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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?
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%
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