Calculate the pH for the titration of 50.0 mL 0.100 M HC2H3O2with 0.200 M NaOH at each of the following points

ph-for-the-titration

#1

Calculate the pH for the titration of 50.0 mL 0.100 M HC2H3O2with 0.200 M NaOH at each of the following points. 0.00mL, 12.5 mL, 25.0 mL and 50 mL of NaOH added. (The Kafor HC2H3O2is 1.8 X 10-5)

Answer:

at 0ml of NaOH
[H+]=ka∗M−−−−−−√
[H+]=((1.8∗10−5)(.1))−−−−−−−−−−−−−−√
[H+]=.00134
-log(.00134)=2.87
PH=2.87

at 12.5 ml, we should write out the equation to understand what is going on.
NaOH + HC2H3O2 ----> NaC2H3O2 + H2O
This can be treated as a buffer solution since we have a weak acid (HC2H3O2) and its conjugate base(NaC2H3O2).
Essentially we are adding 0.0125L of 0.2M NaOH… which is the equivalent of 0.0025 mol NaOH.
We also have 0.05L of HC2H3O2 at 0.1M, equivalent to 0.005mol HC2H3O2.

take 0.005mol HC2H3O2, subtract 0.0025 NaOH from that because they react 1 for 1.
0.005-0.0025= 0.0025 moles of (HC2H3O2)
and from this, 0.0025 moles of the conjugate base (NaC2H3O2) are produced.
We then use the Henderson Hasselbach equation.
PH=pka + log [base]/[acid]
PH= 4.744+log((0.0025)/(0.0025))
PH= 4.744

At 25 ml or 0.025L.
0.025L0.2M= 0.005 moles of NaOH
since we had 0.005 moles of HC2H3O2 , this is the equivalence point.
At the equivalence point all the HC2H3O2 is being consumed along with the NaOH, but the salt solution will possibly break apart and create more [OH-]
So this is treated as a hydrolyses calculation.
We first must calculate the total volume after adding the NaOH solution. 0.025L + 0.05L= 0.75L (total volume)
then we must use this to calculate the Molarity of the of the conjugate base (NaHC2H3O2)… All of the HC2H3O2 is reacted and converted to the conjugate base, therefore the moles of NaC2H3O2 is 0.005.
(0.005 moles)/(0.75L)= 0.00666 Molarity of NaC2H3O2
find the Kb of the conjugate base (NaC2H3O2)
Kw=Kb
Ka
(110^-14)=Kb(1.810^-5)
solve for Kb = 5.55
10^-10

[OH-]=Ka∗M−−−−−−√
[OH-]= 1.92*10^-6
PH= 8.28

When 50 ml or 0.05L of NaOH is added it is considered an excess amount of OH- ion.
0.05L*0.2M= 0.01moles of NaOH
we still have only 0.005 moles of HC2H3O2
to calculate the excess of NaOH we subtract
0.01- 0.005= 0.005 moles of NaOH left unreacted with the HC2H3O2
this is the same thing as saying there are 0.005 moles of OH-
therefore if we find the concentration of [OH-], we can find the final PH.
total volume = 0.05+0.05= 0.1L
0.005moles/ 0.1L= 0.5M=[OH-]
-log(0.5)= 1.3
POH=1.3
PH=12.7