Answer:
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
![K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cdfrac%7B%5BC_2H_5O%5E-_2%5D%5BH_3O%5E%2B%5D%7D%7B%5BHC_2H_3O_2%5D%7D)

Recall that the ka for 
Then;


By solving the above mathematical expression;
x = 0.00137 M
![H_3O^+ = x = 0.00137 \ M \\ \\ pH = - log [H_3O^+] \\ \\ pH = - log ( 0.00137 )](https://tex.z-dn.net/?f=H_3O%5E%2B%20%3D%20x%20%3D%200.00137%20%20%5C%20M%20%5C%5C%20%5C%5C%20%20pH%20%3D%20-%20log%20%5BH_3O%5E%2B%5D%20%20%5C%5C%20%5C%5C%20%20pH%20%3D%20-%20log%20%28%200.00137%20%29)
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:
![pH = pKa + log \dfrac{[base]} {[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cdfrac%7B%5Bbase%5D%7D%20%7B%5Bacid%5D%7D)
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
= 

![k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}](https://tex.z-dn.net/?f=k_b%20%3D%20%5Cdfrac%7B%5B%20HC_2H_3O_2%5D%20%5BOH%5E-%5D%7D%7B%5BC_2H_3O%5E-_2%5D%7D)

![x = [OH^-] = 5.6 \times 10^{-6} \ M](https://tex.z-dn.net/?f=x%20%3D%20%5BOH%5E-%5D%20%3D%205.6%20%5Ctimes%2010%5E%7B-6%7D%20%5C%20M)
![[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%20%5Cdfrac%7B1.0%20%5Ctimes%2010%5E%7B-14%7D%20%7D%7B5.6%20%5Ctimes%2010%5E%7B-6%7D%20%7D)
![[H_3O^+] =1.77 \times 10^{-9}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D1.77%20%5Ctimes%2010%5E%7B-9%7D)
![pH =-log [H_3O^+] \\ \\ pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }](https://tex.z-dn.net/?f=pH%20%3D-log%20%20%5BH_3O%5E%2B%5D%20%20%20%5C%5C%20%5C%5C%20%20pH%20%3D-log%20%281.77%20%5Ctimes%2010%5E%7B-9%7D%29%20%5C%5C%20%5C%5C%20%5Cmathbf%7BpH%20%3D%208.75%20%7D)
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
![[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 ) \ L}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%20%3D%20%5Cdfrac%7B0.000625%20%5C%20moles%7D%7B%280.02%20%2B%200.0218%20%29%20%20%5C%20L%7D)
![[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%20%3D%20%5Cdfrac%7B0.000625%20%5C%20moles%7D%7B0.0418%20%5C%20L%7D)
![[OH^-] = 0.0149 \ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%20%3D%20%200.0149%20%5C%20M)
From above; we can determine the concentration of
by using the following method:
![[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%20%5Cdfrac%7B1.0%20%5Ctimes%2010%5E%7B-14%7D%20%7D%7B0.0149%7D)
![[H_3O^+] = 6.7 \times 10^{-13}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%206.7%20%5Ctimes%2010%5E%7B-13%7D)
![pH = - log [H_3O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH_3O%5E%2B%5D)

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