Hey guys,
There's an example in my book of pH calculation in a weak acid-strong base titration, and something in there just doesn't really sit well with me at all.
The question is as follows:
Calculate the pH of a solution where 5ml of 0.15M NaOH is titrated onto 25ml of 0.1M HCOOH. (Ka for HCOOH = 1.8e-4)
The initial pH was calculated in the equilibrium table and after approximations [H
3O
+]=(Ka[HCOOH])
0.5=4.2e-3M ; -log[H
3O
+]=2.37 . This stage is clear to me.
Now calculating the moles for OH
- : 0.005l*0.15M=7.5e-4 mol
The reaction: HCOOH(aq)+OH
-(aq)
HCO
2-(aq)+H
2O(l)
Okay here is what I don't quite get... it says - of 0.750 milimoles of OH-, 0.750 milimoles of HCO
2- are created, and 1.75mmol HCOOH are left.
The book proceeds to calculate the concentrations of the acid and conj. base:
[HCOOH]= 1.75e-3mol/0.03l=0.0583M ; [HCO
2-]=7.5e-4mol/0.03l=0.0250M
Using HH and after approximations: pH=pKa+log(0.025/0.0583)=3.38
My questions are these:
Why does the book seem to ignore [H
3O
+] that was initially created just by the weak deprotonation? (the 4.2e-3 M)
Why from 0.75mmol of OH- is 0.75mmol of conj base made? I thought that the initial conj base concentration was equal to the initial [H
3O
+] and should be accounted for aswell... and I'm not so sure why the stoichiometry here of 0.750mmol OH- = 0.750mmol HCO
2- is happening.
Does the introduction of OH- neutralize the acid itself? doesn't it neutralize its hydronium ions first? I am so so so confused....
I'd really appreciate some input for this... I've been sitting for hours just trying to get it...
Thank you