Question

Determine the ph of 0.57 m methylamine (ch3nh2) with kb = 4.4 x 10-4 : ch3nh2(aq)+ h2o(l) ⇌ ch3nh3+ (aq) + oh- (aq)

Answer 1

The pH value is 12.2

Further explanation

Given:

• 0.57 M methylamine (CH₃NH₂)
• $K_b = 4.4 \times 10^{-4}$

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.

$\boxed{ \ CH_3NH_2_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3NH_3_{(aq)} + OH^-_{(aq)} \ }$

• CH₃NH₂ is the conjugate acid of CH₃NH₂.
• The concentration of hydroxide ions is needed to calculate pH.

Let's prepare the equilibrium system to calculate the concentration of hydroxide ions. In chemical equilibrium, the liquid phase has no effect.

• Initial concentration (in molars): $\boxed{ \ [CH_3 NH_2] = 0.57 \ }$
• Change (in molars): $\boxed{ \ [CH_3NH_2] = -x \ } \boxed{ \ [CH_3NH_3] = +x \ } \boxed{ \ [OH^-] = +x \ }$
• Equilibrium (in molars): $\boxed{ \ [CH_3NH_2] = 0.57 - x \ } \boxed{ \ [CH_3NH_3] = x \ } \boxed{ \ [OH^-] = x \ }$

$\boxed{ \ K_b = \frac{ [CH_3NH_3] [OH^-] }{ [CH_3NH_2] } \ }$

Here Kb acts as Kc or equilibrium constant.

$\boxed{ \ 4.4 \times 10^{-4} = \frac{ x \cdot x }{ 0.57 - x } \ }$

$\boxed{ \ 4.4 \times 10^{-4} = \frac{x^2}{0.57 - x} \ }$

$\boxed{ \ 2.508 \times 10^{-4} - 4.4 \times 10^{-4}x = x^2 \ }$

$\boxed{ \ x^2 + 4.4 \times 10^{-4}x - 2.508 \times 10^{-4} = 0 \ }$

The solution is obtained through the formula of quadratic equations, i.e., $\boxed{ \ x = [OH^-] = 0.0156 \ M \ }$

Next, we calculated the pOH value followed by the pH value.

$\boxed{ \ pOH = -log [OH^-] \ }$

$\boxed{ \ pOH = -log [0.0156] \ }$

We get $\boxed{ \ pOH = 1.81 \ }$

$\boxed{ \ pH + pOH = 14 \ }$

$\boxed{ \ pH = 14 - pOH \ }$

$\boxed{ \ pH = 14 - 1.81 \ }$

Thus $\boxed{\boxed{ \ pH = 12.19 \ rounded \ to \ 12.2 \ }}$

- - - - - - -

Quick Steps

0.57 M methylamine (CH₃NH₂)

$K_b = 4.4 \times 10^{-4}$

We immediately use the formula to calculate the concentration of hydroxide ions for weak bases.

$\boxed{\boxed{ \ [OH^-] = \sqrt{K_b \times base \ concentration} \ }}$

$\boxed{ \ [OH^-] = \sqrt{4.4 \times 10^{-4} \times 0.57} \ }$

$\boxed{ \ [OH^-] = 0.0158 \ M \ }$

Like the steps above, we calculated the pOH value followed by the pH value.

$\boxed{ \ pOH = -log [OH^-] \ }$

$\boxed{ \ pOH = -log [0.0158] \ }$

$\boxed{ \ pOH = 1.8 \ }$

$\boxed{ \ pH = 14 - pOH \ }$

$\boxed{ \ pH = 14 - 1.8 \ }$

Thus $\boxed{\boxed{ \ pH = 12.2 \ }}$

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Keywords: determine, the pH, 0.57 M, methylamine, CH₃NH₂, CH₃NH₃, OH⁻, Kb, Kc, equilibrium constant, weak base

Answer 2

Answer is: pH of methylamine is 12,2.
Chemical reaction: CH₃NH₂(aq)+ H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq).
Kb(CH₃NH₂) = 4,4·10⁻⁴.

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.