AP EAPCET 18 May 2024 Shift 1 Solved Paper

Let's work through the problem step by step.
The dissociation of the weak acid is given by:
HA⇌H++A−.
The initial concentration of HA is 0.5 M . With a degree of dissociation, α=0.01 (or 1\%), the equilibrium concentrations are:
‌[HA]=0.5(1−0.01)=0.5×0.99=0.495M
‌[H+]=0.5×0.01=0.005M
‌[A−]=0.005M
The acid dissociation constant, Ka, is defined as:
Ka=‌

[H+][A−]

[HA]

.
Substitute the equilibrium concentrations:
Ka=‌

(0.005)(0.005)

0.495

=‌

0.000025

0.495

.
Simplifying the expression:
Ka≈5×10−5.
Thus, the approximate value of Ka (or Ke, as referenced) is 5×10−5.
The correct option is Option B.