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2,4-Dichlorophenoxyacetic acid ("2,4-D") is a widely used herbicide with the molecular formula C8H6Cl2O3. The pKa of 2,4-D is 2.73. A certain water-based herbicide product contains 230 gL−1 of 2,4-D.
What is the pH of the 2,4-D solution?

Respuesta :

Answer:

Explanation:

C₈H₆Cl₂O₃

mol weight

= 8 x 12 + 6 x 1 + 35.5 x 2 + 3 x 16

96 + 6 + 71 + 48 = 221 g

230 g = 230 / 221 mol = 1.04 M solution

pKa = 2.73

Ka = 10⁻²°⁷³

= 1.86 x 10⁻³

Let the compound be represented by KH

KH = K⁻ + H⁺

a        0       0

a - x      x        x  

x ² / (a - x)  = Ka

x² / (1.04 - x) = .00186

x is very small so 1.04 - x = 1.04

x² / 1.04 = .00186

x² = .0019344

x = .044

= 44 x 10⁻³

[ H⁺ ] = 44 x 10⁻³

pH = - log ( 44 x 10⁻³)

= 3 - log 44

= 1.35

pstnonsonjoku

The pH of the solution is 1.37.

Let us represent the compound 2,4-Dichlorophenoxyacetic acid using the symbol BH

The molar mass of 2,4-Dichlorophenoxyacetic acid is 221 g/mol

Mass concentration of 2,4-Dichlorophenoxyacetic acid = 230 gL−1

But;

Mass concentration = molar concentration × molar mass

Molar concentration = Mass concentration/molar mass

Molar concentration = 230 gL−1 /221 g/mol

= 1.04 M

We now have to set up an ICE table as follows;

      BH ⇄        B^-(aq)   +    H^+(aq)

I     1.04            0                  0

C    -x                +x                +x

E    1.04 - x       +x                 +x

PKa = -log Ka

Ka =Antilog(-Ka)

Ka = Antilog(- 2.73)

Ka = 1.86 × 10^-3

Ka = [ B^-] [ H^+] / [BH]

1.86 × 10^-3 = [x] [x] / [1.04 - x]

1.86 × 10^-3 = [x]^2 / [1.04 - x]

[x]^2 = 1.86 × 10^-3  [1.04 - x]

[x]^2 = 1.93 × 10^-3 - 1.86 × 10^-3x

We now have the quadratic equation;

x^2  + 1.86 × 10^-3x - 1.93 × 10^-3 = 0

x^2  + 0.00186x - 0.00193 = 0

x = 0.043 M

[H^+] = 0.043 M

pH = - log[H^+]

pH = -[0.043 M]

pH = 1.37

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