Please give your pH answers to two decimal places.
A 100.0mL 0.100M weak acid solution is titrated with a 0.100M NaOH solution. If the acid has a Ka of 3.4 x 10-5, what is the pH of the acid solution...
Before any NaOH is added =
At the equivalence point in the titration =
1)when 0.0 mL of NaOH is added
HA dissociates as:
HA -----> H+ + A-
0.1 0 0
0.1-x x x
Ka = [H+][A-]/[HA]
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 ((3.4*10^-5)*0.1) = 1.844*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)
3.4*10^-5 = x^2/(0.1-x)
3.4*10^-6 - 3.4*10^-5 *x = x^2
x^2 + 3.4*10^-5 *x-3.4*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 3.4*10^-5
c = -3.4*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 = 1.36*10^-5
roots are :
x = 1.827*10^-3 and x = -1.861*10^-3
since x can't be negative, the possible value of x is
x = 1.827*10^-3
use:
pH = -log [H+]
= -log (1.827*10^-3)
= 2.7383
Answer: 2.74
2)
find the volume of NaOH used to reach equivalence point
M(HA)*V(HA) =M(NaOH)*V(NaOH)
0.1 M *100.0 mL = 0.1M *V(NaOH)
V(NaOH) = 100 mL
Given:
M(HA) = 0.1 M
V(HA) = 100 mL
M(NaOH) = 0.1 M
V(NaOH) = 100 mL
mol(HA) = M(HA) * V(HA)
mol(HA) = 0.1 M * 100 mL = 10 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 100 mL = 10 mmol
We have:
mol(HA) = 10 mmol
mol(NaOH) = 10 mmol
10 mmol of both will react to form A- and H2O
A- here is strong base
A- formed = 10 mmol
Volume of Solution = 100 + 100 = 200 mL
Kb of A- = Kw/Ka = 1*10^-14/3.4*10^-5 = 2.941*10^-10
concentration ofA-,c = 10 mmol/200 mL = 0.05M
A- dissociates as
A- + H2O -----> HA + OH-
0.05 0 0
0.05-x x x
Kb = [HA][OH-]/[A-]
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 ((2.941*10^-10)*5*10^-2) = 3.835*10^-6
since c is much greater than x, our assumption is correct
so, x = 3.835*10^-6 M
[OH-] = x = 3.835*10^-6 M
use:
pOH = -log [OH-]
= -log (3.835*10^-6)
= 5.4163
use:
PH = 14 - pOH
= 14 - 5.4163
= 8.5837
Answer: 8.58
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