Question

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.120 M KOH(aq), with 0.120 M HBr(aq).

(a) before addition of any HBr

(b) after addition of 13.5 mL of HBr

(c) after addition of 23.5 mL of HBr

(d) after the addition of 35.0 mL of HBr

(e) after the addition of 45.5 mL of HBr

(f) after the addition of 50.0 mL of HBr

Answer 1

1)when 0.0 mL of HBr is added
Given:
M(HBr) = 0.12 M
V(HBr) = 0 mL
M(KOH) = 0.12 M
V(KOH) = 35 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.12 M * 0 mL = 0 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.12 M * 35 mL = 4.2 mmol


We have:
mol(HBr) = 0 mmol
mol(KOH) = 4.2 mmol
0 mmol of both will react

remaining mol of KOH = 4.2 mmol
Total volume = 35 mL

[OH-]= mol of base remaining / volume
[OH-] = 4.2 mmol/35 mL
= 0.12 M


use:
pOH = -log [OH-]
= -log (0.12)
= 0.9208


use:
PH = 14 - pOH
= 14 - 0.9208
= 13.0792
Answer: 13.0792

2)when 13.5 mL of HBr is added
Given:
M(HBr) = 0.12 M
V(HBr) = 13.5 mL
M(KOH) = 0.12 M
V(KOH) = 35 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.12 M * 13.5 mL = 1.62 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.12 M * 35 mL = 4.2 mmol


We have:
mol(HBr) = 1.62 mmol
mol(KOH) = 4.2 mmol
1.62 mmol of both will react

remaining mol of KOH = 2.58 mmol
Total volume = 48.5 mL

[OH-]= mol of base remaining / volume
[OH-] = 2.58 mmol/48.5 mL
= 5.32*10^-2 M


use:
pOH = -log [OH-]
= -log (5.32*10^-2)
= 1.2741


use:
PH = 14 - pOH
= 14 - 1.2741
= 12.7259
Answer: 12.7259

3)when 23.5 mL of HBr is added
Given:
M(HBr) = 0.12 M
V(HBr) = 23.5 mL
M(KOH) = 0.12 M
V(KOH) = 35 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.12 M * 23.5 mL = 2.82 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.12 M * 35 mL = 4.2 mmol


We have:
mol(HBr) = 2.82 mmol
mol(KOH) = 4.2 mmol
2.82 mmol of both will react

remaining mol of KOH = 1.38 mmol
Total volume = 58.5 mL

[OH-]= mol of base remaining / volume
[OH-] = 1.38 mmol/58.5 mL
= 2.359*10^-2 M


use:
pOH = -log [OH-]
= -log (2.359*10^-2)
= 1.6273


use:
PH = 14 - pOH
= 14 - 1.6273
= 12.3727
Answer: 12.3727

4)when 35.0 mL of HBr is added
Given:
M(HBr) = 0.12 M
V(HBr) = 35 mL
M(KOH) = 0.12 M
V(KOH) = 35 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.12 M * 35 mL = 4.2 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.12 M * 35 mL = 4.2 mmol


We have:
mol(HBr) = 4.2 mmol
mol(KOH) = 4.2 mmol
4.2 mmol of both will react to form neutral solution
hence pH of solution will be 7
Answer: 7

5)when 45.5 mL of HBr is added
Given:
M(HBr) = 0.12 M
V(HBr) = 45.5 mL
M(KOH) = 0.12 M
V(KOH) = 35 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.12 M * 45.5 mL = 5.46 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.12 M * 35 mL = 4.2 mmol


We have:
mol(HBr) = 5.46 mmol
mol(KOH) = 4.2 mmol
4.2 mmol of both will react
remaining mol of HBr = 1.26 mmol
Total volume = 80.5 mL

[H+]= mol of acid remaining / volume
[H+] = 1.26 mmol/80.5 mL
= 1.565*10^-2 M


use:
pH = -log [H+]
= -log (1.565*10^-2)
= 1.8054
Answer: 1.8054

6)when 50.0 mL of HBr is added
Given:
M(HBr) = 0.12 M
V(HBr) = 50 mL
M(KOH) = 0.12 M
V(KOH) = 35 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.12 M * 50 mL = 6 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.12 M * 35 mL = 4.2 mmol


We have:
mol(HBr) = 6 mmol
mol(KOH) = 4.2 mmol
4.2 mmol of both will react
remaining mol of HBr = 1.8 mmol
Total volume = 85.0 mL

[H+]= mol of acid remaining / volume
[H+] = 1.8 mmol/85 mL
= 2.118*10^-2 M


use:
pH = -log [H+]
= -log (2.118*10^-2)
= 1.6741
Answer: 1.6741