There is a titration with 100 ml of 0.0349M HClO4 with 0.0175M NaOH. In order to draw the titration curve, the pH at various points when NaOH was added need to be calculated.

Question

A. What is the pH when 0 ml of NaOH was added?

B. What is the pH after 100ml of NaOH has been added?

C. What is the pH after 210 ml of NaOH?

D. Determine the volume at the halfway point and equivalence point, what is the pH at the equivalence point?

Answer 1

The titration reaction between HClO4 and NaOH

A) HClO4 + NaOH --> NaClO4 + H2O

when 0 ml of NaOH is added, only acid is present so the pH will be as follows

pH= -log [H+]= -log(0.0349)= 1.457

B) pH when 100mL of 0.0175M NaOH added

mol of HClO4= M1V1 = 0.1L x 0.0349 mol/L = 0.00349 mol

mol of NaOH= M2V2 = 0.1L x 0.0175 mol/L = 0.00175 mol

Here there are more moles of acid than base so the solution is acidic

[H+]= (M1V1 - M2V2)/(V1+V2)= (1.74x10^-3)/0.200= 0.0087

pH= -log (0.0087)= 2.06

C) pH when 210mL of 0.0175M NaOH added

mol of HClO4= M1V1 = 0.00349 mol

mol of NaOH= M2V2 = 0.210L x 0.0175 mol/L = 0.00368 mol

Here there are more moles of base than acid so the solution is basic

[OH-]= (M2V2 - M1V1)/(V1+V2)= (1.9x10^-4)/.310= 0.000612

pOH= -log (0.000612)= 3.21

pH= 14 - pOH= 14 - 3.21= 10.78

D) V at mid-point

M1V1=M2V2

0.0349 x 0.100 = 0.0175 x V NaOH

V NaOH= 0.199 L (this is the Volume at the equivalence point)

V at halfway point would be half of that so 0.0997 L

For the titration of a strong acid and a strong base, the pH at the equivalence point is 7. A neutral salt, which is NaCl in this example, will be formed.