c₀(CH₃NH₂) = 0,57 M.
c(CH₃NH₃⁺) = c(OH⁻) = x.
c(NH₂OH) = 0,57 M - x.
Kb = c(CH₃NH₃⁺) · c(OH⁻) / c(CH₃NH₂).
0,00044 = x² / (0,57 M - x).
Solve quadratic equation: x = c(OH⁻) = 0,0156 mol/L.
pOH = -log(0,0156 mol/L.) = 1,80.
pH = 14 - 1,80 = 12,2.
The pH value is 12.2
Given:
Question:
The pH value of methylamine
The Process:
Methylamine is a weak base. When a weak base reacts with water, it produces its conjugate acid and hydroxide ions.
[tex]\boxed{ \ CH_3NH_2_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3NH_3_{(aq)} + OH^-_{(aq)} \ }[/tex]
Let's prepare the equilibrium system to calculate the concentration of hydroxide ions. In chemical equilibrium, the liquid phase has no effect.
[tex]\boxed{ \ K_b = \frac{ [CH_3NH_3] [OH^-] }{ [CH_3NH_2] } \ }[/tex]
Here Kb acts as Kc or equilibrium constant.
[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{ x \cdot x }{ 0.57 - x } \ }[/tex]
[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{x^2}{0.57 - x} \ }[/tex]
[tex]\boxed{ \ 2.508 \times 10^{-4} - 4.4 \times 10^{-4}x = x^2 \ }[/tex]
[tex]\boxed{ \ x^2 + 4.4 \times 10^{-4}x - 2.508 \times 10^{-4} = 0 \ }[/tex]
The solution is obtained through the formula of quadratic equations, i.e., [tex]\boxed{ \ x = [OH^-] = 0.0156 \ M \ }[/tex]
Next, we calculated the pOH value followed by the pH value.
[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]
[tex]\boxed{ \ pOH = -log [0.0156] \ }[/tex]
We get [tex]\boxed{ \ pOH = 1.81 \ }[/tex]
[tex]\boxed{ \ pH + pOH = 14 \ }[/tex]
[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]
[tex]\boxed{ \ pH = 14 - 1.81 \ }[/tex]
Thus [tex]\boxed{\boxed{ \ pH = 12.19 \ rounded \ to \ 12.2 \ }}[/tex]
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Quick Steps
0.57 M methylamine (CH₃NH₂)
[tex]K_b = 4.4 \times 10^{-4}[/tex]
We immediately use the formula to calculate the concentration of hydroxide ions for weak bases.
[tex]\boxed{\boxed{ \ [OH^-] = \sqrt{K_b \times base \ concentration} \ }}[/tex]
[tex]\boxed{ \ [OH^-] = \sqrt{4.4 \times 10^{-4} \times 0.57} \ }[/tex]
[tex]\boxed{ \ [OH^-] = 0.0158 \ M \ }[/tex]
Like the steps above, we calculated the pOH value followed by the pH value.
[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]
[tex]\boxed{ \ pOH = -log [0.0158] \ }[/tex]
[tex]\boxed{ \ pOH = 1.8 \ }[/tex]
[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]
[tex]\boxed{ \ pH = 14 - 1.8 \ }[/tex]
Thus [tex]\boxed{\boxed{ \ pH = 12.2 \ }}[/tex]
Keywords: determine, the pH, 0.57 M, methylamine, CH₃NH₂, CH₃NH₃, OH⁻, Kb, Kc, equilibrium constant, weak base
