Question

Consider \(1000 . \mathrm{mL}\) of a \(1.00 \times 10^{-4}-M\) solution of a certain acid HA that has a \(K_{\mathrm{a}}\) value equal to \(1.00 \times 10^{-4}\). How much water was added or removed (by evaporation) so that a solution remains in which \(25.0 \%\) of HA is dissociated at equilibrium? Assume that HA is nonvolatile.

Short answer

A 1000 mL solution of acid HA with a molarity of \(1.00 \times 10^{-4}M\) and Ka value equal to \(1.00 \times 10^{-4}\), 666.7 mL of water must be removed by evaporation to achieve a solution in which 25% of HA is dissociated at equilibrium.

Step-by-step solution

Write the equilibrium expression for the acid dissociation

\ First, we write the dissociation reaction for the acid HA: \[ \textnormal{HA} \rightleftharpoons \textnormal{H}^+ + \textnormal{A}^-\] Now, let's write the equilibrium expression: \[K_a = \frac{[\textnormal{H}^+][\textnormal{A}^-]}{[\textnormal{HA}]}\] Here, \(K_a = 1.00 \times 10^{-4}\).

Set up initial and equilibrium concentrations for the reaction

\ Initial concentrations, in moles per liter: \[ [\textnormal{HA}]_0 = 1.00 \times 10^{-4} M \] \[ [\textnormal{H}^+]_0 = [\textnormal{A}^-]_0 = 0 \] At equilibrium, 25% of HA dissociates, therefore: \[ [\textnormal{HA}]_{eq} = 0.75 \times [\textnormal{HA}]_0 \] \[ [\textnormal{H}^+]_{eq} = [\textnormal{A}^-]_{eq} = 0.25 \times [\textnormal{HA}]_0 \]

Use the equilibrium concentrations to set up a Ka equation

\ Now, let's insert the equilibrium concentrations into the Ka equation: \[1.00 \times 10^{-4} = \frac{[\textnormal{H}^+_{eq}][\textnormal{A}^-_{eq}]}{[\textnormal{HA}_{eq}]}\] \[1.00 \times 10^{-4} = \frac{(0.25 \times [\textnormal{HA}_0])^2}{0.75 \times [\textnormal{HA}_0]}\]

Solve for the new molar concentration of HA

\ Now, we can solve the equation for the new molar concentration of HA: \[1.00 \times 10^{-4} = \frac{1.00 \times 10^{-4} \times 0.25^2}{0.75}\] New molarity of HA, \( [ \textnormal{HA}_{f}] = 3.00 \times 10^{-4} M\)

Calculate the new volume of the solution and the amount of water added or removed

\ Finally, we can calculate the new volume of the solution since the number of moles of HA remains constant in this reaction: \[V_f = \frac{n_0}{[\textnormal{HA}_{f}]} = \frac{[\textnormal{HA_0}] \times V_0}{[\textnormal{HA}_{f}]}\] \[V_f = \frac{1.00 \times 10^{-4} \times 1000}{3.00 \times 10^{-4}} = 333.3 \: \textnormal{mL}\] As the initial volume was 1000 mL and the final volume is 333.3 mL, it means that \(1000 - 333.3 = 666.7 \: \textnormal{mL}\) of water has been removed by evaporation.

Key concepts

Equilibrium Expression

When dealing with acid dissociation, the equilibrium expression is a crucial step. An equilibrium expression relates the concentrations of reactants and products at equilibrium for a chemical reaction.

In the case of an acid dissociation, the reaction involves the dissociation of an acid HA into its ions, specifically \\[ \textnormal{HA} \rightleftharpoons \textnormal{H}^+ + \textnormal{A}^- \]\For this reaction, the equilibrium constant, known as the acid dissociation constant (\(K_a\)), can be expressed as:

\[ K_a = \frac{[\textnormal{H}^+][\textnormal{A}^-]}{[\textnormal{HA}]} \]

This expression provides valuable information about the extent to which an acid can dissociate in solution. A larger \(K_a\) value indicates stronger acid, as more of the HA is dissociated into \( \textnormal{H}^+\) and \( \textnormal{A}^- \).

Initial Concentrations

Before the chemical reaction reaches equilibrium, initial concentrations are established. These concentrations are the starting amounts of each substance involved.

For our acid dissociation scenario, we have the following initial concentrations:

• The acid HA starts with \([\textnormal{HA}]_0 = 1.00 \times 10^{-4} \text{ M}\).
• Since no ions have dissociated yet, \([\textnormal{H}^+]_0 = [\textnormal{A}^-]_0 = 0 \text{ M}\).

Knowing the initial concentrations helps set up calculations for what occurs as the reaction proceeds to equilibrium.

Ka Equation

The \(K_a\) equation is a more detailed calculation using the equilibrium concentration values. By understanding this equation, you can determine the extent of an acid’s dissociation.

In our case, given that 25% of HA is dissociated, the equilibrium concentrations are:

• \([\textnormal{HA}]_{eq} = 0.75 \times [\textnormal{HA}]_0 \): Because only 75% of the initial HA remains undissociated.
• \([\textnormal{H}^+]_{eq} = [\textnormal{A}^-]_{eq} = 0.25 \times [\textnormal{HA}]_0 \): Due to 25% dissociation into \( \textnormal{H}^+ \) and \( \textnormal{A}^- \).

Plug these into the \( K_a \) equation:

\[ 1.00 \times 10^{-4} = \frac{(0.25 \times [\textnormal{HA}_0])^2}{0.75 \times [\textnormal{HA}_0]} \]

This equation validates the relationship between dissociated ions and remaining undissociated molecules.

Molar Concentration

Molar concentration, or molarity, is a measure of the concentration of a solute in a solution. It tells you how much of a substance is present in a specific volume.

After solving the \(K_a\) equation, new molar concentration of HA at equilibrium can be derived, which is \([ \textnormal{HA}_{f}] = 3.00 \times 10^{-4} \text{ M}\).

This calculation is essential for determining the changes in volume or concentration necessary to achieve a desired degree of dissociation.

• The higher molarity in the final solution indicates that some water has been removed to concentrate the acid.
• Tracking these changes helps chemists adjust conditions to achieve target reactions.

Understanding molarity provides insight into how reactions balance and can intuitively prepare you for practical chemical computations.