Chemical Equillibria Comparisons In which of the following reactions is the equilibrium farthest to the left?

a) $\ce{P4O10 + $x$ H2O &lt;=&gt; 4 H3PO4(aq)}$

b) $\ce{AlCl3 + $x$ H2O &lt;=&gt; [Al(H2O)6]^3+ + 3 Cl- (aq)}$

c) $\ce{Li2O + $x$ H2O &lt;=&gt; 2 Li+(aq) + 2 OH- (aq)}$

d) $\ce{H2S + H2O &lt;=&gt; H3O+ + HS- (aq)}$

e) $\ce{CaC2 + $x$ H2O &lt;=&gt; Ca^2+(aq) + 2 OH- (aq) + C2H2(g)}$

The correct answer is d, but I'm not sure how to compare these equilibria.  I know that the answer can't be e, since the evolution and purging of gas will drive the equilibrium forward.  Also, the dissociation of ions in water is favorable due to solvation, so that rules out c and b.

But I'm not sure how to compare the remaining two reactions.

Any help would be greatly appreciated.
You have already ruled out the equations b, c, and e.


  But I'm not sure how to compare the remaining two reactions.


Have a look at equation a again.

The reaction of $\ce{P4O10}$ with water generates phosphoric acid ($\ce{H3PO4}$), which is a fairly strong three-protonic acid ($\mathrm{pK_{a1} = 2.16; pK_{a2} = 7.21}$). 

As a result, subsequent deprotonation of phosphoric acid
$$\ce{H3PO4 + H2O &lt;=&gt; H3O+ + H2PO4-}$$ 

will pull the equilibrium in equation a to the right.

Hydrogen sulfide ($\ce{H2S}$), on the other hand, is a rather weak acid with $\mathrm{pK_{a} = 7.0}$. Consequently, you will find a significant amount of undissociated hydrogen sulfide in aqueous solution. Therefore, the correct answer in the comparison is indeed d.