1)when 0.0 mL of HBrO4 is added
HONH2 dissociates as:
HONH2 +H2O -----> HONH3+ + OH-
8*10^-2 0 0
8*10^-2-x x x
Kb = [HONH3+][OH-]/[HONH2]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.1*10^-8)*8*10^-2) = 2.966*10^-5
since c is much greater than x, our assumption is correct
so, x = 2.966*10^-5 M
So, [OH-] = x = 2.966*10^-5 M
use:
pOH = -log [OH-]
= -log (2.966*10^-5)
= 4.5278
use:
PH = 14 - pOH
= 14 - 4.5278
= 9.4722
Answer: 9.47
2)when 4.0 mL of HBrO4 is added
Given:
M(HBrO4) = 0.1 M
V(HBrO4) = 4 mL
M(HONH2) = 0.08 M
V(HONH2) = 20 mL
mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.1 M * 4 mL = 0.4 mmol
mol(HONH2) = M(HONH2) * V(HONH2)
mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol
We have:
mol(HBrO4) = 0.4 mmol
mol(HONH2) = 1.6 mmol
0.4 mmol of both will react
excess HONH2 remaining = 1.2 mmol
Volume of Solution = 4 + 20 = 24 mL
[HONH2] = 1.2 mmol/24 mL = 0.05 M
[HONH3+] = 0.4 mmol/24 mL = 0.0167 M
They form basic buffer
base is HONH2
conjugate acid is HONH3+
Kb = 1.1*10^-8
pKb = - log (Kb)
= - log(1.1*10^-8)
= 7.959
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 7.959+ log {1.667*10^-2/5*10^-2}
= 7.481
use:
PH = 14 - pOH
= 14 - 7.4815
= 6.5185
Answer: 6.52
3)when 8.0 mL of HBrO4 is added
Given:
M(HBrO4) = 0.1 M
V(HBrO4) = 8 mL
M(HONH2) = 0.08 M
V(HONH2) = 20 mL
mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.1 M * 8 mL = 0.8 mmol
mol(HONH2) = M(HONH2) * V(HONH2)
mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol
We have:
mol(HBrO4) = 0.8 mmol
mol(HONH2) = 1.6 mmol
0.8 mmol of both will react
excess HONH2 remaining = 0.8 mmol
Volume of Solution = 8 + 20 = 28 mL
[HONH2] = 0.8 mmol/28 mL = 0.0286 M
[HONH3+] = 0.8 mmol/28 mL = 0.0286 M
They form basic buffer
base is HONH2
conjugate acid is HONH3+
Kb = 1.1*10^-8
pKb = - log (Kb)
= - log(1.1*10^-8)
= 7.959
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 7.959+ log {2.857*10^-2/2.857*10^-2}
= 7.959
use:
PH = 14 - pOH
= 14 - 7.9586
= 6.0414
Answer: 6.04
4)when 12.0 mL of HBrO4 is added
Given:
M(HBrO4) = 0.1 M
V(HBrO4) = 12 mL
M(HONH2) = 0.08 M
V(HONH2) = 20 mL
mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.1 M * 12 mL = 1.2 mmol
mol(HONH2) = M(HONH2) * V(HONH2)
mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol
We have:
mol(HBrO4) = 1.2 mmol
mol(HONH2) = 1.6 mmol
1.2 mmol of both will react
excess HONH2 remaining = 0.4 mmol
Volume of Solution = 12 + 20 = 32 mL
[HONH2] = 0.4 mmol/32 mL = 0.0125 M
[HONH3+] = 1.2 mmol/32 mL = 0.0375 M
They form basic buffer
base is HONH2
conjugate acid is HONH3+
Kb = 1.1*10^-8
pKb = - log (Kb)
= - log(1.1*10^-8)
= 7.959
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 7.959+ log {3.75*10^-2/1.25*10^-2}
= 8.436
use:
PH = 14 - pOH
= 14 - 8.4357
= 5.5643
Answer: 5.56
5)when 16.0 mL of HBrO4 is added
Given:
M(HBrO4) = 0.1 M
V(HBrO4) = 16 mL
M(HONH2) = 0.08 M
V(HONH2) = 20 mL
mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.1 M * 16 mL = 1.6 mmol
mol(HONH2) = M(HONH2) * V(HONH2)
mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol
We have:
mol(HBrO4) = 1.6 mmol
mol(HONH2) = 1.6 mmol
1.6 mmol of both will react to form HONH3+ and H2O
HONH3+ here is strong acid
HONH3+ formed = 1.6 mmol
Volume of Solution = 16 + 20 = 36 mL
Ka of HONH3+ = Kw/Kb = 1.0E-14/1.1E-8 = 9.091*10^-7
concentration ofHONH3+,c = 1.6 mmol/36 mL = 0.0444 M
HONH3+ + H2O -----> HONH2 + H+
4.444*10^-2 0 0
4.444*10^-2-x x x
Ka = [H+][HONH2]/[HONH3+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((9.091*10^-7)*4.444*10^-2) = 2.01*10^-4
since c is much greater than x, our assumption is correct
so, x = 2.01*10^-4 M
[H+] = x = 2.01*10^-4 M
use:
pH = -log [H+]
= -log (2.01*10^-4)
= 3.6968
Answer: 3.70
6)when 20.8 mL of HBrO4 is added
Given:
M(HBrO4) = 0.1 M
V(HBrO4) = 20.8 mL
M(HONH2) = 0.08 M
V(HONH2) = 20 mL
mol(HBrO4) = M(HBrO4) * V(HBrO4)
mol(HBrO4) = 0.1 M * 20.8 mL = 2.08 mmol
mol(HONH2) = M(HONH2) * V(HONH2)
mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol
We have:
mol(HBrO4) = 2.08 mmol
mol(HONH2) = 1.6 mmol
1.6 mmol of both will react
excess HBrO4 remaining = 0.48 mmol
Volume of Solution = 20.8 + 20 = 40.8 mL
[H+] = 0.48 mmol/40.8 mL = 0.0118 M
use:
pH = -log [H+]
= -log (1.176*10^-2)
= 1.9294
Answer: 1.93
Consider the titration of 20.0 mL of 0.0800 M HONH, (a weak base; Kb = 1.10e-08)...
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