Question

What is the concentration of acetic acid and the pH in a solution that is 0.25% ionized?

Answer 1

Answer on Question #44006 - Chemistry - Inorganic Chemistry

Question:

What is the concentration of acetic acid and the pH in a solution that is 0.25% ionized?

Answer:

The acetic acid dissociates according to the equation:

$$\mathrm{CH_3COOH} \rightarrow \mathrm{CH_3COO^-} + \mathrm{H^+}$$

The constant of the dissociation:

$$K_a = \frac{[\mathrm{CH_3COO^-}] [\mathrm{H^+}]}{[\mathrm{CH_3COOH}]}$$

For the weak acid, such as acetic, Ostwald's law of dilution can be written in the next form:

$\alpha = \sqrt{\frac{K_a}{c_0}}$, where $c_0$ is for the initial concentration of acetic acid in the solution and $\alpha$ is for degree of dissociation. In this case the degree of dissociation is the same as degree of ionization, $\alpha = 0.25\%$.

$K_a$ for acetic acid is $1.76 \times 10^{-5}$. In this case, $c_0$:

$$c_0 = \frac{K_a}{\alpha^2} = \frac{1.76 \times 10^{-5}}{(0.0025)^2} = \frac{1.76 \times 10^{-5}}{6.25 \times 10^{-6}} = 0.282 \times 10 = 2.82 \, (\mathrm{mol/L})$$

pH is the negative logarithm of the concentration of $\mathrm{H^+}$:

$$\mathrm{pH} = -\log([\mathrm{H^+}]);$$

Concentration of $\mathrm{H^+}$:

$$[\mathrm{H^+}] = \sqrt{K_a \times c_0} = \sqrt{1.76 \times 10^{-5} \times 2.82} = 7.04 \times 10^{-3}$$

And pH equals:

$$\mathrm{pH} = -\log([\mathrm{H^+}]) = -\log(7.04 \times 10^{-3}) = 2.15$$

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