Is sulfide ion a stronger base than hydroxide ion? I received these three responses. They are all, except the last response, incorrect to varying degrees. I am, however, unsure about how to grade these responses because the question itself, in my opinion, misleads the students by saying that both have the same pH and to use this information to determine the strength of the bases. Their responses also raised a few questions of my own. What are your opinions?

Question
It is an experimental fact that the pH of 1 M $\ce{Na2S}$ is essentially the same as the pH of 1 M $\ce{NaOH}$. Based on this information, is $\ce{S^{2-}}$ or $\ce{HO-}$ the stronger Bronsted-Lowry base? How can you tell?

Student 1's response:

A base's strength is measured by how much it can increase the hydroxide ion concentration in a solvent relative to its own initial concentration.
Given that both the 1 M $\ce{Na2S}$ and 1M $\ce{NaOH}$ solutions have the same pH, they both have the same $\ce{[H3O+]}$ concentration and therefore the same $\ce{[HO^{-}]}$ concentration. Therefore, the two bases are the same strength as each creates a $\ce{[HO^{-}]}$ concentration equal to their initial molarities in water.

I think this response could be improved if the student mentioned the leveling effect and recognized how all bases stronger than hydroxide ion are &quot;leveled.&quot; It's just like the high striker game at the fair; you can hit it hard enough to make the puck go all the way to the top, and so can someone else, but that doesn't necessarily mean you guys are equal in strength; it just means that the device quantifying strength doesn't differentiate between &quot;strong&quot; and &quot;super strong.&quot;
Student 2's response

$\ce{S^{2-} + H2O -&gt;HS- + HO-}$
From this equation and the problem statement, it is clear that $\ce{S^{2-}}$ hydrolyzes water to create a hydroxide ion concentration in water equal to its own initial molarity. Hydroxide ion isn't hydrolyzing water to create hydroxide; the below equation is non-sensical. Therefore, sulfide ion is the stronger base.
$\ce{HO^{-} + H2O -&gt;H2O + HO-}$

Is the second equation really non-sensical?
Student 3's response:

The stronger base has the weaker conjugate acid.
Sulfide ion's conjugate acid is $\ce{HS-}$.
Hydroxide ion's conjugate acid is $\ce{H2O}$.
From the $\ce{K_{a}}$ table in the beginning of the lab manual, one knows that $\ce{HS-}$ is a weaker acid than $\ce{H2O}$. Therefore, sulfide is the stronger base.
Equilibrium Expressions Related to the Solubility of the Sour Corrosion Product
Mackinawite Ind. Eng. Chem. Res. 2008, 47, 1738-1742 compiles 18 different studies concerning the second Ka of H2S and shows that they vary over 7 orders of magnitude from $10^{-19}$ to $10^{-12}$

So firstly it isn't accepted whether or not Na2S is stronger than NaOH.

Putting that aside, the only way that 1M sulfide could yield 1 M hydroxide (and hence pH be equal to pH of 1 M hydroxide) is if sulfide was infinitely strong.  If they were equally strong, it would only yield 0.5M hydroxide.  

Student 1: clearly wrong.

Student 2: correct (based upon the information given)

Student 3: can't say without seeing the table, but if the table actually says that SH- is weaker than water than the answer is true, but not correct in the sense that the instructions say "Based on this information".