Question

5) A 100.0 mL sample of 0.18 M HClO4 is titrated with 0.27 M LiOH. Determine the pH of the solution

a) before the addition of any LiOH.

b) after the addition of 30.0 mL of LiOH.

c) after the addition of 50.0 mL of LiOH.

d) after the addition of 66.67 mL of LiOH (this is the equivalence point).

e) after the addition of 75.0 mL of LiOH.

Answer 1

1)when 0.0 mL of LiOH is added
Given:
M(HClO4) = 0.18 M
V(HClO4) = 100 mL
M(LiOH) = 0.27 M
V(LiOH) = 0 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.18 M * 100 mL = 18 mmol

mol(LiOH) = M(LiOH) * V(LiOH)
mol(LiOH) = 0.27 M * 0 mL = 0 mmol


We have:
mol(HClO4) = 18 mmol
mol(LiOH) = 0 mmol
0 mmol of both will react
remaining mol of HClO4 = 18 mmol
Total volume = 100.0 mL

[H+]= mol of acid remaining / volume
[H+] = 18 mmol/100.0 mL
= 0.18 M


use:
pH = -log [H+]
= -log (0.18)
= 0.7447
Answer: 0.745

2)when 30.0 mL of LiOH is added
Given:
M(HClO4) = 0.18 M
V(HClO4) = 100 mL
M(LiOH) = 0.27 M
V(LiOH) = 30 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.18 M * 100 mL = 18 mmol

mol(LiOH) = M(LiOH) * V(LiOH)
mol(LiOH) = 0.27 M * 30 mL = 8.1 mmol


We have:
mol(HClO4) = 18 mmol
mol(LiOH) = 8.1 mmol
8.1 mmol of both will react
remaining mol of HClO4 = 9.9 mmol
Total volume = 130.0 mL

[H+]= mol of acid remaining / volume
[H+] = 9.9 mmol/130.0 mL
= 7.615*10^-2 M


use:
pH = -log [H+]
= -log (7.615*10^-2)
= 1.1183
Answer: 1.12

3)when 50.0 mL of LiOH is added
Given:
M(HClO4) = 0.18 M
V(HClO4) = 100 mL
M(LiOH) = 0.27 M
V(LiOH) = 50 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.18 M * 100 mL = 18 mmol

mol(LiOH) = M(LiOH) * V(LiOH)
mol(LiOH) = 0.27 M * 50 mL = 13.5 mmol


We have:
mol(HClO4) = 18 mmol
mol(LiOH) = 13.5 mmol
13.5 mmol of both will react
remaining mol of HClO4 = 4.5 mmol
Total volume = 150.0 mL

[H+]= mol of acid remaining / volume
[H+] = 4.5 mmol/150.0 mL
= 3*10^-2 M


use:
pH = -log [H+]
= -log (3*10^-2)
= 1.5229
Answer: 1.52

4)when 66.67 mL of LiOH is added
Given:
M(HClO4) = 0.18 M
V(HClO4) = 100 mL
M(LiOH) = 0.27 M
V(LiOH) = 66.67 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.18 M * 100 mL = 18 mmol

mol(LiOH) = M(LiOH) * V(LiOH)
mol(LiOH) = 0.27 M * 66.67 mL = 18.0009 mmol


We have:
mol(HClO4) = 18 mmol
mol(LiOH) = 18 mmol
18 mmol of both will react

remaining mol of LiOH = 9*10^-4 mmol
Total volume = 166.67 mL

[OH-]= mol of base remaining / volume
[OH-] = 9*10^-4 mmol/166.67 mL
= 5.4*10^-6 M


use:
pOH = -log [OH-]
= -log (5.4*10^-6)
= 5.2676


use:
PH = 14 - pOH
= 14 - 5.2676
= 8.7324
Answer: 8.73

5)when 75.0 mL of LiOH is added
Given:
M(HClO4) = 0.18 M
V(HClO4) = 100 mL
M(LiOH) = 0.27 M
V(LiOH) = 75 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.18 M * 100 mL = 18 mmol

mol(LiOH) = M(LiOH) * V(LiOH)
mol(LiOH) = 0.27 M * 75 mL = 20.25 mmol


We have:
mol(HClO4) = 18 mmol
mol(LiOH) = 20.25 mmol
18 mmol of both will react

remaining mol of LiOH = 2.25 mmol
Total volume = 175.0 mL

[OH-]= mol of base remaining / volume
[OH-] = 2.25 mmol/175.0 mL
= 1.286*10^-2 M


use:
pOH = -log [OH-]
= -log (1.286*10^-2)
= 1.8909


use:
PH = 14 - pOH
= 14 - 1.8909
= 12.1091
Answer: 12.11