Question

Calculate the concentration at which a monoprotic acid with \(K_{\mathrm{a}}=4.5 \times 10^{-5}\) will be 2.5 percent ionized.

Short answer

The concentration needed is approximately 0.070 M.

Step-by-step solution

Define the Variables

First, let's clarify what we know. We have a monoprotic acid with an ionization constant \(K_{\mathrm{a}}=4.5 \times 10^{-5}\). We want to find the concentration \([HA]\) at which the acid will be 2.5% ionized.

Express 2.5% Ionization Mathematically

If the acid is 2.5% ionized, this means the concentration of hydrogen ions, \([H^+]\), formed is 2.5% of the initial concentration of the acid, \([HA]\). Thus, \([H^+] = 0.025 \times [HA]\).

Set Up the Ionization Expression

For a monoprotic acid \(HA\), the ionization reaction is \(HA \rightleftharpoons H^+ + A^-\). The expression for \(K_{\mathrm{a}}\) is:\[K_{\mathrm{a}} = \frac{[H^+][A^-]}{[HA]-[H^+]}\]Since \([H^+] = [A^-]\), the expression simplifies to:\[ K_{\mathrm{a}} = \frac{([H^+])^2}{[HA]-[H^+]}\]

Substitute Known Values

Substitute \([H^+] = 0.025[HA]\) into the ionization expression:\[4.5 \times 10^{-5} = \frac{(0.025[HA])^2}{[HA] - 0.025[HA]}\]Simplify the denominator: \([HA] - 0.025[HA] = 0.975[HA]\).

Simplify and Solve for [HA]

Plug into the equation:\[4.5 \times 10^{-5} = \frac{(0.025[HA])^2}{0.975[HA]}\]Solving further gives:\[4.5 \times 10^{-5} = \frac{0.000625[HA]^2}{0.975[HA]}\]Cancel \([HA]\) on the numerator and denominator:\[4.5 \times 10^{-5} = \frac{0.000625[HA]}{0.975}\]Solve it:\[\ 4.5 \times 10^{-5} \times 0.975 = 0.000625[HA]\]\[\ 4.3875 \times 10^{-5} = 0.000625[HA]\]Finally, divide both sides by 0.000625:\[ [HA] = \frac{4.3875 \times 10^{-5}}{0.000625} \approx 0.070 \text{ M} \]

Recheck the Calculation

It's always good practice to recheck the calculation for any errors or miscalculations. You'll find that the concentration of the acid \([HA]\) needed for it to be 2.5% ionized is approximately 0.070 M.

Key concepts

Monoprotic Acid

A monoprotic acid is a type of acid that can donate a single proton (hydrogen ion, \(H^+\)) per molecule to an aqueous solution. Understanding monoprotic acids is fundamental in chemistry, especially when calculating their ionization in solution.
In chemical terms, an example of a monoprotic acid dissociation can be represented as \(HA \rightleftharpoons H^+ + A^-\), where \(HA\) is the acid, \(H^+\) is the hydrogen ion, and \(A^-\) is the conjugate base.

Key characteristics of monoprotic acids include:

• Only one hydrogen ion is released in the solution.
• They typically have similar ionization behavior.
• Paving way for an understanding of how to calculate ionization constants and related concentrations.

Monoprotic acids are found everywhere, from biological systems to industrial processes, making them essential for many fields of study.

Ionization Constant

The ionization constant of an acid, denoted as \(K_{\mathrm{a}}\), is a reflection of the acid's strength in solution. It indicates how well an acid can donate its hydrogen ion to the surrounding solution.
For monoprotic acids, \(K_{\mathrm{a}}\) can be expressed using the following equation:
\[ K_{\mathrm{a}} = \frac{[H^+][A^-]}{[HA]} \]
where:

• \([H^+]\) is the concentration of hydrogen ions.
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the non-ionized acid.

In essence, \(K_{\mathrm{a}}\) provides insight into the degree of ionization, with higher values indicating stronger acids.

Understanding and accurately using \(K_{\mathrm{a}}\) is vital in predicting the behavior of a given acid in a solution.

Concentration Calculation

Calculating concentration is a critical skill in understanding acid ionization. When dealing with acids, especially in terms of ionization, knowing how to calculate the concentration of solutions is vital.
For a monoprotic acid, concentration calculation involves determining the amount of the acid or its ionized components present in a given volume of solution. In our exercise, we specifically calculate the initial concentration \([HA]\) that results in a given percent of ionization, leveraging known values such as the ionization constant \(K_{\mathrm{a}}\).

Here's how it works: derive the expression:\[ K_{\mathrm{a}} = \frac{(0.025[HA])^2}{0.975[HA]} \]
and solve for \([HA]\), using the equation provided. This manipulation reveals that with specified \(K_{\mathrm{a}}\) and percent ionization, you can determine the required concentration for the desired ionization. It's a method letting chemists control reactions carefully.

Percent Ionization

Percent ionization is an important concept while studying acid behavior and describes the fraction of the total acid that ionizes in an aqueous solution, expressed as a percentage.
This calculation helps understand how much of an acid's molecules disassociate to release hydrogen ions in a solution.

Mathematically, it can be expressed as:

• \(\text{Percent Ionization} = \left( \frac{[H^+]}{[HA]} \right) \times 100\%\)

In the context of our calculations, percent ionization aids in establishing how strong an acid solution is, and what ratio of hydrogen ions are released for a given initial concentration.
This insight is crucial for certain applications such as buffering, pH adjustments, and other processes requiring precise acidic behavior.