Question

Consider the titration of 42.5 mL of 0.215 M HClO4 with 0.124 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. (Assume that all solutions are at 25°C.) (a) 0.0 mL WebAssign will check your answer for the correct number of significant figures. (b) 10.0 mL WebAssign will check your answer for the correct number of significant figures. (c) 40.0 mL WebAssign will check your answer for the correct number of significant figures. (d) 80.0 mL WebAssign will check your answer for the correct number of significant figures. (e) 100.0 mL WebAssign will check your answer for the correct number of significant figures.

Answer 1

1)when 0.0 mL of KOH is added
Given:
M(HClO4) = 0.215 M
V(HClO4) = 42.5 mL
M(KOH) = 0.124 M
V(KOH) = 0 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.215 M * 42.5 mL = 9.1375 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.124 M * 0 mL = 0 mmol


We have:
mol(HClO4) = 9.137 mmol
mol(KOH) = 0 mmol
0 mmol of both will react
remaining mol of HClO4 = 9.137 mmol
Total volume = 42.5 mL

[H+]= mol of acid remaining / volume
[H+] = 9.137 mmol/42.5 mL
= 0.215 M


use:
pH = -log [H+]
= -log (0.215)
= 0.67

2)when 10.0 mL of KOH is added
Given:
M(HClO4) = 0.215 M
V(HClO4) = 42.5 mL
M(KOH) = 0.124 M
V(KOH) = 10 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.215 M * 42.5 mL = 9.1375 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.124 M * 10 mL = 1.24 mmol


We have:
mol(HClO4) = 9.137 mmol
mol(KOH) = 1.24 mmol
1.24 mmol of both will react
remaining mol of HClO4 = 7.897 mmol
Total volume = 52.5 mL

[H+]= mol of acid remaining / volume
[H+] = 7.897 mmol/52.5 mL
= 0.1504 M


use:
pH = -log [H+]
= -log (0.1504)
= 0.8227
Answer: 0.823

3)when 40.0 mL of KOH is added
Given:
M(HClO4) = 0.215 M
V(HClO4) = 42.5 mL
M(KOH) = 0.124 M
V(KOH) = 40 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.215 M * 42.5 mL = 9.1375 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.124 M * 40 mL = 4.96 mmol


We have:
mol(HClO4) = 9.137 mmol
mol(KOH) = 4.96 mmol
4.96 mmol of both will react
remaining mol of HClO4 = 4.177 mmol
Total volume = 82.5 mL

[H+]= mol of acid remaining / volume
[H+] = 4.177 mmol/82.5 mL
= 5.064*10^-2 M


use:
pH = -log [H+]
= -log (5.064*10^-2)
= 1.2955
Answer: 1.300

4)when 80.0 mL of KOH is added
Given:
M(HClO4) = 0.215 M
V(HClO4) = 42.5 mL
M(KOH) = 0.124 M
V(KOH) = 80 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.215 M * 42.5 mL = 9.1375 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.124 M * 80 mL = 9.92 mmol


We have:
mol(HClO4) = 9.137 mmol
mol(KOH) = 9.92 mmol
9.137 mmol of both will react

remaining mol of KOH = 0.7825 mmol
Total volume = 122.5 mL

[OH-]= mol of base remaining / volume
[OH-] = 0.7825 mmol/122.5 mL
= 6.388*10^-3 M


use:
pOH = -log [OH-]
= -log (6.388*10^-3)
= 2.1947


use:
PH = 14 - pOH
= 14 - 2.1947
= 11.8053
Answer: 11.805

5)when 100.0 mL of KOH is added
Given:
M(HClO4) = 0.215 M
V(HClO4) = 42.5 mL
M(KOH) = 0.124 M
V(KOH) = 100 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.215 M * 42.5 mL = 9.1375 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.124 M * 100 mL = 12.4 mmol


We have:
mol(HClO4) = 9.137 mmol
mol(KOH) = 12.4 mmol
9.137 mmol of both will react

remaining mol of KOH = 3.263 mmol
Total volume = 142.5 mL

[OH-]= mol of base remaining / volume
[OH-] = 3.263 mmol/142.5 mL
= 2.289*10^-2 M


use:
pOH = -log [OH-]
= -log (2.289*10^-2)
= 1.6403


use:
PH = 14 - pOH
= 14 - 1.6403
= 12.3597
Answer: 12.3597