1
$\begingroup$

While deriving the relation between $K_a$ and $K_b$ in ionic equilibrium topic, say I take an example of

$$ \ce{NH4+ + H2O -> NH3 + H3O+} \tag{1}\label{1} $$

The $K_a$ of this reaction is:

$$ \frac{\ce{[NH3][H3O+]}}{\ce{[NH4+][H2O]}} $$

Now $\ce{NH3}$ is a conjugate base of $\ce{NH4+}$. Its $K_b$ can be written as:

$$ \ce{NH3 + H2O -> NH4+ + OH-} \tag{2}\label{2} $$

$K_b$ of the reaction:

$$ \frac{\ce{[NH4+][OH-]}}{\ce{[NH3][H2O]}} $$

In my textbooks while deriving relation between $K_a$ and $K_b$, it added the two equations. In the addition of two equations we get:

$$ \ce{NH4+ + 2H2O + NH3 -> NH3 + NH4+ + H3O+ + OH-} \tag{1 + 2}\label{1p2} $$

My textbook cancelled out both the $\ce{[NH4+]}$ and $\ce{[NH3]}$ concentration terms from the LHS and RHS. $\ce{[NH4+]}$ on the LHS is from equation $\eqref{1}$ and $\ce{[NH4+]}$ on the RHS is from equation $\eqref{2}$.

My question is, how can we cancel them out? How can you say the $\ce{[NH4+]}$ formed from hydrolysis of $\ce{NH3}$ in reaction $\eqref{2}$ is equal to that of $\ce{[NH4+]}$ initially taken in reaction $\eqref{1}$? Even if you initially take same concentration of $\ce{[NH3]}$ for reaction $\eqref{2}$ as formed from reaction $\eqref{1}$, that concentration of $\ce{NH3}$ on hydrolysis can never give same concentration of $\ce{[NH4+]}$ as initially taken in reaction $\eqref{1}$. Then, how can you cancel these different $\ce{[NH4+]}$ concentrations when you are adding reaction $\eqref{1}$ and $\eqref{2}$?

$\endgroup$
1
$\begingroup$
  1. $K_\mathrm{a}$ and $K_\mathrm{b}$ should not have a term for the concentration of water inside.

  2. When you dissolve $x$ moles of $\ce{NH3}$ in water (doesn't matter how much $x$ is), both relations

$$K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]} \tag{1}$$

as well as

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \tag{2}$$

must simultaneously hold true. You agree that both of these relations must independently hold no matter what concentration of $\ce{NH3}$ (or $\ce{NH4+}$, or $\ce{H+}$, or $\ce{OH-}$) there is, right? That is why they are called equilibrium constants, after all.

So, in any given single solution with one well-defined value of $[\ce{NH3}]$, one well-defined value of $[\ce{NH4+}]$, ... both equations must simultaneously hold. You can then cancel out the terms as necessary to obtain

$$K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w} \tag{3}$$


If you use two different solutions, with different concentrations of all the relevant species, the relation still holds true. However, the maths will be much harder because for example if you have solution 1 with pH = 1, and solution 2 with pH = 10, then $[\ce{H+}]_\mathrm{sol,1}[\ce{OH-}]_\mathrm{sol,2}$ is no longer equal to $K_\mathrm{w}$. The relation $[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$ is only necessarily true if we are talking about one single solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.