[SOLVED] 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 molespercentage of dissociation=9.0×10^-4×1×100knowing that x%=x/100,we then say;x/100=9.0×10^-4×1×100therefore, x=100×100×9×10^-4×1x=9x percentage of dissociation=9%

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?

Accepted Answer

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