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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 "leveled." 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 "strong" and "super strong."

Student 2's response

$\ce{S^{2-} + H2O ->HS- + HO-}$

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.

$\ce{HO^{-} + H2O ->H2O + HO-}$

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.

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  • $\begingroup$ @Jan - my dilemma too; 1 starts off okay but makes the wrong conclusion. 3, however, I think is succinct and to the point. $\endgroup$ – Dissenter Jun 7 '16 at 17:02
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    $\begingroup$ 3 is technically correct but it's hardly based on the information given in the question. The point made in 1 is complicated by the fact that the resulting species HS- also displays acid-base properties, in order to do it properly consult an analytical chem textbook, they will show you how to set up a system of equations to solve for the pKa. $\endgroup$ – orthocresol Jun 7 '16 at 17:15
  • $\begingroup$ @orthocresol is the information in the question helpful? I think it's irrelevant at best and misleading at worst. $\endgroup$ – Dissenter Jun 7 '16 at 17:21
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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".

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  • $\begingroup$ what do you mean sulfide has to be infinitely strong? $\endgroup$ – Dissenter Jun 7 '16 at 17:49
  • $\begingroup$ @Dissenter I mean sulfide would need to completely react with water to create hydroxide, with no sulfide remaining, for 1 mole of sulfide to yield 1 mole of hydroxide. $\endgroup$ – DavePhD Jun 7 '16 at 17:51
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    $\begingroup$ oh I see now. The question accounted for that by writing that the two (1 M Na2S and 1 M NaOH) are "essentially" the same in terms of pH. $\endgroup$ – Dissenter Jun 7 '16 at 17:56
  • $\begingroup$ What is the second Ka of water? $\endgroup$ – Dissenter Jun 7 '16 at 18:05
  • $\begingroup$ @Dissenter chemistry.stackexchange.com/questions/24342/… $\endgroup$ – DavePhD Jun 7 '16 at 18:28
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It does depend a lot on your marking scheme, but looking at it zealously I read two partial questions:

  • Is the information given enough to discern the strength of either base? If yes, which parts led to the conclusion? If no, why can we not tell?

  • Which base actually is stronger according to the information given.

The question attempts to ask within its own boundaries, so no previous knowledge of $\mathrm{p}K_\mathrm{a}$ values seems assumed.


A perfect answer should thus include the points:

  • levelling effect of water, i.e. sulphide cannot be the weaker base or its pH value would be lower;

  • It is impossible to tell whether $\ce{S^2-}$ is stronger or whether the two are equally strong.

Additional bonus points could then be given to answers like student 3’s which further add additional information such as the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HS-}$ and $\ce{H2O}$.


Student 1’s explanation of base strength is incorrect, no questions asked. (Base strength is measured by $K_\mathrm{b}$ values.) The levelling effect is not present and apparantly not known. The conclusion is possible but incomplete.

Strictly speaking, this should probably be awarded zero.


Student 2 somehow starts talking about the levelling effect by mentioning that sulphide abstracts a proton from water to generate hydrogen sulphide and hydroxide. However, the water equation is far from nonsensical; it is an equilibrium that actually occurs on a molecular scale. The conclusion drawn may be chemically correct, but it is only semi-correct within the boundary of the question.

Maybe give him half of what you allocated for the levelling effect and nothing for the rest.


Student 3 gives chemically correct facts, but fails to address the problem in the question. Students should be taught to read the question, accept its premises, and then answer according to what is inside unless the question clearly asks for external knowledge.

We can give him a bonus mark for the knowledge to answer the question ‘extra-universally’, but the informatio I would be expecting is simply not there.


Also remember that you should generally give marks towards the lower end of the possible spectrum when correcting a test. Students are likely to come to their TAs to ask for better grading but very unlikely to say ‘but I got this completely wrong, please deduct the points.’

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  • $\begingroup$ The second student's comment about nonsensical is very similar to what the following journal article says: pubs.rsc.org/en/Content/ArticleLanding/1998/AN/… $\endgroup$ – DavePhD Jun 7 '16 at 18:14
  • $\begingroup$ @DavePhD I reject that articles assumptions and conclusions. (Note that I never learnt a $K_\mathrm{a}$ of water being defined by the equation $\ce{H3O+ + H2O <=> H2O + H3O+}$; it was defined with the help of the autoprotolysis that the article later write to ‘correspond to a physical process.’) $\endgroup$ – Jan Jun 7 '16 at 18:39
  • $\begingroup$ the article is saying that equation is for the Ka of H3O+, not the Ka of H2O $\endgroup$ – DavePhD Jun 7 '16 at 19:14

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