Chapter 16: Problem 38
Calculate the \(\mathrm{pH}\) of an aqueous solution at \(25^{\circ} \mathrm{C}\) that is (a) \(0.12 \mathrm{M}\) in \(\mathrm{HCl}\), (b) \(2.4 \mathrm{M}\) in \(\mathrm{HNO}_{3}\), and (c) \(3.2 \times 10^{-4} \mathrm{M}\) in \(\mathrm{HClO}_{4}\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Strong Acids
These acids are characterized by their ability to completely ionize in an aqueous solution. This means that when you dissolve a strong acid in water, you can assume that its concentration is equal to the concentration of hydrogen ions \([\text{H}^+]\).
Understanding this property simplifies the calculation of pH, making it straightforward: we use the initial concentration of the acid to determine that of the hydrogen ions directly.
Hydrogen Ion Concentration
This means the concentration of hydrogen ions is equal to the concentration of the acid itself. For example, if you have a 0.12 M solution of HCl, the hydrogen ion concentration is also 0.12 M.
Measuring \([\text{H}^+]\) lets us calculate the pH by using the formula \(\text{pH} = -\log [\text{H}^+]\). This logarithmic measure gives us an easy-to-understand scale of acidity.
HCl Dissociation
The chemical equation for this process is:
- \(\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-\)
Thus, the pH can be calculated as:
- \(\text{pH} = -\log(0.12) \approx 0.92\)
HNO₃ Dissociation
The dissociation equation is:
- \(\text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^-\)
The calculation of the pH follows:
- \(\text{pH} = -\log(2.4) \approx -0.38\)
HClO₄ Dissociation
The dissociation process can be represented by the equation:
- \(\text{HClO}_4 \rightarrow \text{H}^+ + \text{ClO}_4^-\)
This results in a pH calculation:
- \(\text{pH} = -\log(3.2 \times 10^{-4}) \approx 3.5\)