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?

1 Answer
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 #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#