0.100 M sodium propanoate (NaC3H5O2) (HC3H5O2, Ka= 1.3 10-5)?

Question

(2) 0.100 M HC3H5O2 and 0.100 M NaC3H5O2
Calculate the pH after 0.029 mol HCl is added to 1.00 L
please explain in detail. i need a full understanding
thanks

Answer 1

You have a buffering system so you can use the Henderson-Hasselbalch equation pH=pKa+log[conj.base]/[acid]
and in this case conjugate base is NaC3H5O2 (C3H5O2- to be exact) and acid HC3H5O2.

You added some strong acid (HCl). This will react with the conjugate base producing more acid. So you need to find how much conjugate base remains and how much acid you now have (the amount you had before+what was produced) and plug the values into the Henderson-Hasselbalch equation

moles NaC3H5O2 =M*V= 0.1*1=0.1
moles HC3H5O2=M*V =0.1*1=0.1

HCl + NaC3H5O2 ->HC3H5O2 +NaCl

The stoichiometry is 1:1:1 so 0.029 mole HCl will react with 0.029 mole NaC3H5O2 and produce 0.029 mole HC3H5O2
So you will have remaining mole NaC3H5O2=0.1-0.029= 0.071
and total mole HC3H5O2=0.1+0.029= 0.129
Convert into concentrations
[NaC3H5O2]=0.071/1=0.071
[HC3H5O2]=0.129/1=0.129

So pH=pKa + log( [NaC3H5O2] / [HC3H5O2] )=
= -log(1.3*10^-5) + log(0.071/0.129) = 4.63