Question

The dissociation constant of a weak acid is \(4.9 \times 10^{-8}\), its percentage ionization at \(0.1 \mathrm{M}\) is: (a) \(0.07 \%\) (b) \(0.007 \%\) (c) \(0.7 \%\) (d) \(0.0007 \%\)

Short answer

The percentage ionization is 0.07%, so the answer is (a) 0.07%.

Step-by-step solution

Identify Key Variables and Formula

The dissociation constant is given as \( K_a = 4.9 \times 10^{-8} \). The concentration of the acid is \( C = 0.1 \mathrm{M} \). We need to determine the percentage ionization, which is calculated using the formula: \[ \text{Percentage Ionization} = \left( \frac{\text{Concentration of ionized acid}}{\text{Initial concentration of acid}} \right) \times 100 \]

Write the Ionization Equation

For a weak acid \( HA \) dissociating in water, the equation is \( HA \rightleftharpoons H^+ + A^- \). Let \( x \) be the concentration of \( H^+ \) ions at equilibrium. Because \( HA \) is a weak acid, \( x \ll 0.1 \).

Use the Expression for Acids Equilibrium

The expression for the dissociation of the acid is \( K_a = \frac{[H^+][A^-]}{[HA]} \). Assume the initial concentration of \( HA \) decreases by \( x \): \( K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \), since \( x \ll 0.1 \).

Solve for x

Substitute the values into the equation: \[ 4.9 \times 10^{-8} = \frac{x^2}{0.1} \] \[ x^2 = 4.9 \times 10^{-9} \] \[ x = \sqrt{4.9 \times 10^{-9}} \] \[ x \approx 7 \times 10^{-5} \].

Calculate Percentage Ionization

The percentage ionization is given by: \[ \left( \frac{7 \times 10^{-5}}{0.1} \right) \times 100 = 0.07\% \].

Select the Correct Option

The calculated percentage ionization is \(0.07\%\). The correct answer is option (a) \(0.07\%\).

Key concepts

Dissociation Constant

The dissociation constant, often denoted as \( K_a \), is a crucial parameter for understanding the behavior of weak acids in solution. It indicates the extent to which an acid can ionize or dissociate in water. For a given weak acid \( HA \), the dissociation can be represented by the equilibrium equation:

• \( HA \rightleftharpoons H^+ + A^- \)

The dissociation constant is then described by the equation:

• \( K_a = \frac{[H^+][A^-]}{[HA]} \)

A smaller \( K_a \) value corresponds to a weaker acid, meaning it ionizes to a lesser extent. In the exercise, with \( K_a = 4.9 \times 10^{-8} \), it indicates a very weak acid as it only partially dissociates in the solution. This constant helps chemists predict how much of the acid will exist in ionized form under equilibrium.

Percentage Ionization

Percentage ionization quantifies the fraction of an acid that is ionized in solution. For weak acids, this value can be relatively low. To calculate the percentage ionization, the formula is:

• \( \text{Percentage Ionization} = \left( \frac{\text{Concentration of ionized acid}}{\text{Initial concentration of acid}} \right) \times 100 \)

In practical terms, if you dissolve a weak acid at a concentration of 0.1 M, and after equilibrium, if you find that only a small amount of the acid is ionized, the percentage ionization gives you this ratio as a percentage. For instance, the calculated value of 0.07% in the exercise indicates that only 0.07% of the acid molecules ionize under the given conditions. In weak acid solutions, knowing the percentage ionization is invaluable for understanding how the acid behaves and for reactions that depend on the availability of \( H^+ \) ions.

Equilibrium Expression for Acids

An equilibrium expression for acids is derived from the reaction equation and illustrates how concentrations of reactants and products relate at equilibrium. For a weak acid \( HA \), the dissociation follows:

• \( HA \rightleftharpoons H^+ + A^- \)

At equilibrium, the expression capturing this balance is:

• \( K_a = \frac{[H^+][A^-]}{[HA]} \)

This formula tells us that the product of the concentrations of the ions (\( H^+ \) and \( A^- \)) over the concentration of the non-dissociated acid (\( HA \)) equals the dissociation constant. In this exercise, we start with an initial concentration of the acid and a very small \( x \) representing the concentration of \( H^+ \) at equilibrium. The product \( x^2 \) results from \([H^+]\) and \([A^-]\) being equal in a 1:1 dissociation. Solving the equilibrium equation allows for calculating the ionization ratio, which is crucial in determining the acid's behavior in solution.