Answer :
The equilibrium constant Ka, for the acid = 5.3⋅10^−11
The idea then that you need to use the pH of the result to determine the attention of hydronium ions, H3O that's needed in order for the result to have a pH of Indeed without doing any computations, you can say that the acid dissociation constant, Ka for this monoprotic weak acid will be veritably, veritably small.
The pH of the performing result is fairly close to the pH of pure water at room temperature, which means that a veritably, veritably small number of acid motes will actually ionize.
To find the value of the acid dissociation constant, use an ICE table using a general HA monoprotic acid
HA (aq)+ H2O(l)⇌H3O(aq)^+ +A^−(aq)
By description, Ka will be equal to
Ka = [H3O]^+ [⋅A -]/[HA]
Now use the pH of the result to determine the equilibrium attention of hydronium ions
H3O+ = 10^− pH
H3O+ = 10^−6.37 = 8.91 ⋅ 10^− 7M
As you can see from the ICE table, the equilibrium attention of hydronium ions, which is equal to that of the acid's conjugate base
A^− = x
This means that the expression for Ka becomes
Ka = x ⋅ x / 0.0159 − x = (8.91 ⋅ 10^−7)^2 / 0.0159-(8.91 ⋅ 10^−7)^2
The denominator can be approached with
0.0159-(8.91 ⋅ 10^−7)^2 ≈ 0.159
This means that Ka will be equal to
Ka = 8.91^2 ⋅ 10^ / 0.0159 = 5.13 . 10^-11
Indeed, like we originally prognosticated, you are dealing with a veritably weak acid.
To know more about weak acid
https://brainly.com/question/12811944
#SPJ4
