A benzoic acid solution was prepared by adding 0.488 g benzoicacid to a 100 mL v

Question

A benzoic acid solution was prepared by adding 0.488 g benzoicacid to a 100 mL volumetric flask and then adding enough water tobring the total solution volume to 100.00 mL. How many millilitersof 0.0989 M NaOH are required to titrate the entirebenzoic acid solution to the equivalence-point? What is the pH atthe equivalence-point?

If the experiment were repeated using the same mass of benzoicacid and a 50 mL volumetric flask instead of a 100 mL volumetricflask, how many milliliters of 0.0989 M NaOH would berequired to titrate the entire benzoic acid solution to theequivalence-point?

Explanation / Answer

Part 1) 0.488grams benzoic acid = 3.996X10^-3 moles benzoic acid

3.996X10^-3 moles/(0.0989 moles/L) = 0.0404 L = 40.4 mL

pH at equivalence point will be above 7.0 because strong base (NaOH) converts weak acid (C6H5COOH) to its conjugate base (C6H5COO-).

To find pH, first find concentration of conjugate base:

3.996X10^-3 moles/(0.2 L + 0.0404L) = 0.0166 moles/L

Use ICE table to find OH- concentration (x):

C6H5COO- + H2O --> C6H5COOH + OH-

0.0166                          0                  0

-x                                 x                  x

0.0166 - x                     x                  x

Kb = ([C6H5COOH][OH-])/[C6H5COO-]

Kb = X^2/(0.0166 - X)

X is negligible on bottom of fraction

X = sqrt(0.0166Kb)

Kb = (1X10^-14)/Ka = (1X10^-14)/(6.3X10^-5) = 1.59X10^-10

X = 1.62X10^-6

pOH = -logX = 5.79

pOH + pH = 14

pH = 14 - pOH = 14 - 5.79 = 8.21

Part 2) The same volume would be required because the equivalence point is reached when moles of acid = moles of base

Volume = 40.4 mL

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