When the pH of a solution is 2 units lower than the pKa of the buffer in the solution, the conjugate base/weak?

when the pH = the pKa
the ratio of  (A-/HA)  is 1 to 1

When the pH of a solution is 2 units lower, it is more acidic by 100 times

to shift the equilibrium to become more acidic we lessen the conjugate base [A-] or increase the acid [HA]

your answer is
 the conjugate base/weak acid (A-/HA) is 0.01

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the Henderson Hasselbalch equation
@   http://en.wikipedia.org/wiki/Henderson_Hasselbalch
shows that

pH = pKa + log of [A-] / [HA]

so when (A-/HA) is 0.01
pH = pKa + log [A-] / [HA]

pH = pKa + log 0.01

pH = pKa + (-2)

the pH comes out to be "2 units lower than the pKa"