Answer:
The Ka is 9.11 *10^-8
Explanation:
Step 1: Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
Step 2: Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
Step 3: ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
Step 4: Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
The Ka of HX is mathematically given as
Ka = 9.11 *10^-8
Generally, the equation for the molarity of HX is mathematically given as
M HX = moles HX / volume solution
Therefore
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
Therefore, ICE-chart
[H+] = [H3O+]
10^-3.70 = 10^-3.70 = 1.995 *10^-4
The concentration at the equilibrium is
[HX] = (0.437 - x)M
[H3O+] = 1.995*10^-4 M
x=1.995*10^-4
In conclusion
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
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