Question

A 99.0 mL sample of 0.0200 M HBrO₄ is titrated with 0.0400 M RbOH solution. Calculate the pH after the following volumes of base have been added.

(a) 13.9 mL pH =	(b) 48.0 mL pH =	(c) 49.5 mL pH =
(d) 50.5 mL pH =	(e) 76.2 mL pH =

Answer 1

1)when 13.9 mL of RbOH is added
Given:
M(HBrO4) = 0.02 M
V(HBrO4) = 99 mL
M(RbOH) = 0.04 M
V(RbOH) = 13.9 mL


mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.02 M * 99 mL = 1.98 mmol

mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.04 M * 13.9 mL = 0.556 mmol


We have:
mol(HBrO4) = 1.98 mmol
mol(RbOH) = 0.556 mmol
0.556 mmol of both will react
remaining mol of HBrO4 = 1.424 mmol
Total volume = 112.9 mL

[H+]= mol of acid remaining / volume
[H+] = 1.424 mmol/112.9 mL
= 1.261*10^-2 M


use:
pH = -log [H+]
= -log (1.261*10^-2)
= 1.8992
Answer: 1.90

2)when 48.0 mL of RbOH is added
Given:
M(HBrO4) = 0.02 M
V(HBrO4) = 99 mL
M(RbOH) = 0.04 M
V(RbOH) = 48 mL


mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.02 M * 99 mL = 1.98 mmol

mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.04 M * 48 mL = 1.92 mmol


We have:
mol(HBrO4) = 1.98 mmol
mol(RbOH) = 1.92 mmol
1.92 mmol of both will react
remaining mol of HBrO4 = 6*10^-2 mmol
Total volume = 147.0 mL

[H+]= mol of acid remaining / volume
[H+] = 6*10^-2 mmol/147.0 mL
= 4.082*10^-4 M


use:
pH = -log [H+]
= -log (4.082*10^-4)
= 3.3892
Answer: 3.39

3)when 49.5 mL of RbOH is added
Given:
M(HBrO4) = 0.02 M
V(HBrO4) = 99 mL
M(RbOH) = 0.04 M
V(RbOH) = 49.5 mL


mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.02 M * 99 mL = 1.98 mmol

mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.04 M * 49.5 mL = 1.98 mmol


We have:
mol(HBrO4) = 1.98 mmol
mol(RbOH) = 1.98 mmol
1.98 mmol of both will react to form neutral solution
hence pH of solution will be 7
Answer: 7.00

4)when 50.5 mL of RbOH is added
Given:
M(HBrO4) = 0.02 M
V(HBrO4) = 99 mL
M(RbOH) = 0.04 M
V(RbOH) = 50.5 mL


mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.02 M * 99 mL = 1.98 mmol

mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.04 M * 50.5 mL = 2.02 mmol


We have:
mol(HBrO4) = 1.98 mmol
mol(RbOH) = 2.02 mmol
1.98 mmol of both will react

remaining mol of RbOH = 4*10^-2 mmol
Total volume = 149.5 mL

[OH-]= mol of base remaining / volume
[OH-] = 4*10^-2 mmol/149.5 mL
= 2.676*10^-4 M


use:
pOH = -log [OH-]
= -log (2.676*10^-4)
= 3.5726


use:
PH = 14 - pOH
= 14 - 3.5726
= 10.4274
Answer: 10.43

5)when 76.2 mL of RbOH is added
Given:
M(HBrO4) = 0.02 M
V(HBrO4) = 99 mL
M(RbOH) = 0.04 M
V(RbOH) = 76.2 mL


mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.02 M * 99 mL = 1.98 mmol

mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.04 M * 76.2 mL = 3.048 mmol


We have:
mol(HBrO4) = 1.98 mmol
mol(RbOH) = 3.048 mmol
1.98 mmol of both will react

remaining mol of RbOH = 1.068 mmol
Total volume = 175.2 mL

[OH-]= mol of base remaining / volume
[OH-] = 1.068 mmol/175.2 mL
= 6.096*10^-3 M


use:
pOH = -log [OH-]
= -log (6.096*10^-3)
= 2.215


use:
PH = 14 - pOH
= 14 - 2.215
= 11.785
Answer: 11.78