Homework Answers
1)when 0.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 0 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 0 mL = 0 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 0 mmol
0 mmol of both will react
remaining mol of HCl = 9.5 mmol
Total volume = 50.0 mL
[H+]= mol of acid remaining / volume
[H+] = 9.5 mmol/50.0 mL
= 0.19 M
use:
pH = -log [H+]
= -log (0.19)
= 0.7212
Answer: 0.721
2)when 10.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 10 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 10 mL = 1.9 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 1.9 mmol
1.9 mmol of both will react
remaining mol of HCl = 7.6 mmol
Total volume = 60.0 mL
[H+]= mol of acid remaining / volume
[H+] = 7.6 mmol/60.0 mL
= 0.1267 M
use:
pH = -log [H+]
= -log (0.1267)
= 0.8973
Answer: 0.897
3)when 40.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 40 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 40 mL = 7.6 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 7.6 mmol
7.6 mmol of both will react
remaining mol of HCl = 1.9 mmol
Total volume = 90.0 mL
[H+]= mol of acid remaining / volume
[H+] = 1.9 mmol/90.0 mL
= 2.111*10^-2 M
use:
pH = -log [H+]
= -log (2.111*10^-2)
= 1.6755
Answer: 1.68
4)when 45.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 45 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 45 mL = 8.55 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 8.55 mmol
8.55 mmol of both will react
remaining mol of HCl = 0.95 mmol
Total volume = 95.0 mL
[H+]= mol of acid remaining / volume
[H+] = 0.95 mmol/95.0 mL
= 1*10^-2 M
use:
pH = -log [H+]
= -log (1*10^-2)
= 2
Answer: 2.00
5)when 49.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 49 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 49 mL = 9.31 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 9.31 mmol
9.31 mmol of both will react
remaining mol of HCl = 0.19 mmol
Total volume = 99.0 mL
[H+]= mol of acid remaining / volume
[H+] = 0.19 mmol/99.0 mL
= 1.919*10^-3 M
use:
pH = -log [H+]
= -log (1.919*10^-3)
= 2.7169
Answer: 2.72
6)when 50.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 50 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 50 mL = 9.5 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 9.5 mmol
9.5 mmol of both will react to form neutral solution
hence pH of solution will be 7
Answer: 7.00
7)when 51.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 51 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 51 mL = 9.69 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 9.69 mmol
9.5 mmol of both will react
remaining mol of NaOH = 0.19 mmol
Total volume = 101.0 mL
[OH-]= mol of base remaining / volume
[OH-] = 0.19 mmol/101.0 mL
= 1.881*10^-3 M
use:
pOH = -log [OH-]
= -log (1.881*10^-3)
= 2.7256
use:
PH = 14 - pOH
= 14 - 2.7256
= 11.2744
Answer: 11.27
8)when 55.0 mL of NaOH is added
Given:
M(HCl) = 0.19 M
V(HCl) = 50 mL
M(NaOH) = 0.19 M
V(NaOH) = 55 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.19 M * 50 mL = 9.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.19 M * 55 mL = 10.45 mmol
We have:
mol(HCl) = 9.5 mmol
mol(NaOH) = 10.45 mmol
9.5 mmol of both will react
remaining mol of NaOH = 0.95 mmol
Total volume = 105.0 mL
[OH-]= mol of base remaining / volume
[OH-] = 0.95 mmol/105.0 mL
= 9.048*10^-3 M
use:
pOH = -log [OH-]
= -log (9.048*10^-3)
= 2.0435
use:
PH = 14 - pOH
= 14 - 2.0435
= 11.9565
Answer: 11.96
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