Question

If the concentration of acetic acid is 0.10 M, what is the pH of the resulting solution?

Answer 1

Answer:

Option D) 2.89.

Explanation:

Look up the acid dissociation constant of acetic acid:

$K_{\rm a} \approx \rm 1.75\times 10^{-5}$

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

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

$\rm CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$.

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

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

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

At equilibrium:

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

By the definition of the acid dissociation constant, $K_{\rm a}$:

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

That is:

$\displaystyle \frac{x^{2}}{0.10 - x} = \rm 1.75\times 10^{-5}$.

Rearrange and solve for $x$:

$x^{2} + 1.75\times 10^{-5} ~x - 1.75 \times 10^{-6} = 0$.

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

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 $x$ will thus be approximately $0.00131$.

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

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

Note that depending on the $K_{\rm a}$ value, the final result might slightly vary.