A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
a) A buffer solution exists in the flask after 5.00 mL of NaOH have been added. Write equations to show the action
of each component of the buffer when acid or base is added from an external source.
b) Calculate the moles of lactic acid present in the flask before the titration
c) Calculate the pH of the contents of the Erlenmeyer flask at the following points during the titration.
i) After 0.00 mL of NaOH have been added
ii) After 5.00 mL of NaOH have been added
iii) After 10.0 mL of NaOH have been added
iv) After 20.0 mL of NaOH have been added
v) After 25.0 mL of NaOH have been added

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

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