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

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.

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

Here Kb acts as Kc or equilibrium constant.

The solution is obtained through the formula of quadratic equations, i.e.,

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

We get

Thus

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Quick Steps

0.57 M methylamine (CH₃NH₂)

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

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

Thus

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

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