Question

What are the major species present in \(0.250 \mathrm{M}\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Short answer

The major species present in the 0.250 M solutions are: a. For \(\mathrm{HNO}_{2}\) solution: \(\mathrm{HNO}_{2}\), \(\mathrm{H}_{2}\mathrm{O}\), \(\mathrm{H}^{+}\), and \(\mathrm{NO}_{2}^{-}\). b. For $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\right)$ solution: \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), \(\mathrm{H}_{2}\mathrm{O}\), \(\mathrm{H}^{+}\), and \(\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}\). The pH values for these solutions are: a. For \(\mathrm{HNO}_{2}\) solution: pH ≈ 1.88. b. For \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) solution: pH ≈ 3.02.

Step-by-step solution

Identify major species present in the solution

First, we need to identify the major species present in each of the 0.250 M solutions: a. For the \(\mathrm{HNO}_{2}\) solution, the major species are \(\mathrm{HNO}_{2}\), \(\mathrm{H}_{2}\mathrm{O}\), \(\mathrm{H}^{+}\), and \(\mathrm{NO}_{2}^{-}\). b. For the $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3}\mathrm{O}_{2}\right)\( solution, the major species are \)\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\(, \)\mathrm{H}_{2}\mathrm{O}\(, \)\mathrm{H}^{+}\(, and \)\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}$.

Write the equilibrium expression for the acid dissociation

Now, we should express the dissociation equilibrium constants for both acids: a. For \(\mathrm{HNO}_{2}\), the acid dissociation constant is \(K_{a1}\): \[K_{a1} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\] b. For $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3}\mathrm{O}_{2}\right)\(, the acid dissociation constant is \)K_{a2}$: \[K_{a2}= \frac{[\mathrm{H}^{+}][\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}]}{[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}]}\]

Calculate the pH of the acid solution

We need to find the concentration of \(\mathrm{H}^{+}\) ions from the equilibrium constants before we can calculate the pH values. Typically, we should be given the values for \(K_{a1}\) and \(K_{a2}\) or refer to a table of dissociation constants. In this case, we will use hypothetical values, viz.: - For \(\mathrm{HNO}_{2}\), let \(K_{a1} = 7.1 \times 10^{-4}\). - For \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), let \(K_{a2} = 1.8 \times 10^{-5}\). Now, we will solve for the concentration of \(\mathrm{H}^{+}\) for each solution: a. For \(\mathrm{HNO}_{2}\): \[7.1 \times 10^{-4} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]} = \frac{[\mathrm{H}^{+}]^2}{0.250}\] \[ [\mathrm{H}^{+}] = \sqrt{7.1 \times 10^{-4} \times 0.250} = 0.0133 \mathrm{M}\] b. For \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\): \[1.8 \times 10^{-5} = \frac{[\mathrm{H}^{+}][\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}]}{[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}]} = \frac{[\mathrm{H}^{+}]^2}{0.250}\] \[[\mathrm{H}^{+}] = \sqrt{1.8 \times 10^{-5} \times 0.250} = 0.000948 \mathrm{M}\] Lastly, we can calculate the pH for each solution: a. For \(\mathrm{HNO}_{2}\): \[\mathrm{pH} = -\log([\mathrm{H}^{+}]) = -\log(0.0133) \approx 1.88\] b. For \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\): \[\mathrm{pH} = -\log([\mathrm{H}^{+}]) = -\log(0.000948) \approx 3.02\] So, the pH of the \(\mathrm{HNO}_{2}\) solution is approximately 1.88, and that of the \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) solution is approximately 3.02.

Key concepts

pH Calculation

Understanding how to calculate pH is fundamental in chemistry, as it allows us to gauge the acidity or basicity of a solution. pH represents the negative logarithm of the hydrogen ion concentration, written as \( \text{pH} = -\log([\text{H}^+]) \). In simple terms, it transforms the potentially very small concentration value into a more manageable number.

When working with weak acids like \( \text{HNO}_2 \) and \( \text{HC}_2\text{H}_3\text{O}_2 \), an important point to remember is that they do not completely dissociate in solution. This partial dissociation leads to an equilibrium state where undissociated acid molecules coexist with their dissociated ions and water. The extent of dissociation is given by the acid dissociation constant (Ka), and it is what we use to determine the concentration of \( \text{H}^+ \) ions in a solution before computing the pH.

Chemical Equilibrium

The concept of chemical equilibrium is pivotal when dealing with reactions, such as the dissociation of acids in water. An equilibrium is established when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. It's essential to comprehend that even though the concentrations don't change at equilibrium, the reactions are still occurring; they're just balanced.

In the context of weak acids, the equilibrium expression based on the law of mass action provides a way to relate the concentrations of all species involved. For a weak acid \( HA \), the equilibrium expression is given by \( K_a = \frac{[\text{H}^+][A^-]}{[HA]} \). Here \( K_a \) is a specific type of equilibrium constant that reflects the strength of the acid: the higher the \( K_a \) value, the stronger the acid and the more it dissociates in solution.

Weak Acid Behavior

Weak acids are characterized by their incomplete dissociation in water, which sets them apart from strong acids that fully dissociate. This weak dissociation results from a reversible reaction where the acid (HA) partly loses a proton, converting into its conjugate base (A-), and releases hydrogen ions (\text{H}^+).

The degree of dissociation is governed by the acid dissociation constant, \( K_a \) and is indicative of the acid's strength. The smaller the value of \( K_a \) is, the less the acid dissociates, thus making it a weaker acid. Most weak acids require an equilibrium calculation to determine the concentrations of \( \text{H}^+ \) ions for pH calculations since they do not fully ionize. In these calculations, the initial concentration of the acid and the value of \(K_a\) are used to find the equilibrium concentrations, often requiring the use of approximations or the quadratic formula when the values of \(K_a\) are not significantly smaller than the initial concentration of the acid.