Question

Using the average concentration of NaOH found above, calculate the theoretical pH after 2.50 mL and 9.50 mL of NaOH has been added in both the titration of HCl and of HC2H3O2. Indicate if the volume of NaOH is before or after the equivalence point. Compare these Titration Curves theoretical values with the actual values found on the titration curves created in lab. (Hint: Use total volume of titrant (the initial and final volume on the buret) in conjunction with the time required to deliver that volume to convert the volumes given above to time.) Show a sample of all necessary calculations.

Average NaOH: 0.0984 M

HCl: 0.11M

actual inititial V: 2.7 mL

actual final V: 31.1 mL

I'm not sure how to start to find the theoretical pH, if someone could help explain it for the 2.50 mL added to HCl

Answer 1

pH=-log [H+]

where [H+] is the concentration of remaining HCl after incomplete neutralisation.

If complete neutralisation occurs according to the equation,

HCl+NaOHNaOH+H2O

then no H+ in the solution so pH cant be finite as the logarithm will be infinity.

Here, in titration

Volume of NaOH added=2.50 ml

Molarity of NaOH=0.0984M

volume of HCl (after dilution)=100.00 ml

Molarity of HCl=0.11M=0.11mol/L

using formula, Molarity1 * volume1=Molarity2 * volume2

After dilution, Molarity of HCl=0.11 M * 8.00 /100 =0.008M

for incomplete neutraisation,

M1*V1=M2(Vi -Vf)    Vi=initial HCL volume ,Vf= HCl not neutralised

0.0984 M* 2.50ml =0.00815M(108- Vf)

0.246=0.8802-0.00815Vf

solving this we get, Vf=77.82ml

The volume 2.50 ml is before equivalence point.

now the concentration of remaining HCl=0.008

Theoretical pH of HCl=-log(0.008)=2.09

Theoretical pH of NaOH=14 -log(OH-)=14-log(0.0984)=14-1.007=12.993