Question

In: Chemistry

Consider the titration of 100.0 mL of 1.00 M HA (Ka=1.0*10^-6) with 2.00 M NaOH. A....

Consider the titration of 100.0 mL of 1.00 M HA (Ka=1.0*10^-6) with 2.00 M NaOH.

A. Determine the pH before the titration begins

B. Determine the volume of 2.00 M NaOH required to reach the equivalence point

C. Determine the pH after a total of 25.0 mL of 2.00 M NaOH has been added

D. Determine the pH at the equivalence point of the titration.

Thank you in advance!

Expert Solution

A) Before the titration

pH = 1/2[pKa -log C]

= 1/2[ 6 -log 1.0]

=3.0

B) At equivalence

mmoles of acid = mmoles of base

100x1.00 = V x2.00

Thus V = 50mL

volume of 2.00M NaOH = 50mL

C) When 25mL of NaOH is added, it is hhalf equivalence point

At half equivalence

HA + NaOH ----------------> NaA + H2O

100x1.00 25x2 0 0 iniital mmoles

50 0 50 - after reaction

So the solution is a buffer whose pH is given by Hendersen equation

pH = pka + log [conjugate base] / [acid]

= 6 + log 50/50

=6

D) ph at equivalence point

HA + NaOH ----------------> NaA + H2O

100x1.00 50x2 0 0 iniital mmoles

0 0 100 - after reaction

Concentration of salt = mmoles/volume

= 100/150

The solutiion is now of a salt of weak acid with strong base which is basic and whose pH is calculated as

pH = 1/2[pKw + pKa + logC]

= 1/2 [ 14 +6+ log 10/15]

=9.912

queen_honey_blossom answered 2 years ago

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