a)when 0.0 mL of HIO4 is added
C6H5NH2 dissociates as:
C6H5NH2 +H2O -----> C6H5NH3+ + OH-
3*10^-2 0 0
3*10^-2-x x x
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
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 ((4.3*10^-10)*3*10^-2) = 3.592*10^-6
since c is much greater than x, our assumption is correct
so, x = 3.592*10^-6 M
So, [OH-] = x = 3.592*10^-6 M
use:
pOH = -log [OH-]
= -log (3.592*10^-6)
= 5.4447
use:
PH = 14 - pOH
= 14 - 5.4447
= 8.5553
Answer: 8.56
b)when 5.3 mL of HIO4 is added
Given:
M(HIO4) = 0.1 M
V(HIO4) = 5.3 mL
M(C6H5NH2) = 0.03 M
V(C6H5NH2) = 70 mL
mol(HIO4) = M(HIO4) * V(HIO4)
mol(HIO4) = 0.1 M * 5.3 mL = 0.53 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.03 M * 70 mL = 2.1 mmol
We have:
mol(HIO4) = 0.53 mmol
mol(C6H5NH2) = 2.1 mmol
0.53 mmol of both will react
excess C6H5NH2 remaining = 1.57 mmol
Volume of Solution = 5.3 + 70 = 75.3 mL
[C6H5NH2] = 1.57 mmol/75.3 mL = 0.0208 M
[C6H5NH3+] = 0.53 mmol/75.3 mL = 0.007 M
They form basic buffer
base is C6H5NH2
conjugate acid is C6H5NH3+
Kb = 4.3*10^-10
pKb = - log (Kb)
= - log(4.3*10^-10)
= 9.367
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 9.367+ log {7.039*10^-3/2.085*10^-2}
= 8.895
use:
PH = 14 - pOH
= 14 - 8.8949
= 5.1051
Answer: 5.10
c)when 10.5 mL of HIO4 is added
Given:
M(HIO4) = 0.1 M
V(HIO4) = 10.5 mL
M(C6H5NH2) = 0.03 M
V(C6H5NH2) = 70 mL
mol(HIO4) = M(HIO4) * V(HIO4)
mol(HIO4) = 0.1 M * 10.5 mL = 1.05 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.03 M * 70 mL = 2.1 mmol
We have:
mol(HIO4) = 1.05 mmol
mol(C6H5NH2) = 2.1 mmol
1.05 mmol of both will react
excess C6H5NH2 remaining = 1.05 mmol
Volume of Solution = 10.5 + 70 = 80.5 mL
[C6H5NH2] = 1.05 mmol/80.5 mL = 0.013 M
[C6H5NH3+] = 1.05 mmol/80.5 mL = 0.013 M
They form basic buffer
base is C6H5NH2
conjugate acid is C6H5NH3+
Kb = 4.3*10^-10
pKb = - log (Kb)
= - log(4.3*10^-10)
= 9.367
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 9.367+ log {1.304*10^-2/1.304*10^-2}
= 9.367
use:
PH = 14 - pOH
= 14 - 9.3665
= 4.6335
Answer: 4.63
d)when 15.8 mL of HIO4 is added
Given:
M(HIO4) = 0.1 M
V(HIO4) = 15.8 mL
M(C6H5NH2) = 0.03 M
V(C6H5NH2) = 70 mL
mol(HIO4) = M(HIO4) * V(HIO4)
mol(HIO4) = 0.1 M * 15.8 mL = 1.58 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.03 M * 70 mL = 2.1 mmol
We have:
mol(HIO4) = 1.58 mmol
mol(C6H5NH2) = 2.1 mmol
1.58 mmol of both will react
excess C6H5NH2 remaining = 0.52 mmol
Volume of Solution = 15.8 + 70 = 85.8 mL
[C6H5NH2] = 0.52 mmol/85.8 mL = 0.0061 M
[C6H5NH3+] = 1.58 mmol/85.8 mL = 0.0184 M
They form basic buffer
base is C6H5NH2
conjugate acid is C6H5NH3+
Kb = 4.3*10^-10
pKb = - log (Kb)
= - log(4.3*10^-10)
= 9.367
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 9.367+ log {1.841*10^-2/6.061*10^-3}
= 9.849
use:
PH = 14 - pOH
= 14 - 9.8492
= 4.1508
Answer: 4.15
e)when 21.0 mL of HIO4 is added
Given:
M(HIO4) = 0.1 M
V(HIO4) = 21 mL
M(C6H5NH2) = 0.03 M
V(C6H5NH2) = 70 mL
mol(HIO4) = M(HIO4) * V(HIO4)
mol(HIO4) = 0.1 M * 21 mL = 2.1 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.03 M * 70 mL = 2.1 mmol
We have:
mol(HIO4) = 2.1 mmol
mol(C6H5NH2) = 2.1 mmol
2.1 mmol of both will react to form C6H5NH3+ and H2O
C6H5NH3+ here is strong acid
C6H5NH3+ formed = 2.1 mmol
Volume of Solution = 21 + 70 = 91 mL
Ka of C6H5NH3+ = Kw/Kb = 1.0E-14/4.3E-10 = 2.326*10^-5
concentration ofC6H5NH3+,c = 2.1 mmol/91 mL = 0.0231 M
C6H5NH3+ + H2O -----> C6H5NH2 + H+
2.308*10^-2 0 0
2.308*10^-2-x x x
Ka = [H+][C6H5NH2]/[C6H5NH3+]
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 ((2.326*10^-5)*2.308*10^-2) = 7.326*10^-4
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
2.326*10^-5 = x^2/(2.308*10^-2-x)
5.367*10^-7 - 2.326*10^-5 *x = x^2
x^2 + 2.326*10^-5 *x-5.367*10^-7 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 2.326*10^-5
c = -5.367*10^-7
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 2.147*10^-6
roots are :
x = 7.21*10^-4 and x = -7.443*10^-4
since x can't be negative, the possible value of x is
x = 7.21*10^-4
[H+] = x = 7.21*10^-4 M
use:
pH = -log [H+]
= -log (7.21*10^-4)
= 3.142
Answer: 3.14
f)when 23.1 mL of HIO4 is added
Given:
M(HIO4) = 0.1 M
V(HIO4) = 23.1 mL
M(C6H5NH2) = 0.03 M
V(C6H5NH2) = 70 mL
mol(HIO4) = M(HIO4) * V(HIO4)
mol(HIO4) = 0.1 M * 23.1 mL = 2.31 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.03 M * 70 mL = 2.1 mmol
We have:
mol(HIO4) = 2.31 mmol
mol(C6H5NH2) = 2.1 mmol
2.1 mmol of both will react
excess HIO4 remaining = 0.21 mmol
Volume of Solution = 23.1 + 70 = 93.1 mL
[H+] = 0.21 mmol/93.1 mL = 0.0023 M
use:
pH = -log [H+]
= -log (2.256*10^-3)
= 2.6467
Answer: 2.65
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