[orgoclear]

I was revising some of my concepts on Ka, Kb etc (I am basically a grade 10 student).
I had the following doubts:

1. When we define Ka of a substance, what does it mean? I mean I know it is the equilibrium const.

But if we consider the eqbm const. for this reaction,
X^- + H₂O  ::equil::HX + OH^-
Can we write the equilibrium constant as Ka?
i.e. can we write Ka = [OH^-][HX]/[X^-] ?

Does Ka necessarily have to be the equilibrium constant for the reaction of a substance with water to give its conjugate base only?


I've got more doubts which I will ask successively after each doubt gets cleared.

[renge ishyo]

I got confused about this way back when as well, so I will do my best below to straighten things out for you.

The great secret to figure out whether or not you should use K_a or K_b as the equilibrium constant for a reaction is to look at the species being added to water in the balanced chemical equation. It is a little known fact that you can only directly use the equilibrium constant for the species being added to water when you do these calculations (the reason being that the concentration of water was actually multiplied  into the value of K_a and K_b as part of its definition which is why you don't include the concentration of water as one of the reactants when you write any of the equilibrium expressions).

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Here's a few examples showing you how this works.

#1: Acid being added to water on the reactants side.

HX + H₂0  H₃0⁺ + X^-

The species being added to water is the conjugate acid, so use K_a as the equilibrium constant:

K_a = [H₃0⁺][X^-]/[HX]

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#2. Base being added to water on the reactants side.

X^- + H₂0  HX + OH^-

The base is being added to water here, so use K_b:

K_b = [OH^-][HX]/[X^-]

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#3. Base being added to water on the *products* side.

HX + OH^-  H₂0 + X^-

The rule still holds...you can only directly use the K for the species being added to water, which is Kb. But the water is on the wrong side of the equation? No problem, the K for the forward reaction here is just the inverse of Kb:

K = 1/Kb = [X^-]/[OH^-][HX]

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If you are given the wrong K value, such as if you were given K_a for example #2, you can always find the K_b that you need to use from:

K_w = (K_a)(K_b)

There, that's all my secrets on this topic 

[orgoclear]

renge ishyo: thanks for clearing that doubt of mine

now my second doubt:

Q2. For a strong acid why is the value of Keq more while for a strong base the value of Keq is less i.e. a strong acid tends to ionise while for a strong base the equilibrium lies towards the base side?

[Borek]

Q2. For a strong acid why is the value of Keq more while for a strong base the value of Keq is less i.e. a strong acid tends to ionise while for a strong base the equilibrium lies towards the base side?

It is not, high Ka and high Kb means strong acid/base.

Are you sure you have used correct Keq? Perhaps you are trying to compare Ka values for acid and base - if so, you are trying to use the same equlibrium to describe different reactions, no wonder you find it confusing.

[orgoclear]

regarding this question I have a general query:

An acid is a strong acid if it dissociates more (i.e. has the ability to accept the lone pair from a donor atom). So consequently Ka will be more

A base is strong if the lone pair or negative charge on it is more localised so that it can donate the lone pair to the acceptor molecule.

Am I right? Is there any other definition for acids to be strong or bases to be strong?

I know one other.
An acid is strong if its conjugate base is more stabilised. In context of this what is the equivalent statement for bases to be strong?

I am sorry that I replied late to your views

[Arctic-Nation]

You have to be careful to not mix the different concepts of acids and bases.
In Bronsted-Lowry acidity, an acid is strong if it dissociates more (releases protons), and a base is strong if it accepts protons more easily. This theory is solely based on protonation and de-protonation. In Lewis acidity, however, the idea of dissociation of acid is not valid, as it describes acid-base interactions through donation (base) or acceptance (acid) of electron pairs.
As such, Bronsted-Lowry acidity can be considered a special case of Lewis acidity, in which the Lewis acid is the proton (H+) itself.

Your second definition is correct, and its equivalent is: a base is stronger if its conjugate acid is more stabilized. (Though mind 'stronger', rather than 'strong'.)