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Calculate the ph of a 0.40 m solution of ethylamine(c2h5nh2, kb = 5.6 x 10-4.)

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Final answer:

To calculate the pH of a 0.40 M solution of ethylamine, we solve for the concentration of hydroxide ions, then find pOH and subtract from 14 to get pH, resulting in a pH of approximately 12.18.

Step-by-step explanation:

To calculate the pH of a 0.40 M solution of ethylamine, which is a weak base with a base dissociation constant (Kb) of 5.6 x 10-4, we first need to set up the chemical equilibrium that occurs when ethylamine is dissolved in water:

C2H5NH2 + H2O → C2H5NH3+ + OH-

The base dissociation constant expression for ethylamine can be written as:

Kb = [C2H5NH3+][OH-] / [C2H5NH2]

Assuming that the base undergoes a small extent of ionization, we can approximate the concentration of ethylamine that ionizes (x) and rewrite the expression as:

Kb = x^2 / (0.40 - x) ≈ x^2 / 0.40

Solving for x (which represents the concentration of OH- ions):

x = √(Kb × 0.40) = √(5.6 x 10-4 × 0.40) = √(2.24 x 10-4)

x ≈ 0.015

Now, we can calculate the pOH, which is the negative logarithm of the hydroxide ion concentration:

pOH = -log(0.015)

pOH ≈ 1.82

Finally, since pH + pOH = 14 (at 25°C), we can find the pH:

pH = 14 - pOH

pH = 14 - 1.82 = 12.18

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