Question

The pH of a 0.10 M HCN solution is 5.2. a. What is \(\left[\mathrm{H}^{+}\right]\) in that solution?______\(M\) b. What is \(\left[\mathrm{CN}^{-}\right] ?\) What is [HCN]? (Where do the \(\mathrm{H}^{+}\) and \(\mathrm{CN}^{-}\) ions come from?)_____\(M;\)_____\(M\) c. What is the value of \(K_{\mathrm{a}}\) for HCN? What is the value of \(\mathrm{pK}_{\mathrm{a}}\) ?_____ _____

Short answer

a. \(\left[\mathrm{H}^{+}\right] \approx 6.31 × 10^{-6}\) M b. \(\left[\mathrm{CN}^{-}\right] \approx 6.31 × 10^{-6}\) M; [HCN] ≈ 0.09999369 M c. \(K_{\mathrm{a}} \approx 3.97 × 10^{-10}\); \(\mathrm{pK}_{\mathrm{a}} \approx 9.4\)

Step-by-step solution

a. Calculate \(\left[\mathrm{H}^{+}\right]\) from the given pH

To find the \(\left[\mathrm{H}^{+}\right]\) concentration from the given pH value, utilize the pH formula: pH = - log\(\left[\mathrm{H}^{+}\right]\) To find the \(\left[\mathrm{H}^{+}\right]\) concentration, rewrite the formula as: \(\left[\mathrm{H}^{+}\right]\) = 10\(^{-\text{pH}}\) Now plug in the given pH value: \(\left[\mathrm{H}^{+}\right]\) = 10\(^{-5.2}\) Calculate the concentration to obtain: \(\left[\mathrm{H}^{+}\right]\) ≈ 6.31 × 10\(^{-6}\) M

b. Calculate \(\left[\mathrm{CN}^{-}\right]\) and [HCN] concentrations

In an HCN solution, the following equilibrium is established: HCN \(\rightleftharpoons\) H\(^+\) + CN\(^-\) From the given information, the initial concentrations are as follows: [HCN] = 0.10 M \(\left[\mathrm{H}^{+}\right]\) = 6.31 × 10\(^{-6}\) M (calculated in part a) Since the amount of H\(^+\) produced is equal to the amount of CN\(^-\) produced, \(\left[\mathrm{CN}^{-}\right]\) will also be 6.31 × 10\(^{-6}\) M. To find the final concentration of HCN, we need to subtract the \(\left[\mathrm{H}^{+}\right]\) concentration from the initial concentration of HCN: [HCN] \(= 0.10 - 6.31 × 10^{-6}\) [HCN] ≈ 0.09999369 M Thus, we have the concentrations: \(\left[\mathrm{CN}^{-}\right] \approx 6.31 × 10^{-6}\) M [HCN] ≈ 0.09999369 M

c. Calculate the \(K_{\mathrm{a}}\) and \(\mathrm{pK}_{\mathrm{a}}\) values for HCN

To determine the value of \(K_\text{a}\) for HCN, use the equilibrium equation: \(K_a = \frac{\left[\mathrm{H}^{+}\right] \left[\mathrm{CN}^{-}\right]}{\text{[HCN]}}\) Plug in the calculated concentrations: \(K_a = \frac{(6.31 × 10^{-6})(6.31 × 10^{-6})}{0.09999369}\) Calculate the \(K_\text{a}\) value to obtain: \(K_\text{a} \approx 3.97 × 10^{-10}\) Now, to find the \(\mathrm{pK}_{\mathrm{a}}\) value, use the formula: \(\mathrm{pK}_{\mathrm{a}} = -\log(K_\text{a})\) Plug in the calculated \(K_\text{a}\) value: \(\mathrm{pK}_{\mathrm{a}} = -\log{(3.97 × 10^{-10})}\) Calculate the \(\mathrm{pK}_{\mathrm{a}}\) value to obtain: \(\mathrm{pK}_{\mathrm{a}} \approx 9.4\) Thus, the value of \(K_{\mathrm{a}}\) for HCN is ≈ 3.97 × 10\(^{-10}\), and the value of \(\mathrm{pK}_{\mathrm{a}}\) is ≈ 9.4.

Key concepts

pH Calculation

Understanding the pH of a solution is crucial for predicting its acidity or alkalinity. The pH is a logarithmic scale used to specify the concentration of hydrogen ions \(\left[\mathrm{H}^{+}\right]\) in a solution. The lower the pH, the higher the acidity, and vice versa. To calculate the pH, the formula provided is:

pH = -log\(\left[\mathrm{H}^{+}\right]\)

This formula establishes a inverse relationship between the concentration of hydrogen ions and pH value. For example, a 0.10 M solution of HCN with a pH of 5.2 translates to a hydrogen ion concentration of approximately \(6.31 \times 10^{-6} M\) by applying the anti-logarithm operation:

\(\left[\mathrm{H}^{+}\right]\) = 10^{(-\text{pH})}

Thus, \(\left[\mathrm{H}^{+}\right]\) = 10^{-5.2} = 6.31 \times 10^{-6} M.

This basic concept enables students to toggle between pH value and hydrogen ion concentration clarifying how the two are interlinked.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of the reactants and products that remain constant over time in a reversible chemical reaction at equilibrium. In the context of acid dissociation, the equilibrium expression is:

HCN \(\rightleftharpoons\) H\(^+\) + CN\(^-\)

Given initial concentrations and changes due to the reaction, we can establish equilibrium concentrations for each species involved. The concentration of HCN decreases by the amount of H\(^+\) and CN\(^-\) produced since each HCN molecule dissociates into one H\(^+\) and one CN\(^-\) ion.

Assuming that the dissociation of HCN is small due to its weak acid nature, we approximate the equilibrium concentration of HCN as its initial concentration minus the concentration of H\(^+\), which is identical to the concentration of CN\(^-\), derived from the pH calculation. This subtle yet important concept enables us to solve the equilibrium concentrations of all species in the solution.

Ka and pKa Values

The acid dissociation constant (Ka) provides a measure of the strength of an acid in solution. It reflects the equilibrium between the undissociated and dissociated forms of the acid. The equation for Ka in the case of HCN is:

Ka = \(\frac{\left[\mathrm{H}^{+}\right]\ \times \left[\mathrm{CN}^{-}\right]}{\text{[HCN]}}\)

The smaller the Ka value, the weaker the acid. To make handling these small numbers easier, the pKa is commonly used, which is the negative logarithm of the Ka:

pKa = -log(Ka)

As with the pH scale, pKa provides a logarithmic scale to express acid strengths. The higher the pKa, the weaker the acid. These values are essential in understanding acid-base properties in different environments. For instance, an acid with a pKa of 9.4, such as HCN, is considered a weak acid because it does not dissociate extensively in solution — its Ka being around \(3.97 \times 10^{-10}\).