Why does dissociation of carbonic acid cause a decrease in pH?
Any acid that releases a proton when it dissociates in water will also produce a conjugate base. We can conceptualize this as $\ce{HA &lt;=&gt; H+ + A-} $. In this particular example, $\ce{HCO3-}$ is $\ce{A-}$.  The strength of the acid is determined by the extent to which the acid, $\ce{HA}$, dissociates. This can be expressed using the equilibrium constant for this reaction, which has a particular name, $\ce{K_a = \frac{[H^+][A^-]}{[HA]}}$, (where $\ce{[X]}$ is the concentration of $\ce{X}$ at equilibrium). The larger $\ce{K_a}$ is, the further the reaction moves to the right (or, as it is often expressed, the smaller the $\ce{pK_a}$, which is $\mathrm{-log_{10}}$  $\ce{K_a}$). 

Here, the addition of any $\ce{HA}$ will produce $\ce{H^+}$ and $\ce{A^-}$ in an aqueous solution (the blood), and decrease the $\ce{pH}$. $\ce{A^-}$ is only a base insomuch as it can form $\ce{HA}$ by combining with $\ce{H^+}$. This can occur if you add (or, as occurs in the kidney, reabsorb) $\ce{A^-}$ with some counter ion other than a proton. Then you can drive the equilibrium to to the left, towards $\ce{HA}$, decreasing $\ce{[H^+]}$. Otherwise, $\ce{A^-}$ derived from $\ce{HA}$ will not drive the equilibrium one way or the other. $\ce{H^+}$ will still dissociate as determined by the $\ce{K_a}$.