= 3.56 × 10^-5
A monoprotic acid is represented as HA,
Thus; HA →H+ + A-
Therefore; Ka = [H+] × [A-] / [HA]
[H+] = [A-] = Ca ×α
[HA] = Ca ×(1 - α)
α = dissociation % / 100 = 0.0155
Thus;
Ka = (0.146 × 0.0155)^2 / (0.146 × (1 - 0.0155))
= 3.56 × 10^-5
The acid ionization constant (ka) for the acid 3.51×10^-5.
First step is to calculate the degree of dissociation
Degree of dissociation=Concentration of the dissociated portion of the acid/100
Degree of dissociation=1.55/100
Degree of dissociation=0.0155
Second step is to calculate the Acid dissociation constant (Ka) for weak acid is:
Ka=cα²
Ka=0.146×(0.0155)²
Ka=3.51×10^-5
Inconclusion the acid ionization constant (ka) for the acid 3.51×10^-5.
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