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Sagot :
Potassium trifluoroacetate is the salt of a weak acid (trifluoroacetic acid) and a strong base (potassium hydroxide). In water it will dissociate into its ions.
KC₂F₃O₂ ------> K⁺ + C₂F₃O₂⁻
The trifluoroacetate ion is the conjugate base of a weak acid:
C₂F₃O₂⁻ + H₂O <-----> C₂F₃O₂H + OH⁻
The dissociation constant of this conjugate base will be:
Kb = [C₂F₃O₂H] * [OH⁻]/[C₂F₃O₂⁻]
I searched for the value of the Ka for the trifluoroacetatic acid. With this value we can find the Kb for the trifluoroacetate.
Ka = 1
Ka * Kb = 1 * 10^(-14)
Kb = 1 * 10^(-14)/1
Kb = 1 * 10^(-14)
So first we will have to find the concentration of OH⁻ and then we will be able to find the concentration of H+ and finally the pH.
We have to use the ICE table.
C₂F₃O₂⁻ + H₂O <-----> C₂F₃O₂H + OH⁻
I 5.43 0 0
C -x +x +x
E 5.43 - x x x
The initial concentration of the trifluoroacetate ion is 5.43, then some moles of it will convert into moles of C₂F₃O₂H and OH⁻. And finally we found the equilibrium concentrations that we can replace of the expression of the Kb.
Kb = [C₂F₃O₂H] * [OH⁻]/[C₂F₃O₂⁻]
Kb = x * x /(5.43 - x)
Kb = x²/(5.43 - x)
We can replace the Kb for the value that we found and solve this quadratic equation for x.
1 * 10^(-14) = x²/(5.43 - x)
1 * 10^(-14) * (5.43 - x) = x²
x² + 1 * 10^(-14) x - 5.43 * 10^(-14) = 0
This quadratic equation has two roots. We will use the positive one since it is a concentration.
x₁ = 2.33 * 10^(-7) x₂ = - 2.33 * 10^(-7)
The concentration of OH- in the equilibrium is x. We can determine the pOH.
[OH-]eq = x₁
[OH-]eq = 2.33 * 10^(-7)
pOH = - log [OH-]
pOH = - log (2.33 * 10^(-7))
pOH = 6.63
Finally we can find the pH using that value:
pH + pOH = 14
pH = 14 - pOH = 14 - 6.63
pH = 7.37
Answer: the pH of a 5.43 M solution of potassium trifluoroacetate is 7.37 supposing that ka of trifluoroacetic acid is 1.
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