Question

Calculation of the \(\mathrm{pH}\) of a Strong Acid or Base a. Write out the acid dissociation reaction for hydrochloric acid. b. Calculate the \(\mathrm{pH}\) of a solution of \(5 \times 10^{-4} \mathrm{M}\) hydrochloric acid at \(25^{\circ} \mathrm{C}\). c. Write out the acid dissociation reaction for sodium hydroxide. d. Calculate the \(\mathrm{pH}\) of a solution of \(7 \times 10^{-5} \mathrm{M}\) sodium hydroxide at \(25^{\circ} \mathrm{C}\).

Short answer

For HCl, pH is approximately 3.30; for NaOH, pH is approximately 9.85.

Step-by-step solution

Acid Dissociation Reaction - Hydrochloric Acid

The dissociation of hydrochloric acid (HCl) in water is expressed as: \[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]. HCl completely dissociates into hydrogen ions (\(\text{H}^+\)) and chloride ions (\(\text{Cl}^-\)) in an aqueous solution.

Calculate pH for Hydrochloric Acid

For strong acids like hydrochloric acid, the concentration of \(\text{H}^+\) ions is equal to the concentration of the acid. Given, \([\text{HCl}] = 5 \times 10^{-4} \text{ M}\), the \([\text{H}^+]\) is also \(5 \times 10^{-4} \text{ M}\). The \text{pH} is calculated using the formula: \( \text{pH} = -\log_{10} \left([\text{H}^+ ] \right) \). Substituting the values, we get \( \text{pH} = -\log_{10} (5 \times 10^{-4}) \approx 3.30 \).

Base Dissociation Reaction - Sodium Hydroxide

The dissociation of sodium hydroxide (NaOH) in water is expressed as: \[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]. NaOH completely dissociates into sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)) in aqueous solution.

Calculate pH for Sodium Hydroxide

For strong bases like sodium hydroxide, the concentration of \(\text{OH}^-\) is equal to the concentration of the base. Given \([\text{NaOH}] = 7 \times 10^{-5} \text{ M}\), \([\text{OH}^-] = 7 \times 10^{-5} \text{ M}\). The \text{pOH} is calculated using the formula: \( \text{pOH} = -\log_{10} \left([\text{OH}^- ] \right) \). Substituting the values, \( \text{pOH} = -\log_{10} (7 \times 10^{-5}) \approx 4.15 \). Then, use \( \text{pH} + \text{pOH} = 14 \) to find \text{pH}. Thus, \( \text{pH} = 14 - 4.15 = 9.85 \).

Key concepts

Acid Dissociation

Understanding the dissociation of acids in water is crucial in chemistry. When hydrochloric acid (HCl) is dissolved in water, it completely dissociates into its constituent ions. The reaction can be written as: \( \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \). Here, HCl is a strong acid, which means it fully breaks into hydrogen ions (\(\text{H}^+\)) and chloride ions (\(\text{Cl}^-\)) when in solution. This complete dissociation makes it possible to directly equate the concentration of the original acid to the concentration of hydrogen ions in solution. Remember, for strong acids in water, dissociation happens so thoroughly that the reverse reaction (ions recombining to form HCl) does not occur to any significant degree.

Unlike weak acids, which only partially dissociate in water, strong acids provide a straightforward calculation path for their pH. The concept of acid dissociation is pivotal when considering reactions in chemical equilibrium and helps predict and understand their behavior in different scenarios.

Strong Acids

Strong acids like hydrochloric acid (HCl) are characterized by their ability to fully dissociate in water. This means they relinquish their hydrogen ions completely, making them a key focus when calculating pH. For HCl, the dissociation reaction is clear-cut: \( \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \). The important takeaway is that the concentration of \(\text{H}^+\) is equivalent to the original concentration of HCl. This makes calculating the pH straightforward.

The pH is calculated using the formula: \[ \text{pH} = -\log_{10}([\text{H}^+]) \] For solutions where \( [\text{HCl}] = 5 \times 10^{-4} \text{ M} \), \( \text{pH} = -\log_{10}(5 \times 10^{-4}) \approx 3.30 \).

• Complete dissociation means their pH can be directly calculated from their molarity.
• Evaluation of pH offers insights into the solution's potential acidity and its strength relative to other solutions.

Strong acids are frequently encountered in various applications ranging from laboratory experiments to industrial processes.

Strong Bases

Strong bases like sodium hydroxide (NaOH) behave similarly to strong acids in terms of dissociation but with a different product. When NaOH is dissolved in water, it dissociates completely to yield sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)). The reaction looks like this: \( \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \).In this context, calculating the pH of such a solution involves first finding the pOH. This is determined by the formula: \[ \text{pOH} = -\log_{10}([\text{OH}^-]) \] Given \( [\text{NaOH}] = 7 \times 10^{-5} \text{ M} \), \( \text{pOH} = -\log_{10}(7 \times 10^{-5}) \approx 4.15 \).Similarly, \[ \text{pH} = 14 - \text{pOH} = 9.85 \].

• Strong bases also fully dissociate in solution, allowing accurate calculation of their pH via their pOH.
• This property is critical in contexts where neutralization reactions or specific pH levels are desired.

By manipulating these calculations, one can accurately predict the behavior and attributes of strong base solutions.

Chemical Equilibrium

In the context of strong acids and bases, chemical equilibrium plays a somewhat straightforward role. For weak acids and bases, equilibrium involves a balance point where the forward and reverse reactions occur at the same rate, and not all reactants are converted to products. However, for strong acids and bases, complete dissociation means that chemical equilibrium is highly skewed, with virtually no backward reaction. In solutions with strong acids or bases, once dissolved, all components shift quickly and fully to products (ions in solution), creating a system that is effectively at equilibrium but different from an ongoing dynamic balance typically expected in traditional chemical equilibrium scenarios.

• Complete dissociation means equilibrium lies entirely on the products' side.
• Important for understanding how acid and base reactions drive processes toward completion.

Applications of this simplified chemical equilibrium are far-reaching, including processes like acid-base titration where complete reaction is assumed for strong acids and bases.