Ka of water?

[MedPR]

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What is the Ka of water?

A. 10^-14
B. 2*10^-16
C. 1
D. 10

Answer is B. "The Ka is just Kw divided by 55."

Is this a definitional thing or what? I've never heard of this before.

 

[pm1]

What is the Ka of water?

A. 10^-14
B. 2*10^-16
C. 1
D. 10

Answer is B. "The Ka is just Kw divided by 55."

Is this a definitional thing or what? I've never heard of this before.

I just found this online:

H2O ======> H+ + OH-

Ka = [H+][OH-]/[H2O]

Ka = [10^-7][10^-7]/[55 M]

55 M is standard concentration of water at standard room temp or whatever. Just a known fact

 

[typicalindian]

The 55M comes from mols/L

1L water = 1000g water... Water is 18g/mol which gives 55.5mol for 1000g of water
So 1L water has a molarity of 55.5mol/L

Hope that made sense!

 

[pm1]

The 55M comes from mols/L

1L water = 1000g water... Water is 18g/mol which gives 55.5mol for 1000g of water
So 1L water has a molarity of 55.5mol/L

Hope that made sense!

nice. just the basics huh? :thumb up:
thank you!

 

[MrNeuro]

The 55M comes from mols/L

1L water = 1000g water... Water is 18g/mol which gives 55.5mol for 1000g of water
So 1L water has a molarity of 55.5mol/L

Hope that made sense!

isnt the water just liquid?
if so it shouldn't even be included in the equilibrium constant formula...right?

 

[phaloz]

isnt the water just liquid?
if so it shouldn't even be included in the equilibrium constant formula...right?

I was thinking this, but H2O can react with itself in a process called autoionization. When you calculate the Ka for H2O you're using the autoionization reaction
2H2O ----> H3O+ + OH-. Correct me if I'm wrong.

 

[milski]

isnt the water just liquid?
if so it shouldn't even be included in the equilibrium constant formula...right?

That would be the case if we were talking about partial pressures of gases. When using concentrations, all of them are included.


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[MedPR]

The 55M comes from mols/L

1L water = 1000g water... Water is 18g/mol which gives 55.5mol for 1000g of water
So 1L water has a molarity of 55.5mol/L

Hope that made sense!

Yea, I get that part, but I don't understand what you do with the 55mol/L

 

[gettheleadout]

The easier way to answer this question is to recognize A as the Kw, C & D as unreasonable answers, and B as the only remaining choice.

That would be the case if we were talking about partial pressures of gases. When using concentrations, all of them are included.

I'm positive that pure liquids and solids are excluded from even concentration-based equilibrium constant calculations. The exception here for water is something I haven't seen before (I'm guessing it's because we're looking at water specifically), but you definitely exclude H2O(l) from calculations when you have another equilibrium predominating (acid/base in aqueous solution, for example.)

 

[milski]

The easier way to answer this question is to recognize A as the Kw, C & D as unreasonable answers, and B as the only remaining choice.
I'm positive that pure liquids and solids are excluded from even concentration-based equilibrium constant calculations. The exception here for water is something I haven't seen before (I'm guessing it's because we're looking at water specifically), but you definitely exclude H2O(l) from calculations when you have another equilibrium predominating (acid/base in aqueous solution, for example.)

You would exclude the non-reacting solvents. Water is typically just a solvent and is not part of the reaction but that is not the case here.

 

[gettheleadout]

You would exclude the non-reacting solvents. Water is typically just a solvent and is not part of the reaction but that is not the case here.

Right, right, that makes sense for explaining why water is included here. However, even if you had a concentration-based equilibrium constant to derive for a reaction occurring in aqueous solution and none of the components were gaseous, you would still exclude a pure solid wouldn't you?

 

[MedPR]

The easier way to answer this question is to recognize A as the Kw, C & D as unreasonable answers, and B as the only remaining choice.
I'm positive that pure liquids and solids are excluded from even concentration-based equilibrium constant calculations. The exception here for water is something I haven't seen before (I'm guessing it's because we're looking at water specifically), but you definitely exclude H2O(l) from calculations when you have another equilibrium predominating (acid/base in aqueous solution, for example.)

Yes I understand how to do this by process of elimination, but I would also like to know how to find the answer as if it was open ended 😉

 

[milski]

Right, right, that makes sense for explaining why water is included here. However, even if you had a concentration-based equilibrium constant to derive for a reaction occurring in aqueous solution and none of the components were gaseous, you would still exclude a pure solid wouldn't you?

Short answer - yes, I would.

Long answer - the numbers are not really concentrations but something called activity numbers/coefficients. For a lot of situations they are the same or very close to the numerical values of the concentration. For liquids/solids in gas mixtures or solids in liquid solutions they are 1, which means that you can drop that reagent/product from the equilibrium quotient.

 

[MedPR]

Ka=[H+][A-]/[HA]?

So for water it's [H3O+][OH-]/[H2O]. Molarity of [H2O] is 55mol/L so Ka=[H+][OH-]/55

And since Kw=10^-14=[H+][OH-]

Ka = (10^-14)/55

Ka = (10^-14)/(5*10^1) = 1/5*10^-15 = 0.2*10^-15 = 2*10^-16.

Thanks all.

 

[gettheleadout]

Short answer - yes, I would.

Long answer - the numbers are not really concentrations but something called activity numbers/coefficients. For a lot of situations they are the same or very close to the numerical values of the concentration. For liquids/solids in gas mixtures or solids in liquid solutions they are 1, which means that you can drop that reagent/product from the equilibrium quotient.

Interesting, I've never heard about that before. Have you taken analytical?

 

[milski]

Interesting, I've never heard about that before. Have you taken analytical?

No, my gen chem professor brought it up when we were talking about equilibriums. It came up again later with free Gibbs energy. Using these takes care of the units - the coefficients have the appropriate units so that K ends up dimensionless. They can also be used to adjust for temperature so that you can treat K as a constant independent of temperature. Not sure what's so great about that though - you still need to adjust the coefficients.