Question

Applications of Aqueous Equilibria: A 25.0 mL sample of 0.100 M lactic acid (HC3H5O3, Ka = 1.38 X 10^-4) is titrated with a 0.100 M barium hydroxide. Calculate the pH after the addition of 0.0 mL, 2.0 mL, 4.0 mL, 6.25 mL, 10.0 mL, 12.0 mL, 12.25 mL, 12.5 mL, 13.0 mL, 14.0 mL, and 15.0 mL of the Ba(OH)2

Answer 1

Lactic acid (HC3H5O3) is a weak acid and let it be represented as HA.

Ka of lactic acid (HA) = 1.38 x 10^-4

Now,   

HA    H⁺ + A^-
IC:   0.1 0 0
C: - x + x + x
EC: 0.1 - x x x

So, Ka = [H⁺] [A^-] / [HA]
or, 1.38 x 10^-4 = (x) (x) / (0.1 - x)
or, 1.38 x 10^-4 = x² / (0.1 - x)
or, 1.38 x 10^-4 = x² / 0.1 [since, ka is very small, x term in the denominator can be neglected]
or, x² = 1.38 x 10^-5
or, x = 3.71 x 10^-3

So, x =  [H⁺] = 3.71 x 10^-3
pH = - log [H⁺]
= - log (3.71 x 10^-3)
= 2.43

(b) 2.0 mL of 0.100 M Ba(OH)2

25.0 mL sample of 0.100 M lactic acid
Moles of lactic acid = 0.100 M x 0.025 L = 0.0025 moles

2.0 mL of 0.100 M Ba(OH)2
Moles of Ba(OH)2 = 0.100M x 0.002 L = 0.0002 moles

Now, Ba(OH)2 produces two moles of OH^-.
1 mole of Ba(OH)2 reacts with 2 moles of lactic acid
or, 0.0002 moles of Ba(OH)2 reacts with 2 x 0.0002 (= 0.0004) moles of lactic acid.

After adding 2.0 mL of 0.100 M Ba(OH)2, moles of lactic acid remains = 0.0025 moles - 0.0004 moles
  = 0.0021 moles

Total volume = 25 mL + 2 mL = 27 mL = 0.027 L

[HA] = 0.0021 moles / 0.027 L = 0.08 M

Now,   

HA    H⁺ + A^-
IC:   0.08 0 0
C: - x + x + x
EC: 0.08 - x x x

So, Ka = [H⁺] [A^-] / [HA]
or, 1.38 x 10^-4 = (x) (x) / (0.08 - x)
or, 1.38 x 10^-4 = x² / (0.08 - x)
or, 1.38 x 10^-4 = x² / 0.08 [since, ka is very small, x term in the denominator can be neglected]
or, x² = 1.10 x 10^-5
or, x = 3.32 x 10^-3

So, x =  [H⁺] = 3.32 x 10^-3
pH = - log  [H⁺]
= - log (3.32 x 10^-3)
= 2.48

(c) 4.0 mL of 0.100 M Ba(OH)2

25.0 mL sample of 0.100 M lactic acid
Moles of lactic acid = 0.100 M x 0.025 L = 0.0025 moles

4.0 mL of 0.100 M Ba(OH)2
Moles of Ba(OH)2 = 0.100M x 0.004 L = 0.0004 moles

Now, Ba(OH)2 produces two moles of OH^-.
1 mole of Ba(OH)2 reacts with 2 moles of lactic acid
or, 0.0004 moles of Ba(OH)2 reacts with 2 x 0.0004 (= 0.0008) moles of lactic acid.

After adding 4.0 mL of 0.100 M Ba(OH)2, moles of lactic acid remains = 0.0025 moles - 0.0008 moles
  = 0.0017 moles

Total volume = 25 mL + 4 mL = 29 mL = 0.029 L

[HA] = 0.0017 moles / 0.029 L = 0.059 M

Now,   

HA    H⁺ + A^-
IC: 0.059 0 0
C: - x + x + x
EC: 0.059 - x x x

So, Ka = [H⁺] [A^-] / [HA]
or, 1.38 x 10^-4 = (x) (x) / (0.059 - x)
or, 1.38 x 10^-4 = x² / (0.059 - x)
or, 1.38 x 10^-4 = x² / 0.0059 [since, ka is very small, x term in the denominator can be neglected]
or, x² = 8.14 x 10^-7
or, x = 9.02 x 10^-4

So, x =  [H⁺] = 9.02 x 10^-4
pH = - log  [H⁺]
= - log (9.02 x 10^-4)
= 3.04

(d) 15.0 mL of 0.100 M Ba(OH)2

25.0 mL sample of 0.100 M lactic acid
Moles of lactic acid = 0.100 M x 0.025 L = 0.0025 moles

15.0 mL of 0.100 M Ba(OH)2
Moles of Ba(OH)2 = 0.100M x 0.015 L = 0.0015 moles

Now, Ba(OH)2 produces two moles of OH^-.
1 mole of Ba(OH)2 reacts with 2 moles of lactic acid
or, 0.0015 moles of Ba(OH)2 reacts with 2 x 0.0015 (= 0.0030) moles of lactic acid.

After adding 15.0 mL of 0.100 M Ba(OH)2, all lactic acid will neutralize
So, moles of Ba(OH)2 remains = 0.0030 moles - 0.0025 moles
= 0.0005 moles

Total volume = 25 mL + 15 mL = 40 mL = 0.040 L

[Ba(OH)2] = 0.0005 moles / 0.040 L = 0.0125 M
So, [OH^-] = 2 x 0.0125 M =0.0250 M

So, pOH = - log  [OH^-]
= - log (0.0250)
  = 1.60

pH = 14 - pOH
= 14 - 1.60
= 12.4