Question

Consider the titration of 50.0ml of 0.300 M NaOH by 0.100 M HClO₄^-. Calculate the pH of the solution at the following points in the titration:

a) no HClO₄ added

b) 50.0ml of HClO₄ added

c) 100.0ml of HClO₄ added

d) 150.0ml of HClO₄ added

e) 200.0ml of HClO₄ added

Answer 1

a)
when no acid is added:
[OH-]= 0.300 M
use:
pOH = -log [OH-]
= -log (0.3)
= 0.5229


use:
PH = 14 - pOH
= 14 - 0.5229
= 13.4771
Answer: 13.48

b)
Given:
M(HClO4) = 0.1 M
V(HClO4) = 50 mL
M(NaOH) = 0.3 M
V(NaOH) = 50 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.1 M * 50 mL = 5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.3 M * 50 mL = 15 mmol


We have:
mol(HClO4) = 5 mmol
mol(NaOH) = 15 mmol
5 mmol of both will react

remaining mol of NaOH = 10 mmol
Total volume = 100.0 mL

[OH-]= mol of base remaining / volume
[OH-] = 10 mmol/100.0 mL
= 0.1 M


use:
pOH = -log [OH-]
= -log (0.1)
= 1


use:
PH = 14 - pOH
= 14 - 1
= 13
Answer: 13

c)
Given:
M(HClO4) = 0.1 M
V(HClO4) = 100 mL
M(NaOH) = 0.3 M
V(NaOH) = 50 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.1 M * 100 mL = 10 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.3 M * 50 mL = 15 mmol


We have:
mol(HClO4) = 10 mmol
mol(NaOH) = 15 mmol
10 mmol of both will react

remaining mol of NaOH = 5 mmol
Total volume = 150.0 mL

[OH-]= mol of base remaining / volume
[OH-] = 5 mmol/150.0 mL
= 0.0333 M


use:
pOH = -log [OH-]
= -log (3.333*10^-2)
= 1.4771


use:
PH = 14 - pOH
= 14 - 1.4771
= 12.52
Answer: 12.52

d)
Given:
M(HClO4) = 0.1 M
V(HClO4) = 150 mL
M(NaOH) = 0.3 M
V(NaOH) = 50 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.1 M * 150 mL = 15 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.3 M * 50 mL = 15 mmol


We have:
mol(HClO4) = 15 mmol
mol(NaOH) = 15 mmol
15 mmol of both will react
solution will be neutral and pH will be 7
Answer: 7.00

e)
Given:
M(HClO4) = 0.1 M
V(HClO4) = 200 mL
M(NaOH) = 0.3 M
V(NaOH) = 50 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.1 M * 200 mL = 20 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.3 M * 50 mL = 15 mmol


We have:
mol(HClO4) = 20 mmol
mol(NaOH) = 15 mmol
15 mmol of both will react
remaining mol of HClO4 = 5 mmol
Total volume = 250.0 mL

[H+]= mol of acid remaining / volume
[H+] = 5 mmol/250.0 mL
= 0.02 M


use:
pH = -log [H+]
= -log (2*10^-2)
= 1.699
Answer: 1.70