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