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

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.