Question

Predict whether the following salt solutions are neutral or acidic, and calculate the \(\mathrm{pH}\) of each: (a) \(0.25 \mathrm{M} \mathrm{NH}_{4} \mathrm{Br}\) (b) \(0.20 \mathrm{M} \mathrm{NaNO}_{2} ; K_{a}\) for \(\mathrm{HNO}_{2}\) is \(4.6 \times 10^{-4}\).

Short answer

NH₄Br is acidic with pH ≈ 4.93; NaNO₂ is basic with pH ≈ 7.82.

Step-by-step solution

Identify Components of the Salt

First, determine the identity of the ions that make up each salt given: (a) In \(\text{NH}_4\text{Br}\), the ions are \(\text{NH}_4^+\) (ammonium) and \(\text{Br}^-\) (bromide).(b) In \(\text{NaNO}_2\), the ions are \(\text{Na}^+\) (sodium) and \(\text{NO}_2^-\) (nitrite).

Determine the Effect of the Ions on pH

Analyze whether each ion affects the pH:(a) \(\text{NH}_4^+\) can donate a proton (acting as a weak acid) while \(\text{Br}^-\) is inert (does not affect pH). Therefore, \(\text{NH}_4\text{Br}\) solution will be acidic.(b) \(\text{Na}^+\) is inert, having no effect on pH, while \(\text{NO}_2^-\) can accept a proton (acting as a weak base). Therefore, \(\text{NaNO}_2\) solution will be basic.

Calculate pH for \(\text{NH}_4\text{Br}\)

Since \(\text{NH}_4^+\) comes from the weak base \(\text{NH}_3\), its conjugate \(\text{NH}_4^+\) is a weak acid. Use the formula for the pH of a weak acid:\[K_{a} = \frac{K_w}{K_b}\]The \(K_b\) for \(\text{NH}_3\) (\(1.8 \times 10^{-5}\)) gives \(K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\).Next, use the ionization equation:\[\text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+\]The expression is:\[K_a = \frac{x^2}{C - x} \approx \frac{x^2}{C}, \text{ since } x << C\]Plugging values:\[5.56 \times 10^{-10} = \frac{x^2}{0.25}\]Solving for \(x\):\[x = \sqrt{5.56 \times 10^{-10} \times 0.25} \approx 1.18 \times 10^{-5}\]\(pH = -\log(1.18 \times 10^{-5}) \approx 4.93\). Thus, the solution is acidic.

Calculate pH for \(\text{NaNO}_2\)

As \(\text{NO}_2^-\) is a weak base, use the relationship: \[K_b = \frac{K_w}{K_a}, \text{ where } K_a = 4.6 \times 10^{-4}\]Calculate \(K_b\):\[ K_b = \frac{1.0 \times 10^{-14}}{4.6 \times 10^{-4}} = 2.17 \times 10^{-11}\]Set up the ionization equation:\[\text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^-\]The expression is:\[K_b = \frac{x^2}{C - x} \approx \frac{x^2}{C}, \text{ since } x << C\]Plugging in values:\[2.17 \times 10^{-11} = \frac{x^2}{0.20}\]Solving for \(x\):\[x = \sqrt{2.17 \times 10^{-11} \times 0.20} \approx 6.58 \times 10^{-7}\]Find \(pOH\):\[pOH = -\log(6.58 \times 10^{-7}) \approx 6.18\]Then, calculate \(pH\):\[pH = 14 - 6.18 \approx 7.82\]. The solution is basic.

Key concepts

Salt Solutions

Salt solutions are formed when salts dissolve in water, breaking into their constituent ions. How these ions interact with water can affect the solution's pH, making it neutral, acidic, or basic.
For example:

• A neutral salt solution forms when the ions do not interact significantly with water, like sodium chloride (\( ext{NaCl}\)).
• An acidic salt solution forms if the cation donates protons to water, such as ammonium bromide (\( ext{NH}_4 ext{Br}\)).
• A basic salt solution arises when the anion accepts protons from water, like sodium nitrite (\( ext{NaNO}_2\)).

Understanding the nature of the ions derived from a salt helps predict the acid-base characteristics of the solution.

pH Calculation

Calculating the pH involves determining the concentration of hydrogen ions (\( ext{H}^+\)) in a solution. For solutions with weak acids or bases, this can be more intricate due to partial ionization.
For a weak acid such as \( ext{NH}_4^+\), the pH is calculated using the ionization constant (\(K_a\)). For a weak base like \( ext{NO}_2^-\), the pH calculation involves first finding the \(pOH\), then using the relationship: \(pH = 14 - pOH\).
Setting up equations based on equilibrium expressions is crucial. These equations connect the ionization constants to the concentrations, allowing you to solve for the concentration of \( ext{H}^+\) or \( ext{OH}^-\), which are pivotal in determining the pH or pOH.

Weak Acids and Bases

Weak acids and bases partially ionize in water, meaning they do not fully dissociate into ions. This characteristic affects the pH of solutions they form.

• A weak acid like \( ext{NH}_4^+\) releases few \( ext{H}^+\) ions.
• A weak base such as \( ext{NO}_2^-\) does not produce many \( ext{OH}^-\) ions.

The extent of ionization is indicated by the \(K_a\) (acid ionization constant) for acids or \(K_b\) (base ionization constant) for bases. Lower values of \(K_a\) or \(K_b\) signify weaker acids or bases, as they correlate with less ionization. Understanding these constants helps predict how a solution will behave and what pH range it might exhibit.

Ionization Equilibrium

Ionization equilibrium is the state where the rate of ion formation equals the rate of recombination back into the molecular form. This dynamic balance is crucial in weak acid and base reactions.
For example, in the case of \( ext{NH}_4^+\) and \( ext{NH}_3\), the equilibrium equation establishes this balance:\( ext{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+\).
Similarly, for \( ext{NO}_2^-\), the equilibrium is:\( ext{NO}_2^- + ext{H}_2 ext{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^-\).

• The position of the equilibrium, guided by \(K_a\) or \(K_b\), indicates how much of the weak acid or base remains ionized.
• This knowledge helps predict how the solution's concentration changes with altering the system conditions.

Understanding ionization equilibrium is necessary to predict and calculate pH accurately in mix solutions of weak acids and bases.