hydronium ion concentration?

Question

Calculate the hydronium ion concentration and pH of the the solution that results when 22.0mL of 0.15M acetic acid is mixed with 22.0mL 0.15M NaOH.

Answer 1

This is a complete neutralization of both acid and base. However, since you have a weak acid at the start, the final product, NaOAc, is somewhat basic. You can start an equilibrium process as if that salt was initially there:
OAc- + H2O -> HOAc + OH-
We start with the acetate ion conc based on 0.15M solution from 22 mL, which is 0.00375 moles. Since the total liquid volume is now 0.044L, the initial conc= 0.00375 moles/0.044 L or 0.08 mol/L. Now, at equilibrium X mol/L if OAc- have reacted as noted above. So, we have
[HOAc][OH-]/[OAc-] = Kb.
The initial solution will be for [OH-]. Once we have that, conversion to hydronium ion and pH is easy.

Normally, Ka is supplied for the acid dissociation of HOAc. But Kb can be easily computed:
Kb = 10x10-15/Ka.
Use Ka=1.8x10-5, which makes Kb=5.6x10-10
Then we substitute in our numbers:
[X][X]/[0.08-X] = 5.6x10-10
Compared to 0.08, X will be considerably smaller, so we can neglect it in the denominator. Then, X^2= 0.08x5.6x10-10=4.5x10-12 and
X=2.2x10-6 mol/liter.
This is [OH-]. Hydronium ion is then:
[H+]= 10x10-15/2.2x10-6 = 4.5x10-9
and pH is 9-log 4.5= 8.34 appx.

Answer 2

Technically, it somewhat is not one of the above. you may make the argument for b (H+) in spite of the undeniable fact that it somewhat is particularly H3O+ (an H+ ion solvated via an H2O molecule). One consumer-friendly way this could ensue is whilst an acid is dissociated in water. This reaction yields the anions of the acid (Cl-, case in point) and the cation/proton (H+) the two of their aqueous state. As H2O is a polar molecule, the detrimental/oxygen end attracts the surely charged H+, forming the hydronium ion.