Respuesta :

jacob193

Answer:

Option D) 2.89.

Explanation:

Look up the acid dissociation constant of acetic acid:

[tex]K_{\rm a} \approx \rm 1.75\times 10^{-5}[/tex]

(CRC Handbook of Chemistry and Physics, 84th Edition (2004).)

Acetic acid partially dissociate to produce acetate ions and hydrogen ions:

[tex]\rm CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}[/tex].

Let the final concentration of [tex]\rm H^{+}[/tex] in the solution be [tex]x\; \rm M[/tex]. The concentration of acetic acid would have dropped by [tex]x\; \rm M[/tex] and the concentration of acetate ions would have increased by [tex]x\; \rm M[/tex]. The initial concentration of [tex]\rm H^{+}[/tex] in pure water is [tex]1\times 10^{-7}\;\rm M[/tex] and will barely influence the outcome.

Construct a RICE table for this reaction: (all values here are in [tex]M[/tex], which stands for concentrations in moles per liter.)

[tex]\begin{array}{c|ccccc}\textbf{R}& \mathrm{CH_3COOH} & \rightleftharpoons & \mathrm{CH_3COO^{-}} & + & \mathrm{H^{+}}\\\textbf{I} & 0.10\\ \textbf{C} & -x & & +x & & +x \\\textbf{E} & 0.10 - x & & x & & x \end{array}[/tex].

At equilibrium:

  • [tex][\mathrm{CH_3COOH}] = (0.10 - x) \; \rm M[/tex];
  • [tex][\mathrm{CH_3COO^{-}}] = x\; \rm M[/tex];
  • [tex][\mathrm{H^{+}}] = x\; \rm M[/tex].

By the definition of the acid dissociation constant, [tex]K_{\rm a}[/tex]:

[tex]\displaystyle K_{\rm a}(\mathrm{CH_3COOH}) = \frac{[\mathrm{CH_3COO^{-}}]\cdot [\mathrm{H^{+}}]}{[\mathrm{CH_3COOH}]}[/tex].

That is:

[tex]\displaystyle \frac{x^{2}}{0.10 - x} = \rm 1.75\times 10^{-5}[/tex].

Rearrange and solve for [tex]x[/tex]:

[tex]x^{2} + 1.75\times 10^{-5} ~x - 1.75 \times 10^{-6} = 0[/tex].

[tex]\displaystyle x = \frac{-1.75\times 10^{-5} \pm \sqrt{{\left(1.75\times 10^{-5}\right)^{2}}- 4\times \left(-1.75\times 10^{-6}\right)}}{2}[/tex].

There might be more than one solution to this equation. However, keep in mind that all concentration should be positive (at least non-negative.) The only possible value of [tex]x[/tex] will thus be approximately [tex]0.00131[/tex].

In other words, at equilibrium [tex][\mathrm{H^{+}}] \approx 0.00131 \; \rm M[/tex]. By the definition of pH,

[tex]\begin{aligned} \rm pH &= -\log_{10}{[\mathrm{H^{+}]}\\&= - \log_{10}{0.00131} \\&\approx 2.9\end{aligned}[/tex].

Note that depending on the [tex]K_{\rm a}[/tex] value, the final result might slightly vary.

Otras preguntas