Question

Ionization constant of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(1.7 \times 10^{-5}\) and concentration of \(\mathrm{H}^{+}\) ion is \(3.4 \times 10^{-4}\). Then initial concentration of \(\mathrm{CH}_{3} \mathrm{COOH}\) is: (a) \(3.4 \times 10^{-4}\) (b) \(3.4 \times 10^{-3}\) (c) \(6.8 \times 10^{-4}\) (d) \(6.8 \times 10^{-3}\)

Short answer

The initial concentration of \(\mathrm{CH}_3\mathrm{COOH}\) is \(6.8 \times 10^{-3}\) M, answer (d).

Step-by-step solution

Understand the Ionization Equation

Acetic acid (\(\mathrm{CH}_3\mathrm{COOH}\)) dissociates according to the equation: \(\mathrm{CH}_3\mathrm{COOH} \leftrightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+\). We are given the ionization constant \(K_a\) as \(1.7 \times 10^{-5}\) and the concentration of \(\mathrm{H}^+\) as \(3.4 \times 10^{-4}\). We are asked to find the initial concentration of acetic acid.

Write the Expression for Ionization Constant

The ionization constant (\(K_a\)) expression is given by \[ K_a = \frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \]. We will let \([\mathrm{CH}_3\mathrm{COOH}] = x\) as the initial concentration of the acetic acid.

Substitute Given Values into the Expression

Knowing \([\mathrm{H}^+] = 3.4 \times 10^{-4}\) and assuming complete conversion,\([\mathrm{CH}_3\mathrm{COO}^-] = 3.4 \times 10^{-4}\) (because every mole of\(\mathrm{H}^+\) produced comes from\(\mathrm{CH}_3\mathrm{COO}^-\)). Substitute these values in:\[ 1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4})(3.4 \times 10^{-4})}{x - 3.4 \times 10^{-4}} \]

Solve for Initial Concentration x

Rearrange the equation to solve for \(x\):\[ 1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4})^2}{x - 3.4 \times 10^{-4}} \]Multiply through by \(x - 3.4 \times 10^{-4}\) to clear the fraction:\[ 1.7 \times 10^{-5}(x - 3.4 \times 10^{-4}) = (3.4 \times 10^{-4})^2 \]Solve this equation for \(x\).After calculation, \(x = 6.8 \times 10^{-3}\) (d).

Final Answer

The calculations show that the initial concentration of\(\mathrm{CH}_3\mathrm{COOH}\)is \(6.8 \times 10^{-3}\). Therefore, the correct answer is (d).

Key concepts

Ionization Constant

In chemistry, the ionization constant, commonly represented by \( K_a \), is a crucial factor that determines the strength of an acid in solution. It's a value that reflects how well an acid can dissociate into its ions. For a weak acid like acetic acid (\( \mathrm{CH}_3\mathrm{COOH} \)), we use the equation:

• \( \mathrm{CH}_3\mathrm{COOH} \leftrightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+ \). This equation represents the partial ionization of the acid in aqueous solution.

The ionization constant \( K_a \) for acetic acid is provided as \( 1.7 \times 10^{-5} \). This number indicates a relatively small degree of ionization, which is typical for weak acids. By applying this constant, alongside known concentrations in the ionization constant expression \( K_a = \frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \), we can determine unknown quantities such as initial concentrations. Recognizing and using \( K_a \) is crucial in predicting how an acid behaves in various chemical contexts.

Acid-Base Chemistry

Acid-base chemistry is a foundational aspect of understanding reactions in aqueous solutions. Acids like acetic acid donate protons (\( \mathrm{H}^+ \)), while bases accept them. In our equation \( \mathrm{CH}_3\mathrm{COOH} \leftrightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+ \), acetic acid acts as an acid as it donates a hydrogen ion. This reaction is reversible, indicating it does not proceed to completion and reaches an equilibrium.

• When an acid dissolves in water, it typically increases the \( \mathrm{H}^+ \) ion concentration.
• The balance between acid and conjugate base (here \( \mathrm{CH}_3\mathrm{COO}^- \)) is key for understanding the pH of solutions.

Acid-base chemistry not only applies to titration and pH calculations but also extends to buffer solutions, where small amounts of acids or bases are added. Understanding the dissociation of acids allows us to predict changes in equilibrium or pH when acids or bases are added to a solution.

Concentration Changes

When an acid like acetic acid dissolves in water, it affects the equilibrium concentrations of ions in the solution. Let's delve into how these concentration changes occur. Starting with the initial concentration \([\mathrm{CH}_3\mathrm{COOH}] = x\), upon dissociation:

• The concentration of \( \mathrm{H}^+ \) ions increases, equivalent to the concentration of \( \mathrm{CH}_3\mathrm{COO}^- \).
• Given \( [\mathrm{H}^+] = 3.4 \times 10^{-4} \), we presume \( [\mathrm{CH}_3\mathrm{COO}^-] = 3.4 \times 10^{-4} \) due to the 1:1 stoichiometry.

By substituting these values consistently alongside the ionization constant, we solve the expression \[ 1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4})^2}{x - 3.4 \times 10^{-4}} \], revealing that this affects the measure of initial acid present. Successfully rearranging and solving this kind of equation is essential to determining concentrations and understanding equilibria in acid-base contexts.