Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. the volume of added base required to reach the equivalence point c. the pH at 5.0 mL of added base d. the pH at one-half of the equivalence point e. the pH at the equivalence point f. the pH after adding 5.0 mL of base beyond the equivalence point

Explanation:

Given that:

Concentration of = 0.105 M

Volume of   = 20.0 mL

Concentration of = 0.125 M

The  chemical reaction can be expressed as:

Using the ICE Table to determine the equilibrium concentrations.

I            0.105                                     0                  0

C              -x                                         +x                +x

E            0.105 - x                                  x                  x

Recall that the ka for

Then;

By solving the above mathematical expression;

x = 0.00137 M

pH = 2.86

Hence, the initial pH = 2.86

b)  To determine the volume of the added base needed to reach the equivalence point by using the formula:

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of molarity × volume

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH =

After reacting with 5.0 mL NaOH, the number of moles is as follows:

Initial moles                   0                      0

F(moles)    0              

The pH of the solution is then calculated as follows:

Recall that:

pKa for

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

d)

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the with

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of = moles/volume

=

= 0.0571 M

Now, using the ICE table to determine the concentration of ;

I              0.0571                                0                      0

C              -x                                       +x                     +x

E             0.0571 - x                             x                       x

Recall that the Ka for =

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

Before                             0.0021              0.002725         0

After                                   0                     0.000625        0.0021

From above; we can determine the concentration of by using the following method:

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

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