[InsaneDefense]

This is the question:

A 2.50 g tablet of calcium hydroxide is dissolved in 400.0 mL of water. What is the pH of the solution of calcium hydroxide?

I calculated n of calcium hydroxide: 0.0337 mol

and c of calcium hydroxide: 0.0843 mol/L

and now I need to find pH.

Anyone know how I can do this?

[AWK]

The solubility of calcium hydroxide is approximately 1.7 g/L, so even half of this tablet will not dissolve and only the dissolved compound introduces the hydroxide ions into the solution.

[InsaneDefense]

I forget solubility, but I'm assuming that there's a chart for it, which is a constantthat you put in a formula to get the pH?

From my understanding we get pH when we know the concentration of OH or H ions in the solution.

So, the answer is that most likely half of the tablet won't dissolve, but is there a chemical formula for this in which I can relate the mass of Calcium Hydroxide and a volume of solution which is h2O to pH?


Also, I am trying to figure out the conjugate base and acid for this chemical formula. I'm confused because of the subscript 2 after (OH).  I assume only 1 OH is removed, so that would be a conjugate acid of Ca(OH), but what would the conjugate base be? Since OH is removed, is it H3O2? I'm pretty sure I'm wrong.

This is the formula I have:
Ca(OH)2subscript + H2O -> conj. acid + conj. base

[Borek]

If it is a high school level questions most likely whoever asked it wanted you to assume Ca(OH)₂ dissociates completely, producing two OH^-, and to ignore intermediate steps. Hard to be sure though.

Real situation is much more complicated, not only the numbers given are over the solubility of the substance, also there are several dissociation equilibria involved.

[InsaneDefense]

If it is a high school level questions most likely whoever asked it wanted you to assume Ca(OH)₂ dissociates completely, producing two OH^-, and to ignore intermediate steps. Hard to be sure though.

Real situation is much more complicated, not only the numbers given are over the solubility of the substance, also there are several dissociation equilibria involved.

I don't think we are producing two OH yet. I think you're getting at diprotics or something like that.

So, you're telling me there are more then just conj. acid and conj. base involved as the products?

[Borek]

M(OH)₂ MOH⁺ + OH^-

MOH⁺ M²⁺ + OH^-

overall reaction

M(OH)₂ M²⁺ + 2OH^-

[InsaneDefense]

Thanks. The final expression was what I was looking for. I assumed the final products would have 1 OH instead of H3O because it's a base, but the subscript in  front of (OH) confused me.

Also from my understanding, to get pH and pOH we use the concentration of either H or OH ions as a power of 10^-pH or 10^-OH


So, is there a way of finding how how much OH is in the solution?
 

[AWK]

https://www.chemteam.info/Equilibrium/calc-pH-from-Ksp.html

[Borek]

So, is there a way of finding how how much OH is in the solution?

Depends.

I mean - you can follow the AWK's advice about using Ksp. That will give one possible answer.

Then you can do the full equilibrium calculation assuming the solution is saturated. This will produce second answer.

Finally, you can do calculations assuming concentration is given by the amount of substance given in the problem statement and calculate pH assuming the base is fully dissociated. This will produce third answer.

Second answer is the closest to the reality. In HS I would guess third approach is the one you are asked for, but there really is no way to tell which answer and which approach you teacher expects.

[InsaneDefense]

https://www.chemteam.info/Equilibrium/calc-pH-from-Ksp.html

The thing is I'm given a text book to refer to K values and I'm not sure what sp is, but we are using Kb.

And the teacher said the method to find the pOH of the product doesn't require a K value or an ice table. I think she said we find pH first, but that doesn't really help because I don't think there is a way to get pH.


It may have something to do with finding the pOH and pH of Strong Acids and Bases. If you know an answer that comes to you let me know. In the meantime I will rewatch the lecture in regards to finding pOH and pH of Strong Acids and Bases and come back with a solution if I don't find it.

Thanks!
 

[InsaneDefense]

Hopefully my question gets answered and doesn't get spammed by this bot.

[Borek]

Sounds like you overcomplicating things then - it is about using basic principles. Finding concentration of OH^- from the given concentration and with the assumption of full dissociation is trivial, converting it to pOH and then to pH is trivial as well.

Putting reality aside - can you use the data given to calculate the molar concentration of Ca(OH)₂?

If it is fully dissociated - what is the concentration of OH^-?

[InsaneDefense]

Sounds like you overcomplicating things then - it is about using basic principles. Finding concentration of OH^- from the given concentration and with the assumption of full dissociation is trivial, converting it to pOH and then to pH is trivial as well.

Putting reality aside - can you use the data given to calculate the molar concentration of Ca(OH)₂?

If it is fully dissociated - what is the concentration of OH^-?

So the molar concentration based on the values I gave in the beginning of the question is 0.0843 mol/L.

And I'm not sure how I can really find the concentraton of OH from the concentration of Ca(OH)2

So it's

Ca(OH)₂ + H₂O -> Ca(OH) + OH^-

[AWK]

Read and understand the content of the link I provided earlier, Even neglecting Ksp, you will find a solution to all your dilemma there.

[Borek]

So the molar concentration based on the values I gave in the beginning of the question is 0.0843 mol/L.

OK

Ca(OH)₂ + H₂O -> Ca(OH) + OH^-

I told you earlier what is the full dissociation reaction.

Stop thinking in terms of conjugate acids, bases and whatnot, this is just about a following a basic stoichiometry.