wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An unknown volume and unknown concentration of a weak acid HX is titrated with NaOH of unknown concentration. After addition of 10.0 cm3 of the NaOH solution, the pH of the solution is 5.7. After the addition of 20.0 cm3 of the NaOH solution, the pH is 6.3. Calculate the pKa of the weak acid, HX. (Given: log2 = 0.3, log = 0.5)

Open in App
Solution

Let the molarity and volume of HX be M1 and V1 respectively, For NaOH assume the molarity to be M2
Now the reaction involving the titration will be
HX+NaOHNaX+H2O
Since the complete neutralization has not taken place, the solution will act as an acidic buffer:
pH=pKa+log[salt][acid]
When 10 ml of NaOH is used
5.7=pKa+log[10M2][M1V110M2]
Lets say V2 volume of NaOH is needed for complete neutralization and M1V1=M2V2, so
5.7=pKa+log[10M2][M2V210M2]
5.7=pKa+log[10][V210] (i)
similarly for 20 ml NaOH used
6.3=pKa+log[20][V220] (ii)
Now from (i) and (ii) we have
6.35.7=log20(V210)10(V220)
2×0.3=2log2=log4=log2(V210)1(V220)
Hence V2=30ml;
5.7=pKa+log[10][V210]
5.7=pKa+log[10][20]
5.7=pKalog2
pKa=5.7+0.3=6

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Buffer Solutions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon