Respuesta :
Answer:
The dissociation constant for the acid ( experimental ) is 1.45 lit/mol
Explanation:
The value of dissociation constant can be calculated as,
[tex]K_{a}[/tex] = C × ∝²
Where, C = concentration of the solution = 0.329M
∝ = Degree of dissociation
again , Degree of dissociation can be obtained form :
[tex]p_{H}[/tex] = C × ∝
∝ = [tex]\frac{p_{H} }{C}[/tex]
∝ = [tex]\frac{2.327}{0.329}[/tex] = 7.072
So, now [tex]K_{a}[/tex] = C × ∝²
= 0.329 ×( 7.072)²
= 1.45 lit/ mol
Answer:
The Ka = 6.74 * 10^-5
Explanation:
Step 1: Data given
Concentration of benzoic acid = 0.329 M
pH = 2.327
Step 2: Calculate the Ka
pH = -log (√([HA]*Ka))
2.327 = -log (√(0.329*Ka))
10 ^ - 2.327 = √(0.329*Ka))
0.0047098 = √(0.329*Ka))
2.218 * 10^-5 = 0.329 * Ka
Ka = 6.74 * 10^-5
The Ka = 6.74 * 10^-5
