-1

The ionproduct of pure water is well known to be $$ K_{\rm W} = \left[ {\rm OH^-} \right] \left[ {\rm H_3O^+} \right] = 10^{-14} $$ at some standard conditions (I think 25°C).

However having some acid HA in water as far as I'm aware this product is always the same i.e. $pK_{\rm a} + pK_{\rm b} = 14$.

How can this be derived?

Diger
  • 101
  • 2
  • I do not think this question is in any way answered there! Equilibrium constants are defined for pure reactions, for instance $$ {\rm H_2O + H_2O} \quad \substack{k_1 \ \longrightarrow \ \longleftarrow \ k_2} \quad {\rm OH^- + H_3O^+} $$ is such a pure reaction and then $K_{\rm W} = \frac{k_1 \left[{\rm H_2O}\right]^2}{k_2}$. However adding additional reactions of the form $$ {\rm HA + H_2O} \quad \substack{\longrightarrow \ \longleftarrow} \quad {\rm H_3O^+ + A^-} $$ in principle changes the chemistry and thus also the equilibrium constant! – Diger Jul 12 '18 at 19:58
  • 1
    @Diger - You're sort of right but not in the way that your postulating. In reality it is the activity of H+ and OH- that are in equilibrium. In "dilute" solutions the activity is essentially the same as the concentration. However in more concentrated solutions the activity is less than the concentration. – MaxW Jul 13 '18 at 02:28
  • @Diger - An "aqueous solution" should be about 1 kg per liter and have about 55 moles of water per liter for equilibriums like Ka and Ksp to work nicely. (In other words, "dilute solutions.") As a poor example think of putting a drop of sulfuric acid in 100 ml water. That's fine. Now if you put a liter of sulfuric acid in 100 ml of water it isn't really an "aqueous solution" any more. – MaxW Jul 13 '18 at 02:30
  • So in a general set of chemical reactions, does the equilibrium constant $K$ for one specific reaction change when I am adding further reactions to my system? In fact that is what I would expect. – Diger Jul 13 '18 at 04:50
  • For example consider the differential equation for $\ce{OH-}$ only of a dilute solution for the first reaction above in pure water $$ \frac{{\rm d}}{{\rm d}t} \ce{[OH-]} = k_1 \ce{[H2O]^2} - k_2 \ce{[H3O+][OH-]} $$ in equilibrium the derivative is zero and then you precisely get $K_{\rm W} = \frac{k_1 \left[{\rm H_2O}\right]^2}{k_2}$. – Diger Jul 13 '18 at 05:07
  • However if I now add the second equation to the system this equation changes to $$ \frac{{\rm d}}{{\rm d}t} \left[ {\rm OH^-} \right] = k_{1} \left[ {\rm H_2O} \right]^2 + k_{\rm A^- + H_2O} \left[ {\rm H_2O} \right] \left[ {\rm A^-} \right] - \left{ k_{\rm HA + OH^-} \left[ {\rm HA} \right] + k_{2} \left[ {\rm H_3O^+} \right] \right} \left[{\rm OH^-}\right] , . $$ And then $$ \ce{[OH-][H3O+]}=\frac{k_{1} \left[ {\rm H_2O} \right]^2 + k_{\rm A^- + H_2O} \left[ {\rm H_2O} \right] \left[ {\rm A^-} \right] - k_{\rm HA + OH^-} \left[ {\rm HA} \right]\left[ {\rm OH^-} \right]}{k_2} ,. $$ – Diger Jul 13 '18 at 05:18
  • Of course in general/principle I would have to solve the entire system i.e. with the additional reactions $$ \frac{{\rm d}}{{\rm d}t} \left[ {\rm H_3O^+} \right] = k_{1} \left[ {\rm H_2O} \right]^2 + k_{\rm HA + H_2O} \left[ {\rm H_2O} \right] \left[ {\rm HA} \right] - \left{ k_{\rm H_3O^+ + A^-} \left[ {\rm A^-} \right] + k_{2} \left[ {\rm OH^-} \right] \right} \left[{\rm H_3O^+}\right] $$ – Diger Jul 13 '18 at 05:21
  • and $$ \frac{{\rm d}}{{\rm d}t} \left[ {\rm A^-} \right] = -\frac{{\rm d}}{{\rm d}t} \left[ {\rm HA} \right] = \left{ k_{\rm HA + H_2O} \left[ {\rm H_2O} \right] + k_{\rm HA + OH^-} \left[ {\rm OH^-} \right] \right} \left[ {\rm HA} \right] - \left{ k_{\rm A^- + H_2O} \left[ {\rm H_2O} \right] + k_{\rm H_3O^+ + A^-} \left[ {\rm H_3O^+} \right] \right} \left[{\rm A^-}\right] $$ and then create the product $\ce{[OH-][H3O+]}$ with the solutions of steady state, but I do not expect this product to be precisely $K_{\rm W}$. – Diger Jul 13 '18 at 05:22

1 Answers1

1

For some acid $\ce{HA}$, you have

$$\ce{HA + H2O <=> H3O+ + A-}$$

$$\ce{A- + H2O <=> OH- + HA}$$

We then define:

$$K_{\mathrm{a}} = \frac{\ce{[H3O+]\ce{[A-]}}}{\ce{[HA]}}$$

$$K_{\mathrm{b}} = \frac{\ce{[OH-]\ce{[HA]}}}{\ce{[A-]}}$$

Immediately, you see that $K_{\mathrm{a}}K_{\mathrm{b}}=K_{\mathrm{w}}$.

Now, if you take the negative log of both sides:

$$-\log\left(K_{\mathrm{a}}K_{\mathrm{b}}\right)=-\log K_{\mathrm{w}}$$ $$=\mathrm{p}K_{\mathrm{a}}+ \mathrm{p}K_{\mathrm{b}} = 14$$

The 14 comes from the value of $K_{\mathrm{w}}$.

Zhe
  • 17,392
  • 1
  • 38
  • 70
  • I guess the problem I have with this is the way how to view it. Say you start with very little $\ce{HA}$ so that $\ce{[A-]}$ is very small and the equilibrium hardly deviates from $\ce{[OH-]}=10^{-7}$ and $\ce{[H3O+]}=10^{-7}$. In this case you are right about the product being $K_{\rm W}$. Now you continuously increase the acid $\ce{HA}$ while both $\ce{[OH-]}$ and $\ce{[H3O+]}$ individually change their values. Is it because of $K_a K_b$ being a constant only dependent on temperature and also due to continuity, that this cannot change while the acid concentration is changed over magnitudes? – Diger Jul 12 '18 at 20:24
  • Another point worth mentioning which I'm not sure about is that: Aren't $K_a$ and $K_b$ by itsself so by each reaction above alone defined as I noted in the other comment? Do these values not changed upon mutual influence in the reaction? – Diger Jul 12 '18 at 20:47
  • Or put it another way: Does the value of $K$ not changed in equilibrium, regardless of how many reactions I might add to the chemistry which certainly will affect it. – Diger Jul 12 '18 at 20:53