Question

A chemist mixes \(.5\) moles of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) and \(.5\) moles of HCN with enough water to make a one liter solution. Calculate the final concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\). Assume the following constants: \(\mathrm{K}_{\mathrm{HC} 2 \mathrm{H} 3 \mathrm{O} 2}=1.8 \times 10^{-5}, \mathrm{~K}_{\mathrm{HCN}}=4 \times 10^{-10}, \mathrm{~K}_{\mathrm{W}}=1.0 \times 10^{-14}\)

Short answer

The final concentrations of Hydronium ion (H3O+) and Hydroxide ion (OH-) in the solution are approximately: For H3O+: \[x + y \approx 3.78 \times 10^{-3} \text{M}\] For OH-: \[[OH^-] \approx 2.65 \times 10^{-12} \text{M}\]

Step-by-step solution

Write the dissociation reactions for both weak acids

Write the dissociation reactions for acetic acid and HCN in water. For acetic acid: \[HC_2H_3O_2 (aq) + H_2O (l) \rightleftharpoons H_3O^+(aq) + C_2H_3O_2^-(aq)\] For HCN: \[HCN (aq) + H_2O (l) \rightleftharpoons H_3O^+(aq) + CN^-(aq)\]

Set up ICE tables for both weak acids

Initial, Change, and Equilibrium (ICE) tables help to keep track of the changes in concentration upon the dissociation of weak acids. Set up ICE tables for acetic acid and HCN, respectively. Acetic acid ICE table: | | HC2H3O2 | H3O+ | C2H3O2- | |:------:|:---------:|:--------:|:--------:| | Initial| 0.5 M | 0 | 0 | | Change | -x | +x | +x | | Equilibrium | 0.5-x | x | x | HCN ICE table: | | HCN | H3O+ | CN- | |:------:|:---------:|:--------:|:--------:| | Initial| 0.5 M | 0 | 0 | | Change | -y | +y | +y | | Equilibrium | 0.5-y | y | y |

Write the expressions for Ka

Write expressions for the ionization constant (Ka) of each weak acid, using their respective ICE table values. For acetic acid: \[K_a = \frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}\] \[1.8 \times 10^{--5} = \frac{x(x)}{0.5-x}\] For HCN: \[K_a = \frac{[H_3O^+][CN^-]}{[HCN]}\] \[4.0 \times 10^{-10} = \frac{y(y)}{0.5-y}\]

Approximate the concentration of H3O+

Since both Ka values are small, assume that x and y are also very small. Therefore, the equilibrium concentrations of acetic acid and HCN can be approximated as follows: For acetic acid: \[1.8 \times 10^{-5} \approx \frac{x^2}{0.5}\] Solve for x to find the concentration of H3O+ due to acetic acid dissociation. For HCN: \[4.0 \times 10^{-10} \approx \frac{y^2}{0.5}\] Solve for y to find the concentration of H3O+ due to HCN dissociation.

Calculate the total concentration of H3O+

Find the total concentration of H3O+ by adding the contributions from both weak acids. The total concentration of H3O+ is approximately equal to x + y. Calculate x + y. H3O+ concentration: x + y

Calculate the concentration of OH-

Use the ion-product constant for water (Kw) to find the concentration of the hydroxide ion (OH-): \[K_w = [H_3O^+][OH^-]\] \[1.0 \times 10^{-14} = (x+y)([OH^-])\] Solve for [OH-] using the total H3O+ concentration found in Step 5. Finally, you will have obtained the final concentrations of H3O+ and OH- in the solution.

Key concepts

ICE Table

Understanding the ICE Table (Initial, Change, Equilibrium) is crucial for examining chemical equilibria, particularly with weak acids. The ICE table is a straightforward method to keep track of concentrations in a reaction as we shift from initial conditions to equilibrium.
It's set up in three rows, with 'I' for the initial concentrations, 'C' for the changes that occur as the acids dissociate, and 'E' for the equilibrium concentrations. Columns reflect the reactants and products. In our exercise, we would list acetic acid, hydronium ions (H_3O^+), and acetate ions (C_2H_3O_2^-) for acetic acid dissociation, and similarly HCN, (H_3O^+), and cyanide ions (CN^-) for HCN dissociation.
It's important to maintain the stoichiometry, which often means that the change in concentration of HCN will be equal to the change in concentration of (H_3O^+) and (CN^-). This systematic approach provides a clear path to write expression for equilibrium constants and solve for unknown concentrations.

Weak Acid Equilibrium

Weak acids, unlike strong acids, do not fully dissociate in water. This partial dissociation is a dynamic equilibrium between the undissociated acid and its ions. The acid dissociation constant, (Ka), quantifies the extent of dissociation and is specific to each weak acid.
For our exercise, acetic acid (HC_2H_3O_2) and HCN both have their respective (Ka) values. A lower (Ka) signifies a weaker acid. To calculate the concentrations of ions at equilibrium, we take the changes in concentrations from the ICE table and apply them to the (Ka) expression.
Remember, because we're dealing with weak acids, we often assume that the change in concentration will be so small that it won't significantly affect the initial concentration (this is known as the 'small approximation'). This simplification helps students solve for concentrations without complex algebra.

pH Calculation

Calculating pH is the process of finding the acidity or basicity of a solution. The pH scale ranges from 0 to 14, where 7 is neutral, lower values are acidic, and higher values are basic. To calculate the pH, you first need the concentration of hydronium ions ([H_3O^+]) in the solution.
Once you have the concentration ([H_3O^+]), obtain the pH using the negative logarithm:
\[pH = -\log[H_3O^+]\] Applied to the exercise, the final pH would be determined by the total concentration of hydronium ions generated by both weak acids, which is summed up as x + y in the final steps of the ICE table method.

Ion-Product Constant of Water

The ion-product constant of water (Kw) is an often overlooked but significant chemical constant that reflects the autoionization of water—that is, water’s ability to form hydronium [(H_3O^+)] and hydroxide [(OH^-)] ions. At 25°C, (Kw) has a value of (1.0 \times 10^{-14}).
When the concentration of (H_3O^+) is known, you can calculate the concentration of (OH^-) in the solution through the relationship:
\[Kw = [H_3O^+][OH^-]\] For our mixed-acid solution, after finding the concentration of (H_3O^+), we can rearrange this relationship to solve for ([OH^-]), thus finding the solution’s final OH- concentration. This allows students to fully understand the nature of the solution’s acid-base equilibrium.