Question

Consider the titration of 40.0 mL of 0.200 MHCIO by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a. 0.0 ml pH = b. 10.0 ml pH = c. 70.0 ml pH = d. 80.0 mL pH = e. 130.0 ml pH =

Answer 1

1)when 0.0 mL of KOH is added
Given:
M(HClO4) = 0.2 M
V(HClO4) = 40 mL
M(KOH) = 0.1 M
V(KOH) = 0 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.2 M * 40 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 0 mL = 0 mmol


We have:
mol(HClO4) = 8 mmol
mol(KOH) = 0 mmol
0 mmol of both will react
remaining mol of HClO4 = 8 mmol
Total volume = 40.0 mL

[H+]= mol of acid remaining / volume
[H+] = 8 mmol/40.0 mL
= 0.2 M


use:
pH = -log [H+]
= -log (0.2)
= 0.699
Answer: 0.699

2)when 10.0 mL of KOH is added
Given:
M(HClO4) = 0.2 M
V(HClO4) = 40 mL
M(KOH) = 0.1 M
V(KOH) = 10 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.2 M * 40 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 10 mL = 1 mmol


We have:
mol(HClO4) = 8 mmol
mol(KOH) = 1 mmol
1 mmol of both will react
remaining mol of HClO4 = 7 mmol
Total volume = 50.0 mL

[H+]= mol of acid remaining / volume
[H+] = 7 mmol/50.0 mL
= 0.14 M


use:
pH = -log [H+]
= -log (0.14)
= 0.8539
Answer: 0.854

3)when 70.0 mL of KOH is added
Given:
M(HClO4) = 0.2 M
V(HClO4) = 40 mL
M(KOH) = 0.1 M
V(KOH) = 70 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.2 M * 40 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 70 mL = 7 mmol


We have:
mol(HClO4) = 8 mmol
mol(KOH) = 7 mmol
7 mmol of both will react
remaining mol of HClO4 = 1 mmol
Total volume = 110.0 mL

[H+]= mol of acid remaining / volume
[H+] = 1 mmol/110.0 mL
= 9.091*10^-3 M


use:
pH = -log [H+]
= -log (9.091*10^-3)
= 2.0414
Answer: 2.04

4)when 80.0 mL of KOH is added
Given:
M(HClO4) = 0.2 M
V(HClO4) = 40 mL
M(KOH) = 0.1 M
V(KOH) = 80 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.2 M * 40 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 80 mL = 8 mmol


We have:
mol(HClO4) = 8 mmol
mol(KOH) = 8 mmol
8 mmol of both will react to form neutral solution
hence pH of solution will be 7
Answer: 7.00

5)when 130.0 mL of KOH is added
Given:
M(HClO4) = 0.2 M
V(HClO4) = 40 mL
M(KOH) = 0.1 M
V(KOH) = 130 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.2 M * 40 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 130 mL = 13 mmol


We have:
mol(HClO4) = 8 mmol
mol(KOH) = 13 mmol
8 mmol of both will react

remaining mol of KOH = 5 mmol
Total volume = 170.0 mL

[OH-]= mol of base remaining / volume
[OH-] = 5 mmol/170.0 mL
= 2.941*10^-2 M


use:
pOH = -log [OH-]
= -log (2.941*10^-2)
= 1.5315


use:
PH = 14 - pOH
= 14 - 1.5315
= 12.4685
Answer: 12.47