# 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?

Jul 17, 2017

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 $C a \left(O H\right) C l \left(a q\right)$, and thus there is still one equiv of $H {O}^{-}$ in solution.

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

$C a {\left(O H\right)}_{2} \left(s\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

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 ${H}_{3} {O}^{+}$, conceived as the characteristic cation of the water solvent, whereas calcium hydroxide is a base that gives rise to increased concentrations of $H {O}^{-}$, likewise conceived to be the characteristic anion of the water solvent.

A solution that is $1 \cdot m o l \cdot {L}^{-} 1$ with respect to $C a {\left(O H\right)}_{2}$ is $2 \cdot m o l \cdot {L}^{-} 1$ with respect to hydroxide anion. Agreed? We can further quantify the acid-base behaviour by invoking the $p H$ scale. $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, and given water's autoprotolysis,......

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$, where ............

${K}_{w} = \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right] = {10}^{-} 14$

.....if we take ${\log}_{10}$ of both sides, we gets..............

${\log}_{10} {K}_{w} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$

But ${\log}_{10} {K}_{w} = {\log}_{10} \left({10}^{-} 14\right) = - 14$ Do you see this? It is important to appreciate and understand this fact.

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

${\log}_{10} {K}_{w} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = - 14$

On rearrangement.......

$14 = {\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{p H} {\underbrace{- {\log}_{10} \left[H {O}^{-}\right]}}_{p O H}$

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

$p H + p O H = 14$