Q:

Do buffers only work for species of the opposite variety? (I.e. weak acids buffer strong bases and vice versa)

I have looked at several sources including Khan Academy, ChemCollective and the answer to a similar question:
Can someone please explain buffers to me?
However, it is not clear whether buffers can only protect against the opposite species. I.e. a weak acid can only protect against the pH change caused by strong bases, or if it can also protect against the pH change caused by strong acids.
This is my thought:
If we have $\ce {H_2CO_3_{(aq)} + H_2O_{(l)}<=>H_CO^{-}_3_{(aq)} +  H_3O^+_{(aq)}}$
Then we add a strong acid $\ce{HCl + H_2O-> H_3O^+ + Cl^-}$
Then we have $x$ amount of protons in the solution and for every one of the weak acid's conjugate base, there is a proton. So, to me it appears that there is no way for the weak acid to act as a buffer for a strong acid. Is this correct or am I missing something?

A:

Think of it in terms of the equilibrium constant: 
$$K_{c}\; =\; \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{H2CO3}]}$$
Let's assume for a moment that there are 2 mols of $\ce{H3O+}$, 2 mols of $\ce{HCO3-}$, and 1 mol of $\ce{H2CO3}$. In our pretend scenario, $K_c$ is 4 (however in reality the $K_c$ is much lower and is about $2.5 \cdot 10^{-4}$). When you add an acid, the $\ce{H3O+}$ concentration rises. Lets assume there are now 9 mols of $\ce{H3O+}$. So now the reaction quotient is
$$Q\; =\; \frac{9\cdot 2}{1} = 18$$
Because $Q > K$ the reaction has to shift left in order to compensate and go back to the equilibrium state. If 1 mol of $\ce{H3O+}$ recombines with 1 mol of $\ce{HCO3-}$ to give back 1 mols of $\ce{H2CO3}$ now you have
$$Q\; =\; \frac{8\cdot 1}{2} = 4$$
and because $Q = K$ the reaction is now at equilibrium.
So while the total number of mols of $\ce{H3O+}$ did increase, the effect of it was dampened by the buffer (as it only went up to 8 mols instead of 9 mols).

A:

If you just add straight carbonic acid to water, the $\mathrm{p}\ce{H}$ will drop as it is ionized (as represented by your equation). Since it is a weak acid it won't ionize completely and you'll be left with carbonic acid, bicarbonate and hydronium in solution at equilibrium. Now, if you increase the concentration of $\ce{H+}$ (such as by adding an acid), Le Chatelier's principle dictates that the equilibrium will be disturbed and the system will move in the direction that will restore the equilibrium (in this case to the left, as protons react with bicarbonate to produce carbonic acid). Because the new protons are not free in solution, the pH will not drastically change until the bicarbonate is exhausted. Thus the solution is buffered. 
Mathematically, the acid dissociation constant, which is unchanging for a specific species by definition, is given by $K_a=\frac{[\ce{HCO3^{-}}][\ce{H+}]}{[\ce{H2CO3}]}$. If you add more $\ce{H+}$, the ratio of $\ce{HCO3^-}$ to $\ce{H2CO3}$ must decrease for the system to regain equilibrium 

