# What is the significance of pH in a basic solution?

## Main Question or Discussion Point

I am a little confused about the meaning of pH, especially when you have a basic solution. I know that $pH = -log[H^{+}]$, but I don't understand how you can have a pH for a basic solution. Don't basic solutions only produce hydroxide ions?

I know that $pH + pOH = 14$ but why this equation be true for anything other than water?

For example, consider the basic solution of ammonia to produce ammonium ions and hydroxide ions.

$$NH_{3}(g) + H_{2}O(l) \rightleftharpoons NH^{+}_{4}(aq) + OH^{-}(aq)$$

Knowing the concentrations and the base-dissociation constant, we could easily calculate $[OH^{-}]$ and from that we can get pOH, but what is the significance of pH if there is no $H^{+}$ ions?

Why do you say other than water? Water has its own dissociation constant Kw=10-14=[H3O+][OH-]. Taking logarithm on both sides, I think you would get the same result.

And it is due to the fact that at temperature of your room, water dissociate in the aforesaid proportions; therefore, under aquatic, room temperature, it is always true. If any of the ions are in excess, the equilibrium of the dissociation reaction would shift, so that some of those would recombine and be back to equilibrium. Of course you can analyse the free energy and understand more precisely, but that's not quite relevant.

Thank you for your replies everyone. I do know that water dissociates, but I still don't understand the significance of pOH in an acidic solution and the significance of pH in basic solution.

Acid produces hydrogen ions in water, but the hydroxide ions come from the water and not from the acid? What is the point in counting the water's hydroxide ions toward the calculation of pOH for the acid? I'm not sure I get it... all I know is that water dissociates into both hydrogen and hydroxide ions. But acid dissociates ONLY into hydrogen right? So where does the pOH for the acid come from? And pH for the base? Am I missing something?

Thanks!

BiP

Borek
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