Question

In: Chemistry

1) 20.00 mL of a 0.3000 M lactic acid solution is titrated with 0.1500 M NaOH....

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?

Expert Solution

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

queen_honey_blossom answered 2 years ago

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