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Chloroacetic acid, ClCH2COOH, has a pKa of 2.87. What are the concentrations of [H3O+], [ClCH2COO−], and [ClCH2COOH], as well as the pH, in a 1.55 M ClCH2COOH solution?

Answer :

The values of [H₃O⁺], [ClCH₂COO⁻], [ClCH₂COOH], and pH in a solution of 1.55 M of ClCH₂COOH are:

  • [H₃O⁺] = [ClCH₂COO⁻] = 0.045 M
  • [ClCH₂COOH] = 1.505 M
  • pH = 1.35

The dissociation reaction of chloroacetic acid in water is the following:

ClCH₂COOH(aq) + H₂O(l) ⇄ ClCH₂COO⁻(aq) + H₃O⁺(aq)

The acid constant of the above reaction is given by:

[tex] Ka = \frac{[ClCH_{2}COO^{-}][H_{3}O^{+}]}{[ClCH_{2}COOH]} [/tex] (1)

We can find the concentrations of ClCH₂COO⁻, H₃O⁺, and ClCH₂COOH from the equilibrium conditions:

ClCH₂COOH(aq) + H₂O(l) ⇄ ClCH₂COO⁻(aq) + H₃O⁺(aq) (2)

1.55 - x x x

After entering the value of concentrations of reaction (2) into equation (1), we have:

[tex]Ka = \frac{x*x}{(1.55 - x)}[/tex]

[tex] Ka(1.55 - x) - x^{2} = 0 [/tex]

The Ka can be calculated with the pKa:

[tex] pKa = -log(Ka) [/tex]

[tex] Ka = 10^{-2.87} [/tex]

so:

[tex] 10^{-2.87}(1.55 - x) - x^{2} = 0 [/tex]

By solving the above quadratic equation for x, and taking the positive value of x (concentrations cannot be negatives), we have:

[tex] x = 0.045 M = [H_{3}O^{+}] = [ClCH_{2}COO^{-}] [/tex]

The concentration of ClCH₂COOH is:

[tex] [ClCH_{2}COOH] = (1.55 - 0.045) M = 1.505 M [/tex]

And the pH of the solution is:

[tex] pH = -log([H_{3}O^{+}]) = -log(0.045) = 1.35 [/tex]

Therefore, the values of [H₃O⁺], [ClCH₂COO⁻], [ClCH₂COOH], and pH are 0.045 M, 0.045 M, 1.505 M, and 1.35, respectively.

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Rewritten by : Barada

Answer:

See explanation below

Explanation:

first to all, this is an acid base reaction where the chloroacetic acid is being dissociated in water. Therefore, is an equilibrium reaction.

ClCH₂COOH has a pKa of 2.87 so, the Ka would be:

pKa = -logKa ---> Ka = antlog(-pka)

Ka = antlog(-2.87)

Ka = 1.35x10⁻³

Now that we know the Ka, we need to write the chemical reaction and then, an ICE chart:

ClCH₂COOH + H₂O <----------> ClCH₂COO⁻ + H₃O⁺ Ka = 1.35x10⁻³

i) 1.55 0 0

c) -x +x +x

e) 1.55-x x x

Writting now the equilibrium reaction:

Ka = [H₃O⁺] [ClCH₂COO⁻] / [ClCH₂COOH]

Replacing the values of the chart:

1.35x10⁻³ = x² / 1.55-x

1.35x10⁻³(1.55-x) = x²

2.0925x10⁻³ - 1.35x10⁻³x = x²

x² + 1.35x10⁻³x - 2.0925x10⁻³ = 0 --> a = 1; b = 1.35x10⁻³x; c = 2.0925x10⁻³

From here we use the general equation for solve x in a quadratic equation which is:

x = -b±√(b² - 4ac) / 2a

Replacing the values we have:

x = -1.35x10⁻³ ±√(1.35x10⁻³)² - 4*1*(-2.0925x10⁻³) / 2

x = -1.35x10⁻³ ±√(8.37x10⁻³) / 2

x = -1.35x10⁻³ ± 0.091 / 2

x1 = -1.35x10⁻³ + 0.091 / 2 = 0.045 M

x2 = -1.35x10⁻³ - 0.091 / 2 = -0.046 M

In this case, we will take the positive value of x, in this case, x1.

With this value, the equilibrium concentrations are the following:

[H₃O⁺] = [ClCH₂COO⁻] = 0.045 M

[ClCH₂COOH] = 1.55 - 0.045 = 1.505 M

Finally the pH:

pH = -log[H₃O⁺]

pH = -log(0.045)

pH = 1.35