Answer:
The pH is 3.08
Step-by-step explanation:
Step 1:
The equation for the reaction.
AH ⇌ A- + H+
Step 2:
Data obtained from the question.
Initial concentration of AH, [AH] = 0.97 M
Dissociation constant, Ka = 7.23x10^-7
Step 3:
Determination of the concentration of A-, [A-], the concentration of H+, [H+] and the concentration of AH, [AH] after the reaction. This is illustrated below:
Before the reaction:
[AH] = 0.97 M
[A-] = 0
[H+] = 0
During the reaction:
[AH] = -y
[A-] = +y
[H+] = +y
After the reaction
[AH] = 0.97 - y
[A-] = y
[H+] = y
Step 4:
Obtaining the value of y. This is illustrated below:
Ka = [A-] [H+] / [AH]
Ka = 7.23x10^-7
[AH] = 0.97
[A-] = y
[H+] = y
Ka = [A-] [H+] / [AH]
7.23x10^-7 = y x y / 0.97
Cross multiply to express in linear form
7.23x10^-7 x 0.97 = y^2
7.0131x10^-7 = y^2
Take the square root of both side
y = √(7.0131x10^-7)
y = 8.37x10^-4
Therefore [H+] = y = 8.37x10^-4 M
Step 5:
Determination of the pH.
pH = - log [H+]
[H+] = 8.37x10^-4 M
pH = - log [H+]
pH = - log 8.37x10^-4
pH = 3.08