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Chemistry, 25.07.2019 23:30 blessing5266

Calculate the ph of 0.726 m anilinium hydrochloride (c6h5nh3cl) solution in water, given that kb for aniline is 3.83 x 10-4.

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Answer from: psychocatgirl1

The pH of 0.726 M anilinium hydrochloride solution in water is 5.36.

C₆H₅NH₃Cl dissociates completely in water according to the following equation.

C₆H₅NH₃Cl(aq) ⇒ C₆H₅NH₃⁺(aq) + Cl⁻(aq)

Cl⁻ is a very weak base so it can't react with water, whereas C₆H₅NH₃⁺ reacts with water according to the following acid equation.

C₆H₅NH₃⁺(aq) + H₂O(l) ⇒ C₆H₅NH₂(aq) + H₃O⁺

Given Kb for C₆H₅NH₂ is 3.83 × 10⁻⁴, we can calculate Ka for C₆H₅NH₃⁺ using the following expression.

$Ka \times Kb = Kw\\\\Ka = \frac{Kw}{Kb} = \frac{1.00 \times 10^{-14} }{3.83 \times 10^{-4} } = 2.61 \times 10^{-11}$

Given the concentration of C₆H₅NH₃⁺ (Ca) is 0.726 M and the acid dissociation constant (Ka) is 2.61 × 10⁻¹¹, we can calculate [H⁺] using the following expression.

$[H^{+} ] = \sqrt{Ka \times Ca } = \sqrt{(2.61 \times 10^{-11} ) \times 0.726 } = 4.35 \times 10^{-6} M$

The pH of the solution is:

$pH = -log[H^{+} ] = -log (4.35 \times 10^{-6}) = 5.36$

The pH of 0.726 M anilinium hydrochloride solution in water is 5.36.

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Calculate the pH of 0.726 M anilinium hydrochloride (C₆H₅NH₃Cl) solution in water, given that Kb for aniline is 3.83 × 10⁻⁴.

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Answer from: liltonio

First, we have to get Ka:

when Ka = Kw / Kb and we have Kb = 3.83 x 10^-4 & Kw = 1 x 10^-14

∴Ka = (1 x 10^-14) / (3.83 x 10^-4)

= 2.6 x 10^-11

according to the reaction equation:

by using ICE table:

C6H5NH3+ + H2O ↔ C6H5NH2 + H3O+

initial 0.726 0 0

change -X +X +X

Equ (0.726 -X) X X

when Ka = [C6H5NH2][H3O+]/[C6H5NH3+]

by substitution:

2.6 x 10^-11 = X*X / (0.726 -X) by solving for X

∴X = 4.3 x 10^-6

when [H3O+] = X = 4.3 x 10^-6

∴PH = -㏒[H3O+]

= -㏒(4.3 x 10^-6)

= 5.4

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The angle of reflection is 30°

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