Question

Determine the ph of a 0. 35 m aqueous solution of CH3NH2 (methylamine). The kb of methylamine is?

Answer 1

The pH of 0.35 M of CH₃NH₂ is 12.09 and the $k_b$ for methylamine is 4.4 x 10⁻⁴.

What is pH?

pHis the quantitative measure of acidity and basicity of an aqueous or other liquid solution. The scale range goes from 0 to 14. Water has a pH of 7 and is neutral in nature.

What is dissociation constant?

The dissociation constant is an equilibrium constant that describes the dissociation or ionization of a base or an acid

$k_a$ describes the dissociation of an acid

$k_b$ describes the dissociation of a base

For methylamine,

$CH_3NH_2 + H_2O\Leftrightarrow CH_3NH_3^+ + OH^-$

Initial concentration of methylamine = 0.35 M

Initial concentration of products = 0

Let, at equilibrium concentration of CH₃NH₂ = 0.35 - x

Then, concentration of CH₃NH₃⁺ and OH⁻ is x and x respectively

$k_b = \frac{[CH_3NH_3^+] [OH^-]}{[CH_3NH_2]}$

The dissociation constant for methylamine, $k_b$ = 4.4 x 10⁻⁴

$4.4 \times 10^-^4 = \frac{x.x}{0.35} \\x^{2} = 4.4 \times 10^-^4 \times 0.35\\x^{2} = 1.54 \times 10^-^4$

$x =0.0124$

pOH = -log[OH] = -log(0.0124) = 1.91

pH + pOH = 14

pH = 14 - 1.91 = 12.09

Thus, the pH of methylamine is 12.09 and $k_b$ is 4.4 x 10⁻⁴

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