Problem 38
Question
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 \(\mathrm{HClO}_{4}\).
Step-by-Step Solution
Verified Answer
(a) pH ≈ 0.92; (b) pH ≈ -0.38; (c) pH ≈ 3.5.
1Step 1: Understanding Strong Acids
Hydrochloric acid (HCl), nitric acid (HNO₃), and perchloric acid (HClO₄) are all strong acids. This means they dissociate completely in water. For strong acids, the concentration of hydrogen ions [H⁺] is equal to the concentration of the acid.
2Step 2: Calculating pH for 0.12 M HCl
Since HCl is a strong acid and dissociates completely, [H⁺] = [HCl] = 0.12 M. The pH is calculated using the formula: \[\text{pH} = -\log [\text{H}^+]\]Substitute [H⁺] with 0.12:\[\text{pH} = -\log(0.12) \approx 0.92\]
3Step 3: Calculating pH for 2.4 M HNO₃
Similarly, for HNO₃, [H⁺] = [HNO₃] = 2.4 M. Calculate the pH:\[\text{pH} = -\log(2.4) \approx -0.38\]Since negative pH values are possible with strong concentrations, this result is reasonable.
4Step 4: Calculating pH for 3.2 x 10⁻⁴ M HClO₄
For HClO₄, [H⁺] = [HClO₄] = 3.2 \times 10^{-4} M. Calculate the pH:\[\text{pH} = -\log(3.2 \times 10^{-4}) \approx 3.5\]
5Step 5: Interpreting Results
For a strong acid, the lower the concentration, the higher the pH. Even for strong acids, at very low concentrations, pH can be greater than zero. This is reflected in the fact that 3.2 x 10⁻⁴ M HClO₄ results in a pH of about 3.5.
Key Concepts
Strong AcidsHydrogen Ion ConcentrationHCl DissociationHNO₃ DissociationHClO₄ Dissociation
Strong Acids
When we talk about strong acids, we refer to acids that completely dissociate in water, releasing all their hydrogen ions. Three prominent strong acids that often appear in chemistry studies are hydrochloric acid (HCl), nitric acid (HNO₃), and perchloric acid (HClO₄).
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.
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
Hydrogen ion concentration, represented as \([\text{H}^+]\), is a crucial factor in determining the acidity of a solution. For strong acids, calculating \([\text{H}^+]\) becomes a simple task since the acid dissociates completely.
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.
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
Hydrochloric acid (HCl) is a prototypical strong acid, demonstrating the concept of complete dissociation. When HCl dissolves in water, it splits entirely into hydrogen ions and chloride ions.
The chemical equation for this process is:
Thus, the pH can be calculated as:
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
Nitric acid (HNO₃) shares the behavior of strong acids by also fully dissociating in water. In this scenario, it separates into hydrogen and nitrate ions.
The dissociation equation is:
The calculation of the pH follows:
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
Perchloric acid (HClO₄) is another example of a strong acid, dissociating entirely in aqueous solutions into hydrogen ions and perchlorate ions.
The dissociation process can be represented by the equation:
This results in a pH calculation:
The dissociation process can be represented by the equation:
- \(\text{HClO}_4 \rightarrow \text{H}^+ + \text{ClO}_4^-\)
This results in a pH calculation:
- \(\text{pH} = -\log(3.2 \times 10^{-4}) \approx 3.5\)
Other exercises in this chapter
Problem 36
Which of the following statements are true regarding a \(1.0-M\) solution of a strong acid \(\mathrm{HA}\) at \(25^{\circ} \mathrm{C} ?\) (Choose all that apply
View solution Problem 37
Why are ionizations of strong acids and strong bases generally not treated as equilibria?
View solution Problem 39
Calculate the \(\mathrm{pH}\) of an aqueous solution at \(25^{\circ} \mathrm{C}\) that is (a) \(1.02 \mathrm{M}\) in \(\mathrm{HI}\), (b) \(0.035 \mathrm{M}\) i
View solution Problem 40
Calculate the concentration of \(\mathrm{HBr}\) in a solution at \(25^{\circ} \mathrm{C}\) that has a pH of (a) 0.12 , (b) \(2.46,\) and \((\mathrm{c}) 6.27\).
View solution