Given a diprotic acid, H2A, with two ionization constants of Ka1 = 4.5× 10–2 and Ka2 = 5.2× 10–7.
Calculate the pH and molar concentration of each protonated form for a 0.133 M solution of NaHA.
Calculate the pH and molar concentration of each protonated form for a 0.133 M solution of Na2A.
To calculate the pH and molar concentration of each protonated form for a 0.133 M solution of NaHA and Na2A, we need to consider the ionization reactions of the diprotic acid and its corresponding conjugate base.
1. For NaHA solution:
a) Calculate the concentration of HA (unprotonated form):
Since NaHA is a salt of the diprotic acid, it will fully dissociate in water. Therefore, the concentration of HA will be equal to the initial concentration of NaHA, which is 0.133 M.
b) Calculate the concentration of H2A- (first protonated form):
H2A- is formed by the partial dissociation of HA. The reaction can be represented as:
HA ⇌ H+ + A-
Since the concentration of H2A- is small compared to the initial concentration of HA, it can be assumed that x (concentration of H2A-) is negligible compared to 0.133 M. Hence, the concentration of H2A- is approximately 0.
c) Calculate the concentration of H+ (protons):
To calculate the concentration of H+, we need to consider the equilibrium expression for the dissociation of HA:
Ka1 = [H+][A-] / [HA]
Since the concentration of A- is negligible compared to the initial concentration of HA, we can assume [A-] to be 0. Hence:
Ka1 = [H+] / [HA]
Rearranging the equation, we get:
[H+] = Ka1 * [HA]
[H+] = (4.5×10^(-2)) * (0.133 M)
d) Calculate the pH:
By definition, pH is equal to the negative logarithm of the hydrogen ion concentration:
pH = -log[H+]
Substituting the calculated [H+] value, we can determine the pH using a logarithmic calculator or using the logarithm properties.
2. For Na2A solution:
a) Calculate the concentration of A2- (unprotonated form):
Since Na2A is a salt of the diprotic acid's second dissociation product, the concentration of A2- will be equal to twice the initial concentration of Na2A, which is (2 * 0.133 M) = 0.266 M.
b) Calculate the concentration of HA- (first protonated form):
HA- is formed by the partial dissociation of A2-. The reaction can be represented as:
A2- ⇌ H+ + HA-
Similar to the previous case, we assume x (concentration of HA-) to be negligible. Hence, the concentration of HA- is approximately 0.
c) Calculate the concentration of H+ (protons):
To calculate the concentration of H+, we need to consider the equilibrium expression for the dissociation of HA-:
Ka2 = [H+][HA-] / [A2-]
Since the concentration of HA- is negligible compared to the initial concentration of A2-, we can neglect [HA-]. Hence:
Ka2 = [H+] / [A2-]
Rearranging the equation, we get:
[H+] = Ka2 * [A2-]
[H+] = (5.2×10^(-7)) * (0.266 M)
d) Calculate the pH:
Using the same equation as before, substituting the calculated [H+] value, we can determine the pH.