Acid

[unsung]

EK 1001Qs #770 & #771

770. Hydrofluoric acid is a weak acid (pKa = 3.1). In terms of its ionization in water, fluoride is:

A. a weak acid
B. a weak base
C. a strong acid
D. a strong base

The answer is B. Their explanation is that the pKb of fluoride is 11. So what pKb constitutes a weak base exactly ? I know the conjugate base of a weak acid can be either weak or strong, but how do you tell the difference?

771. What is the acid dissociation constant Ka of water?

A. 1.0 x 10^-14
B. 1.8 x 10^-16
C. 1
D. 10

The answer is B. Their explanation is that Ka = (Kw)/55M, but where did this equation comes from? What's the difference between Ka and Kw for water? I don't really get it. If Kw is just based on autoionization of water, what is Ka?

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[ecoli]

770 - I think this might be a case fairly unique to flourine, which is a strange molecule.

771 - 55M is the approximate concentration of water, given that it's molar mass is 18 g/mole.

H20 -> H+ + OH-

[H+][OH-] / [H20] = Ka

Therefore, the top part of the equation is the Kw (this is the dissociation constant, given the part H+ times OH-,) divided by the concentration of water, approx 55M when there's no solutes around.

 

[supafield]

Kw is defined as [H+]*[OH-], and this turns out to be 1.E-14.

For any acid HA in water, though, Ka is defined as [H+]*[A-]/[HA]

So if we think of water as an acid, Ka = [H+] * [OH-] / [H2O] = 1.0E-14 / [H2O]
1 g of water occupies 1 mL, near enough, which means that 18
gram of water = 1 mol of water occupies 18 mL. So we can fit 55.6 mol of pure water in 1.00 L.

It is therefore fair to say that the "concentration" of water is 55.6 M, and so Ka = 1.8E-16

The log of 1.8E-16 to base 10 is -15.75, which is the pKa value

I found that explanation above elsewhere... to make it more relevant to the question

Ka = [H30+][OH-]/[H20]
Ka = 1X10E-14/55.6-1x10E-14
the - on the denominator is negligible
play around with the exponents and get
Ka = 100X10E-16/55.6
Ka =~ 2X10E-16

The question really is just asking if you notice... that the dissociation constant Kw is 1X10E-14 but Acid dissociation constant will include the concentration water in the denominator.... If you know the concentration is more then one then the Ka of water will be even smaller than it's dissociation constant... which avoids lots of the work above... hope that helps...

 

[161927]

I was tempted to say fluoride is a strong base, because it is the conjugate base of a weak acid. Fluoride ion has some structural features that temper its basicity, such as the fact that it is highly electronegative, so it is more able to hold on to the negative charge, and also that fluorine is large enough to be polarizable. That is, it can spread the charge around its surface area to stabilize the charge. These factors allow for the qualitative judgment that fluoride is a weak base.

 

[unsung]

Kw is defined as [H+]*[OH-], and this turns out to be 1.E-14.

For any acid HA in water, though, Ka is defined as [H+]*[A-]/[HA]

So if we think of water as an acid, Ka = [H+] * [OH-] / [H2O] = 1.0E-14 / [H2O]
1 g of water occupies 1 mL, near enough, which means that 18
gram of water = 1 mol of water occupies 18 mL. So we can fit 55.6 mol of pure water in 1.00 L.

It is therefore fair to say that the "concentration" of water is 55.6 M, and so Ka = 1.8E-16

The log of 1.8E-16 to base 10 is -15.75, which is the pKa value

I found that explanation above elsewhere... to make it more relevant to the question

Ka = [H30+][OH-]/[H20]
Ka = 1X10E-14/55.6-1x10E-14
the - on the denominator is negligible
play around with the exponents and get
Ka = 100X10E-16/55.6
Ka =~ 2X10E-16

The question really is just asking if you notice... that the dissociation constant Kw is 1X10E-14 but Acid dissociation constant will include the concentration water in the denominator.... If you know the concentration is more then one then the Ka of water will be even smaller than it's dissociation constant... which avoids lots of the work above... hope that helps...

This may be a weird question, but why is Kw = [H+][OH-] while Ka for water = ( [H+][OH-] )/[H2O] )?

The way I understand equilibrium constant K, we always take the (products^coefficient)/(reactants^coefficients), while any pure elements (such as pure liquid water) are ignored ( = 1) by raising them to the power of 0. So that's why Kw isn't divided by anything, as it's derived from the autoionization of water equation:

H2O + H2O --> H3O+ + OH-

(Where H3O+ is basically the same thing as H+) And in this equation, both waters in the reactants are ignored.

Whereas for an acid, our equation for Ka would come from:

HA + H2O --> H3O+ + A-

And again we ignore water in the reactants to get Ka = ( [H+][A-])/[HA] ).

Right? So that all makes sense to me until it gets to water, which for some reason we ignore water in the denominator for Kw of water, but we don't ignore it for Ka of water.

HELP??!

 

[161927]

HA + H2O --> H3O+ + A-
A- + H2O --> OH- + HA

Ka = [H30+][A-] / [HA]
Kb = [OH-][HA] / [A-]

Ka*Kb = [H30+][A-][OH-][HA] / [HA][A-]
= [H30+][OH-]