Calculate the ph of a polyprotic acid given and sketch the titration curves for the following reaction:

a 20.0-ml aliquot of 0.100m of a tartaric acid with 0.100m naoh
pka1 = 2.3 pka2= 4.3

calculate the ph when:

1- initial, 0ml of naoh added

2-ph at first 1/2 equivalence point

3-ph at first equivalence point

4-ph at second 1/2 equivalenec point

5-ph at second equivalence point

6-ph afteter the second equivalence point

(1). pH = 1.70

(2). pH = 2.3

(3). pH = 3.3

(4). pH = 4.3

(5). pH = 8.41

(6). pH = 10.22

Explanation:

we assume that the formula representation of acid is H₂A

the titration curve has reasonably sharp breaks at both equivalence points, corresponding to the reactions;

H₂A + OH⁻ → HA⁻ + H₂O

HA⁻ + OH⁻ → A²⁻ + H₂O

the volume of NaOH (V₀) at the first equivalent point is,

V₀ = (20.0 mL)(0.100M) / 0.100M = 20.0mL

where volume of NaOH at 1/2 equivalent point is,

V₀/2 = 10.0mL

also Volume of NaOH at the second equivalence (2V₀) point is 40.0mL

the volume of NaOH at 1/2 second equivalent point is,

V₀ + V₀/2 = 30.0mL

Volume of NaOH after second equivalence exceeds 40mL

therefore, at 0 mL NaOH addition;

where the extent of ionization is assumed to be x, we have

                        H₂A   ⇆     HA⁻   +   H⁺

where initial:   0.1 M       -            -

          change:   -x         +x           +x

          Equili:      0.1-x      x             x

Kаl = [HA⁻][H⁺] / [H₂A]

10⁻²³ = (x)(x) / (0.1-x)

x = 0.020

[H⁺] = 0.020 M

pH = -log [H⁺]

pH = -log(0.020)

pH = 1.70

(2). at 10 mL NaOH addition

[H₂A]ini = 0.10 M * 20.0 mL = 2 mmol

[OH⁻] = 0.1 M * 10 mL = 1 mmol

after reaction:

[H₂A] = 1 mmol

[H⁻] = 1 mmol

pH = pKa₁ + log [HA⁻] / [[HA⁻]

pH = 2.3 + log 1mmol / 1mmol

pH = 2.3

(3). pH at the first equivalence point is,

pH = 1/2 (pKa₁ + pKa₂)

pH = 1/2(2.3 + 4.3) = 3.3

pH = 3.3

(4). pH at the second 1/2 equivalence point is

pH = pKa₂ = 4.3

pH = 4.3

(5). pH at the second equivalence point;

all H₂A is converted into A²⁻

[A²⁻] = initial moles of H₂A / total volume = (20.0 mL)(0.10 M) / (20.0 + 40.0) mL = 0.033 M

at equilibrium:

                   A²⁻ + H²O    ⇆   HA⁺ OH⁻

          0.033 - x

from the Kb₁ expression,

Kb₁ = [OH⁻][HA⁻] / [A²]

Kw/Ka₂ = x²/(0.0333 - x)

10⁻¹⁴/10⁻⁴³ = x²/(0.0333 - x)

x = 2.57 * 10⁻⁶

[OH⁻] = 2.57 * 10⁻⁶M

pH = -log Kw/[OH⁻] = 8.41

pH = 8.41

(6). pH after second equivalence point;

assuming the volume of NaOH is 40.10 mL

after second equivalence point OH⁻ in excess

[OH⁻] = 0.10 M * 0.10 mL / (20 + 40.10) mL = 1.66 * 10⁻⁴ M

pH = 0=-log Kw/[OH⁻] = 10.22

pH = 10.22