If you add 0.5321 g of benzoic acid to a 100 ml volumetric flask and then add en

Question

If you add 0.5321 g of benzoic acid to a 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 will be required to completely neutralize the benzoic acid (pK_a = 4.204)? At the equivalence point, will the solution be somewhat acidic, neutral, or somewhat basic? If you redo the experiment in question 1 using a 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? If you accidentally overshoot the neutralization point in the experiment described in question 1 and add 0.40 ml too much of the NaOH solution, what will be the final solution pH? Considering your answers to questions 1 and 3 carefully, which of the stock pH indicators (methyl orange, thymol blue, or phenolphthalein) do you want to use for your titrations with the indicator dye? Explain your reasoning. You need to determine both the molar mass and the pK_a of your unknown acid. On which of these experiments will you use the pH meter? (Remember, you can only use the pH meter on one experiment.) Please explain your decision.

Explanation / Answer

1)

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 equivalence point solution is basic. because it is a salt of strong base and weak acid

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