50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the amino acid alanine are 2.344 ( p K a1 ) and 9.868 ( p K a2 ) , which correspond to the carboxylic acid and amino groups, respectively. a) how do you calculate the PH of the first equivalent and b) the second equivalent? please help

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

     ------>  

1 mole of  alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL  of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via  the following process as shown below;

=

=

=

= 0.08 M

Alanine serves as an intermediary form, however the concentration of and the pH can be determined as follows;

=

=

=

=  

pH = - log

pH =

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

          ----->  

          ----->  

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

=

=

=

= 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

       ⇄              

(0.053 - x)                                  x                             x

=

=

Using quadratic equation formula;

we have:

OR

= 0.0134                    OR                -0.0179

So; we go by the positive integer which says

x = 0.0134

So

pH =

pH =

pH = 1.87

Thus, the pH of the second equivalent point = 1.87