Answer :
The titration between the weak acid (HCOOH) and the strong base (KOH) can be represented like:
HCOOH (aq) + KOH (aq) <----> HCOOK (aq) + H₂O (l)
In the equivalence point we have the same number of moles of acid and base.
The number of moles of Acid that we initially have is:
number of moles of HCOOH = 10.00 mL * 0.1 mmol/mL
number of moles of HCOOH = 1.00 mmol
In the equivalence point we will have the same number of moles of the strong base:
number of moles of KOH = 1.00 mmol
The volume of KOH that we add in the equivalence point is:
volume of KOH = number of moles/molar concentration
volume of KOH = 1.00 mmol / 0.1 mmol/mL
volume of KOH = 10.00 mL
So we reach the equivalence point when 10.00 mL of KOH are added. Unfortunately in the chart that we are given, we are not given the pH when 10.00 mL of KOH are added, so we will have to calculate it.
We have to set up the ICE table.
HCOOH (aq) + OH- (aq) ----> HCOO- (aq) + H₂O (l)
I: 1.00 mmol 1.00 mmol 0 0
C: -1.00 mmol -1.00 mmol +1.00 mmol
E: 1 - 1 = 0 1-1 = 0 1.00 mmol
We reach the equivalence point when we have the same number of moles of the acid and base. That's why the initial amount of them is the same. In the equilbrium the base and the acid will be converted into the conjugate base of the weak acid (HCOO-). So the equilibrium number of moles of HCOOH and KOH are zero, and we produced 1.00 mmol of HCOO--
HCOO- is the conjugate base of the weak acid HCOOH. Since it is the only compound present in the equilibrium, we have to calculate the pH of that weak base.
HCOO- (aq) + H₂O (l) <----> HCOOH + OH-
The Kb of the conjugate base of our weak acid can be calculated using this expression:
Ka * Kb = Kw
Kb = Kw/Ka
We are given the Ka and the Kw is a constant (Kw = 1.0 *10^-14)
Kb =(1.0 * 10^-14)/ 1.8 * 10^-4
Kb = 5.56 *10^-11
Now we have to find the concentration of ion HCOO- in the equivalence point. We know that at the equilibrium we have 1.00 mmol of it. But, we had 10.00 mL of the HCOOH solution and we added 10.00 mL of the KOH solution. So now, the total volume that we have in the erlenmeyer is 20.00 mL. So the concentration of HCOO- is:
Volume of HCOOH = 10.00 mL
Volume of KOH = 10.00 mL
Total volume = 20.00 mL
Concentration at equilibrium of HCOO- = 1.00 mmol/20.00 mL
Concentration at equilibrium of HCOO- = 0.05 M
Finally we have to find the pH of the conjugate base (HCOO-) that is the only compound present at the equivalence point.
HCOO- (aq) + H₂O (l) <----> HCOOH + OH-
I 0.05 M 0 0
C -X +X +X
E 0.05 M - X +X +X
Kb = [HCOOH] * [OH-]/[HCOO-]
Kb = X * X / (0.05 M - X)
Kb = X²/(0.05 M - X)
If we consider that x is too small, we can assume that 0.05 M - X = 0.05 M
Kb = X²/0.05 M
We found that the Kb = 5.56 *10^-11
X² = 5.56 *10^-11 * 0.05
X² = 2.78 *10^-12
X = √2.78 *10^-12
X= 1.67*10^-6 M = [OH-]
We found that the concentration of OH- at the equilibrium is 1.67*10^-6 M. With that value we can find the pOH and finally we can find the pH.
pOH = -log [OH-]
pOH = -log (1.67*10^-6 M)
pOH = 5.78
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 5.78
pH = 8.22
Answer: The thoretical pH at the equivalence point is 8.22
