Respuesta :
Answer:
The change in the pH (ΔpH) is 2,17
Explanation:
The reaction:
CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻
[tex]kb = \frac{[OH^{-}][CH_{3}NH_{3}^+]}{[CH_{3}NH_{2}]}[/tex] (1)
In equilibrium, a solution of CH₃NH₂ 4,7M produces:
[CH₃NH₂] = 4,7 - x
[CH₃NH₃⁺] = x
[OH⁻] = x
Replacing in (1):
[tex]4,38x10^{-4} = \frac{x^2}{4,7-x}[/tex]
x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0
The solutions are:
x = -0,0456 No physical sense. There are not negative concentrations.
x = 0,04515 Real answer.
The concentration of [OH⁻] is 0,04515 M.
As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:
pH = 12,65
The addition of 6,7M produce this changes in concentrations:
[CH₃NH₂] = 4,656 + x
[CH₃NH₃⁺] = 6,74515 - x
[OH⁻] = 0,04515 - x
Replacing in (1) you will obtain:
x² - 6,7907x + 0,3025 = 0
Solving for x:
x = 6,74586 No physical sense
x = 0,04484 Real answer.
Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M
pOH = 3,51.
pH = 10,49
Thus ΔpH is 12,65 - 10,49 = 2,16 ≈ 2,17
I hope it helps!
The change in the pH is:
D. ΔpH=2.17
Ionic equation:
CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻
The equilibrium constant can be written as:
[tex]k_b=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
In equilibrium, a solution of CH₃NH₂ 4.7M produces:
[CH₃NH₂] = 4.7 - x
[CH₃NH₃⁺] = x
[OH⁻] = x
Adding the values in the above equation:
[tex]4.3*10^{-4}=\frac{x^2}{4.7-x}[/tex]
x² + 4.38x10⁻⁴ x - 2.0586 x10⁻³ = 0
x = -0.0456 or x = 0.04515
The concentration of [OH⁻] is 0.04515 M.
As, pOH = -log [OH⁻]
pH + pOH = 14.
The pH of this solution is:
pH = 12.65
The addition of 6.7M produce this changes in concentrations:
[CH₃NH₂] = 4.656 + x
[CH₃NH₃⁺] = 6.74515 - x
[OH⁻] = 0.04515 - x
On adding the values in the above equation:
x² - 6,7907x + 0.3025 = 0
Solving for x:
x = 6.74586 or x = 0.04484
Thus, [OH⁻] = 0.04515 - 0.044842 = 3.08x10⁻⁴M
pOH = 3.51
pH = 10.49
Thus, ΔpH is 12.65 - 10.49 = 2.16 ≈ 2,17
Find more information about pH here:
brainly.com/question/22390063
