# 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 "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.

• @Jan - my dilemma too; 1 starts off okay but makes the wrong conclusion. 3, however, I think is succinct and to the point. Commented Jun 7, 2016 at 17:02
• 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. Commented Jun 7, 2016 at 17:15
• @orthocresol is the information in the question helpful? I think it's irrelevant at best and misleading at worst. Commented Jun 7, 2016 at 17:21
• How about "Is $\ce{Na^17OH}$ or $\ce{Na^16OH}$ as stronger base, given that a 1 M solution of each have essentially the same pH?". Same setup, different answer?
– Karsten
Commented Jan 18, 2020 at 0:10

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".

• what do you mean sulfide has to be infinitely strong? Commented Jun 7, 2016 at 17:49
• @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. Commented Jun 7, 2016 at 17:51
• 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. Commented Jun 7, 2016 at 17:56
• What is the second Ka of water? Commented Jun 7, 2016 at 18:05
• Commented Jun 7, 2016 at 18:28

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.’

• 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/… Commented Jun 7, 2016 at 18:14
• @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.’)
– Jan
Commented Jun 7, 2016 at 18:39
• the article is saying that equation is for the Ka of H3O+, not the Ka of H2O Commented Jun 7, 2016 at 19:14

Thinking that if they were equally strong, 1M $$\ce{S^2-}$$ should give only 0.5M OH- isn't really correct.

Another thing is that the $$\ce{S^2-}$$ is in fact, stronger base than $$\ce{OH-}$$, but weaker base than $$\ce{O^2-}$$.

$$\ce{S^2-}$$ removes 1 H+ from $$\ce{H2O}$$ and forms $$\ce{HS-}$$ and OH- (same quantity for each), then an equilibrium state is reached ($$\ce{HS-}$$ and OH- can't deprotonate each other, even OH- is a little bit stronger than $$\ce{HS-}$$).

1M $$\ce{S^2-}$$ will give exactly 1M OH-, so Na2S will have the same pH than NaOH in water, even the first is stronger base than NaOH. It's also true for other much stronger bases like NaNH2 or NaH which will give the same pH than NaOH, but are much stronger than OH- as a base.

Question

It is an experimental fact that the pH of 1 M Na2S is essentially the same as the pH of 1 M NaOH. Based on this information, is S2− or HO− the stronger Bronsted-Lowry base? How can you tell?

Assumptions

Questions asked as part of a course usually lack context for people not taking the course. The question does not state which solvent was used and at what temperature and pressure measurements were done. It also does not say what level of theory to use in the answer. Lastly, it uses the terms "essentially the same" and "stronger base" without defining them. Here are the assumptions I will make in this answer:

1. The solvent is water.
2. The measurement was taken at room temperature and normal pressure where pKw is 14.
3. Ionic strength differences are ignored. Activity coefficients are set to 1.
4. "essentially the same" means a difference of 2% in concentration is fine.
5. The criterion for "stronger base" is the pKa of the conjugate acid.

I will first compare a 1 M solution of NaOH and a 1 M solution of 17-oxygen NaOH which I will abbreviate as NaQH. To make things easier, I will assume that the NaOH and the water are made with isotopically pure 16-oxygen.

For the NaOH solution, the final concentrations are 1 M $$\ce{OH-}$$ and $$\pu{1e-14 M}$$ $$\ce{H+}$$, and the pH is 14.

For the NaQH solution, I will use 49 M for the concentration of water to simplify the math. So there is 1 M of total 17-oxygen atoms in the system. For simplicity, I am assuming no isotope effects on pKa values. The 17-oxygen will distribute equally among the hydroxide and the water, with 2% of oxygen atoms being 17-oxygen. The final concentrations are 0.02 M $$\ce{QH-}$$, 0.98 M $$\ce{H2Q}$$ and $$\pu{1e-14 M}$$ $$\ce{H+}$$. From this, we can calculate the Ka as:

$$\ce{H2Q <=> QH- + H+}$$

$$K_a = \frac{[\ce{QH-}][\ce{H+}]}{[\ce{H2Q}]} = \frac{1}{49} \cdot \pu{e-14}$$

So the value of the Ka of oxygen-17 hydroxide is that of the Kw divided by 49, and the equilibrium concentrations are given above. The Ka value of hydroxide is the same, of course, because we excluded isotope effects.

What happens if the base has the same strength as hydroxide?

Now we turn the problem around and start with a base $$\ce{B-}$$ that has the same strength as hydroxide. Because it has the same strength, the equilibrium concentrations will be essentially the same, i.e. 0.02 M $$\ce{B-}$$, 0.98 M $$\ce{HB}$$, and $$\pu{1e-14 M}$$ $$\ce{H+}$$. The only difference is in the hydroxide concentration, which is 0.98 M instead of 1 M (well, the hydrogen ion concentration will be slightly different, too, 2% higher, and the pH will be 13.99 instead of 14.00).

Why are hydroxide and $$\ce{B-}$$ not at the same concentration, even though they have the same strength? It is because the conjugate acid (HB or water) are present at different concentrations (about 50-fold difference).

What is the answer to the question the OP cited in the posted question

From the given information (together with the assumptions I had to make), sulfide could be a stronger base than hydroxide, or of the same strength.

What is the answer to the OPs question?

According to a 2018 paper by May et al title "Goodbye to $$\ce{S^2-}$$ in aqueous solution" (https://doi.org/10.1039/C8CC00187A), the sulfide ion has never been detected in aqueous solution. The hydroxide ion has, so sulfide is the stronger base by that argument. As for the OP's question in the body "What are your opinions?", that is mainly opinion-based and should be closed.