In: Chemistry
1) 20.00 mL of a 0.3000 M lactic acid solution is titrated with 0.1500 M NaOH.
a. What is the pH of the initial solution (before any base is added)?
b. What is the pH of the solution after 20.00 mL of the base solution has been added?
c. What is the pH of the solution after 40.00 mL of the base solution has been added?
Given
Lactic acid(C3H6O3)
Volume 0.02 L
[C3H6O3] 0.3 M
Ka =1.40E-04
ICE
C3H6O3+ H2O -- > C3H5O3- + H3O+
I 0.3 0 0
C -x +x +x
E (0.3-x) x x
Ka expression
ka=[C3H5O3-][H3O+]/[C3H6O3]
Lets plug in the ka and equilibrium concentration values.
1.40E-4=x^2 / (0.3-x)
x^2 + 1.40 E-4 x - 4.2E-5=0
Lets solve this quadratic equation
x=0.006411
[H3O+]=x=0.006411
pH=-log[H3O+]
pH=-log (0.006411)= 2.193074223
pH of the solution = 2.20
b)
Addition of base
Calculation of moles of base
Mol of base = Volume of base in L x molarity of base
volume 0.02 L
Molarity 0.15 M
mol 0.003 mol
mol acid 0.006 mol
Reaction between C3H6O3 and NaOH
C3H6O3 (aq) + NaOH (aq) ---- > C3H5O3Na (aq) + H2O
I 0.006 0.003 0
C -0.003 -0.003 +0.003
E 0.003 0 0.003
Moles of both acid and salt are same therefore,
[C3H6O3] = [C3H5O3-] ….(since concentration of C3H5O3Na and C3H5O3- is same)
We know use Henderson equation
pH = - log ka + log ( [ conj. Base ] / [ weak acid])
here conj base is C3H5O3- and weak acid is C3H6O3
Lets plug in the values to get pH
pH = - log ( 1.40E-4 ) + log ([C3H5O-]/[ C3H6O3])
since
[C3H6O3] = [C3H5O3-] , log ([C3H5O-]/[ C3H6O3]) = log 1 = 0
Therefore pH = - log ( 1.40E-4 )
pH = 3.854
c)
mol base 0.006
mol acid 0.006
This is equivalnce point reaction and there is formation 0.006 mol C3H5O3-
[C3H5O-] = mol / total volume
= 0.006 mol / 0.06 L
= 0.1 M
Reaction of C3H5O3- with water
C3H5O3- + H2O -- > C3H6O3 + OH-
I 0.1 0 0
C -x +x +x
E (0.1-x) x x
Kb = x2 / (0.1-x)
Kb = 1.0 E-14 / ka = 1.0 E -14 / 1.40E-4 = 7.14E-11
Since the value of kb is very small we can neglect x in the denominator.
7.14E-11 = x2/ 0.1
x = 2.67 E-6
x = [OH- ] = 2.67 E-6
pOH = -log ( 2.67 E-6) = 5.57
pH = 14 – pOH = 14 -5.57 = 8.4
pH of the solution after addition of 40.00 mL of base = 8.4