Question

What sort of solution will result if one equiv of hydrochloric acid is added to one equiv of calcium hydroxide? And how does `pH` relate to `pOH` in such solutions?

Answer 1

Well hydrochloric acid is an acid that gives one equivalent of hydronium ion.........so you will get a BASIC solution is you add one equiv of calcium hydroxide........

Explanation:

Whereas, calcium hydroxide gives TWO equivalents of hydroxide anion. The solution will thus be stoichiometric in `Ca(OH)Cl(aq)`, and thus there is still one equiv of `HO^-` in solution.

We could represent neutralization of calcium hydroxide by hydrochloric acid by the following stoichiometric equation.....

`Ca(OH)_2(s) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)`

Calcium hydroxide is not particularly soluble in aqueous solution; on t'other hand, calcium chloride is soluble.......

As background, we know that in aqueous solution gives rise to an increase in the concentration of `H_3O^+`, conceived as the characteristic cation of the water solvent, whereas calcium hydroxide is a base that gives rise to increased concentrations of `HO^-`, likewise conceived to be the characteristic anion of the water solvent.

A solution that is `1*mol*L^-1` with respect to `Ca(OH)_2` is `2* mol*L^-1` with respect to hydroxide anion. Agreed? We can further quantify the acid-base behaviour by invoking the `pH` scale. `pH=-log_10[H_3O^+]` and `pOH=-log_10[HO^-]`, and given water's autoprotolysis,......

`2H_2OrightleftharpoonsH_3O^+ + HO^-`, where ............

`K_w=[HO^-][H_3O^+]=10^-14`

.....if we take `log_10` of both sides, we gets..............

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]`

But `log_10K_w=log_10(10^-14)=-14` Do you see this? It is important to appreciate and understand this fact.

And so..........

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]=-14`

On rearrangement.......

`14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)`

And so our defining relationship.......

`pH+pOH=14`