Answer:
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