Question

The \(\mathrm{pH}\) of an HF solution is 6.20 . Calculate the ratio [conjugate base]/[acid] for HF at this pH.

Short answer

The ratio [conjugate base]/[acid] for HF at the pH of 6.20 is approximately \( \frac{10^{-6.20}}{7.2 \times 10^{-4}}\).

Step-by-step solution

Calculate the hydrogen ion concentration

The pH is given as 6.20. Recall that \( \mathrm{pH} = -\log[H^+]\) . To calculate the concentration of hydrogen ions \([H^+]\), rearrange the equation to solve for \([H^+]\) making it as \( [H^+] = 10^{-\mathrm{pH}} \) . Substituting the value of the pH in the equation, we have \( [H^+] = 10^{-6.20} \)

Apply the acid-base equilibrium concept

We know that HF dissociates into H+ and F- ions in solution. The chemical reaction can be written as \[HF \leftrightarrow H^+ + F^- \]. We can set up the expression for the equilibrium constant \(K_a\) of this reaction as \(K_a = \frac{[H^+][F^-]}{[HF]}\). Now, [H+] is essentially equal to [F-] (conjugate base) because for every HF that ionizes, one H+ and one F- are formed. Therefore, we can simplify the \(K_a\) expression to \(K_a = \frac{[H+][H+]}{[HF]} = \frac{[H+]^2}{[HF]}\). Substituting the value of [H+] from the first step, we can rearrange the equation to solve for the ratio [F-]/[HF] which is equal to \([F-]/[HF] = \frac{[H+]}{K_a} \)

Use the Ka Value

For HF, the Ka value (acid dissociation constant) is approximately 7.2 * 10^-4. Substituting the values of [H+] and Ka into \([F-]/[HF] = \frac{[H+]}{K_a} \), we have the ratio is \( \frac{10^{-6.20}}{7.2 \times 10^{-4}}\)

Key concepts

pH Calculation

Understanding the pH calculation is fundamental for studying acid-base chemistry. The pH is a measure of the acidity or basicity of a solution, with a lower pH indicating a more acidic solution and a higher pH indicating a more basic one. It is calculated using the formula \( \mathrm{pH} = -\log[H^+] \) where \( [H^+] \) is the hydrogen ion concentration in moles per liter.

In simple terms, pH quantifies the level of hydrogen ions present in a solution. The logarithmic scale means that each pH unit represents a tenfold change in ion concentration. For example, a pH of 3 is ten times more acidic than a pH of 4. To calculate \( [H^+] \) from a given pH, the inverse logarithmic relationship is used: \( [H^+] = 10^{-\mathrm{pH}} \). Thus, if a solution has a pH of 6.20, its hydrogen ion concentration is \( [H^+] = 10^{-6.20} \), which can then be used in further calculations.

Hydrogen Ion Concentration

The hydrogen ion concentration \( [H^+] \) in a solution is a direct measure of the solution's potential to conduct electricity, engage in chemical reactions, and affect biological systems. This concentration is typically found in units of moles per liter (M). As the concentration of hydrogen ions increases, the solution becomes more acidic, and the pH value decreases accordingly.

As seen in the textbook step-by-step solution, computing the hydrogen ion concentration is crucial for deducing various equilibrium constants in chemical reactions. The relationship between \( [H^+] \) and pH is inverse and logarithmic, meaning that even a small change in \( [H^+] \) can result in a notable shift in pH. Understanding \( [H^+] \) also allows for the calculation of the ratio of a conjugate base to its corresponding acid, as in the case of HF and its dissociation in water.

Acid Dissociation Constant (Ka)

The acid dissociation constant, symbolized by \( K_a \), is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation reaction of the acid to its ions in water. The higher the \( K_a \) value, the stronger the acid, as it indicates a greater propensity for the acid to lose its proton (\( H^+ \) ion).

For the acid HF, understanding \( K_a \) allows us to calculate how much of the acid has dissociated at equilibrium. The equilibrium expression for \( K_a \) is \( K_a = \frac{[H^+][F^-]}{[HF]} \), where \( [H^+] \) and \( [F^-] \) are the concentrations of the dissociated ions and \( [HF] \) is the concentration of the undissociated acid. In the given exercise, by knowing the \( [H^+] \) and \( K_a \) for HF, one can calculate the ratio of \( [F^-] \) to \( [HF] \) using the rearranged expression \( [F^-]/[HF] = \frac{[H^+]}{K_a} \) and thus determine the extent to which the acid has dissociated at a specific pH.