Question

Ka, the acid dissociation constant, for an acid is 9 × 10^{−4} at room temperature. At this temperature, what is the approximate percent dissociation of the acid in a 1.0 M solution?

(A) 0.03%
(B) 0.09%
(C) 3%
(D) 5%
(E) 9%
Can you please explain with some details?

Answer 1

dissociation=k×no of moles
percentage of dissociation=9.0×10^-4×1×100
knowing that x%=x/100,we then say;
x/100=9.0×10^-4×1×100
therefore, x=100×100×9×10^-4×1
x=9
x percentage of dissociation=9%