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Question

1. If you add 0.5321 g of benzoic acid 100 ml volumetric flask and then add enough water t0 bring the total volume to 100.00 ml; how many ml of 0.2154 M NaOH solution will be required completely neutralize the benzoic acid (pKa = 4.204) the equivalence point, will the solution be somewhat acidic, neutral, or somewhat basic?

2. If you redo the experiment in question using 50 ml volumetric flask instead of a 100 ml volumetric flask (but the same amount of benzoic acid), how many ml of 0.2154 M NaOH solution will be required to completely neutralize the benzoic acid?

3. If you accidentally overshoot the neutralization point the experiment described question and add 0.40 ml too much of the NaOH solution; what will be the final solution pH?

4. Considering your answers to questions and 3 carefully, which of the stock pH indicators (methyl orange, thymol blue, phenolphthalein) do you want to use for your titrations with the indicator dye? Explain your reasoning.

5. You need to determine both the molar mass and the pK of your unknown acid. On which these experiments will you use the pH meter? (Remember you can only use the pH meter on one experiment) Please explain your decision.

Answer “If you add 0.5321 g of benzoic acid 100 ml volumetric flask and then add enough water to bring the total volume to 100.00 ml; How many ml of 0.2154 M NaOH solution…”.

Neutralization Reaction:

Acid-base neutralization is the reaction where the acid and base react to each other’s properties and create a salt in their final solution. For example, a salt (NaCl) formed in the solution of an acid (HCl) and a base (NaOH).

HCl+NaOH→NaCl+H₂O

Answer:

Moles of benzoic acid = 0.5321 / 122.12 = 4.357 x 10^-3

Volume = 100 mL

Concentration = 4.357 x 10^-3 / 0.1 = 0.04357 M

At neutralization point :

Moles of acid = Moles of base

0.04357 x 100 = 0.2154 x V

=> V = 20.23 mL

Volume of NaOH added = 20.23 mL

At the equivalence level solution, it is basic. It is a salt of weak acid and strong base.

Last words

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