0047 L

Volume of base = 24.90 - 3.01

= 21.89 mL

= 0.02189 L

2. Molarity of NaOH = 2/3 M

In first titration -

Moles of NaOH = 0.02222 * 2/3

= 0.00370 moles

In second titration -

Moles of NaOH =  0.01759 * 2/3

= 0.0117 moles

In third titration -

Moles of NaOH = 0.02189 * 2/3

= 0.0146 moles

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