Calculate the pH during the titration of 40.0 mL of 0.1000 M propanoic acid Ka = 1.3 X 10^-5 after adding the following volumes of 0.1000 M NaOH A. 0.00 mL B. 7.50 mL C. 20.00 mL D. 40.00 mL E. 45 mL
This is all the information provided. Thank you!
a)when 0.0 mL of NaOH is added
C2H5COOH dissociates as:
C2H5COOH -----> H+ + C2H5COO-
0.1 0 0
0.1-x x x
Ka = [H+][C2H5COO-]/[C2H5COOH]
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 ((1.3*10^-5)*0.1) = 1.14*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
1.3*10^-5 = x^2/(0.1-x)
1.3*10^-6 - 1.3*10^-5 *x = x^2
x^2 + 1.3*10^-5 *x-1.3*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.3*10^-5
c = -1.3*10^-6
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 = 5.2*10^-6
roots are :
x = 1.134*10^-3 and x = -1.147*10^-3
since x can't be negative, the possible value of x is
x = 1.134*10^-3
use:
pH = -log [H+]
= -log (1.134*10^-3)
= 2.9455
Answer: 2.95
b)when 7.5 mL of NaOH is added
Given:
M(C2H5COOH) = 0.1 M
V(C2H5COOH) = 40 mL
M(NaOH) = 0.1 M
V(NaOH) = 7.5 mL
mol(C2H5COOH) = M(C2H5COOH) * V(C2H5COOH)
mol(C2H5COOH) = 0.1 M * 40 mL = 4 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 7.5 mL = 0.75 mmol
We have:
mol(C2H5COOH) = 4 mmol
mol(NaOH) = 0.75 mmol
0.75 mmol of both will react
excess C2H5COOH remaining = 3.25 mmol
Volume of Solution = 40 + 7.5 = 47.5 mL
[C2H5COOH] = 3.25 mmol/47.5 mL = 0.0684M
[C2H5COO-] = 0.75/47.5 = 0.0158M
They form acidic buffer
acid is C2H5COOH
conjugate base is C2H5COO-
Ka = 1.3*10^-5
pKa = - log (Ka)
= - log(1.3*10^-5)
= 4.886
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.886+ log {1.579*10^-2/6.842*10^-2}
= 4.249
Answer: 4.25
c)when 20.0 mL of NaOH is added
Given:
M(C2H5COOH) = 0.1 M
V(C2H5COOH) = 40 mL
M(NaOH) = 0.1 M
V(NaOH) = 20 mL
mol(C2H5COOH) = M(C2H5COOH) * V(C2H5COOH)
mol(C2H5COOH) = 0.1 M * 40 mL = 4 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 20 mL = 2 mmol
We have:
mol(C2H5COOH) = 4 mmol
mol(NaOH) = 2 mmol
2 mmol of both will react
excess C2H5COOH remaining = 2 mmol
Volume of Solution = 40 + 20 = 60 mL
[C2H5COOH] = 2 mmol/60 mL = 0.0333M
[C2H5COO-] = 2/60 = 0.0333M
They form acidic buffer
acid is C2H5COOH
conjugate base is C2H5COO-
Ka = 1.3*10^-5
pKa = - log (Ka)
= - log(1.3*10^-5)
= 4.886
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.886+ log {3.333*10^-2/3.333*10^-2}
= 4.886
Answer: 4.89
d)when 40.0 mL of NaOH is added
Given:
M(C2H5COOH) = 0.1 M
V(C2H5COOH) = 40 mL
M(NaOH) = 0.1 M
V(NaOH) = 40 mL
mol(C2H5COOH) = M(C2H5COOH) * V(C2H5COOH)
mol(C2H5COOH) = 0.1 M * 40 mL = 4 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 40 mL = 4 mmol
We have:
mol(C2H5COOH) = 4 mmol
mol(NaOH) = 4 mmol
4 mmol of both will react to form C2H5COO- and H2O
C2H5COO- here is strong base
C2H5COO- formed = 4 mmol
Volume of Solution = 40 + 40 = 80 mL
Kb of C2H5COO- = Kw/Ka = 1*10^-14/1.3*10^-5 = 7.692*10^-10
concentration ofC2H5COO-,c = 4 mmol/80 mL = 0.05M
C2H5COO- dissociates as
C2H5COO- + H2O -----> C2H5COOH + OH-
0.05 0 0
0.05-x x x
Kb = [C2H5COOH][OH-]/[C2H5COO-]
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 ((7.692*10^-10)*5*10^-2) = 6.202*10^-6
since c is much greater than x, our assumption is correct
so, x = 6.202*10^-6 M
[OH-] = x = 6.202*10^-6 M
use:
pOH = -log [OH-]
= -log (6.202*10^-6)
= 5.2075
use:
PH = 14 - pOH
= 14 - 5.2075
= 8.7925
Answer: 8.79
e)when 45.0 mL of NaOH is added
Given:
M(C2H5COOH) = 0.1 M
V(C2H5COOH) = 40 mL
M(NaOH) = 0.1 M
V(NaOH) = 45 mL
mol(C2H5COOH) = M(C2H5COOH) * V(C2H5COOH)
mol(C2H5COOH) = 0.1 M * 40 mL = 4 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 45 mL = 4.5 mmol
We have:
mol(C2H5COOH) = 4 mmol
mol(NaOH) = 4.5 mmol
4 mmol of both will react
excess NaOH remaining = 0.5 mmol
Volume of Solution = 40 + 45 = 85 mL
[OH-] = 0.5 mmol/85 mL = 0.0059 M
use:
pOH = -log [OH-]
= -log (5.882*10^-3)
= 2.2304
use:
PH = 14 - pOH
= 14 - 2.2304
= 11.7696
Answer: 11.77
Calculate the pH during the titration of 40.0 mL of 0.1000 M propanoic acid Ka =...