A 10.00 mL aliquot of a solution of a weak acid
, HA, whose pK
a
is 4.76 is titrated
with OH
-
generated by electrolysis of water according to th
e following scheme:
2 H
2
O + 2 e
-
⇔
2 OH
-
+ H
2
(a)
[5 points] If the end point in the titration is ob
served after the passage of a
current of 0.0500 amperes for 2 minutes, what is th
e concentration of HA in the
original sample?
To solve this problem, we can use the concept of Faraday's Law of Electrolysis. This law states that the amount of substance produced or consumed during an electrolysis reaction is directly proportional to the amount of electricity passed through the electrolyte.
In this case, OH- ions are generated by electrolysis of water. We know that 2 moles of electrons are required to produce 1 mole of OH- ions. The relationship between moles of electrons and electricity passed is given by the formula:
Q = nF
where Q is the total charge passed in Coulombs, n is the number of moles of electrons, and F is the Faraday constant (96485 C/mol).
In this problem, we are given the current (I = 0.0500 A) and the time (t = 2 minutes). We need to convert the time to seconds for consistency.
2 minutes = 2 * 60 seconds = 120 seconds.
Next, we can calculate the total charge passed:
Q = I * t
Q = 0.0500 A * 120 s
Q = 6.00 C
Now, we can use the relationship between Q and n to find the number of moles of electrons:
n = Q / F
n = 6.00 C / 96485 C/mol
n = 6.217 x 10^-5 mol
Since 2 moles of electrons are required to produce 1 mole of OH- ions, the number of moles of OH- ions generated is also 6.217 x 10^-5 mol.
Now, we can use the stoichiometry of the reaction to determine the number of moles of HA in the original sample.
The balanced equation for the reaction is:
HA + OH- -> A- + H2O
From the equation, we can see that 1 mole of HA reacts with 1 mole of OH- to produce 1 mole of A-. Therefore, the number of moles of HA is also 6.217 x 10^-5 mol.
To determine the concentration of HA in the original sample, we need to know the volume of the original sample.
Let's assume the volume of the original sample is V mL.
Therefore, the concentration of HA in the original sample would be:
Concentration = (number of moles of HA) / (volume of sample in liters)
Concentration = (6.217 x 10^-5 mol) / (V/1000 L)
Since we are given that the aliquot used for titration is 10.00 mL, the concentration of HA in the original sample would be:
Concentration = (6.217 x 10^-5 mol) / (10 mL/1000 L)
Concentration = 6.217 x 10^-3 M
So, the concentration of HA in the original sample is 6.217 x 10^-3 M.