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Magazine Chapter 7: Problem 33
Calculate the \(\mathrm{pH}\) of the solution prepared by adding \(0.10 \mathrm{~mol}\) each of hydroxylamine and hydrochloric acid to \(500 \mathrm{~mL}\) water.
Short Answer
Expert verified
The \(\mathrm{pH}\) of the solution is approximately 3.87.
Step by step solution
01
Understand the reaction
Hydroxylamine, NH鈧侽H, will react with HCl to form NH鈧侽H鈧傗伜 ions since hydroxylamine is a weak base. The reaction can be represented as: \[ \text{NH}_2\text{OH} + \text{HCl} \rightarrow \text{NH}_2\text{OH}_2^+ + \text{Cl}^- \]
02
Calculate initial concentrations
Since 0.10 mol of each solute is dissolved in 0.5 L of solution, the initial concentration of both hydroxylamine and HCl is \( \frac{0.10 \text{ mol}}{0.5 \text{ L}} = 0.20 \text{ M} \).
03
Reaction completion
HCl is a strong acid and will react completely with hydroxylamine. All 0.10 mol of HCl will convert hydroxylamine to its conjugate acid, leaving the solution with 0.20 M NH鈧侽H鈧傗伜.
04
Calculate \\ \mathrm{pH} from \\ [NH鈧侽H鈧傗伜]
The solution now effectively consists of NH鈧侽H鈧傗伜 at a concentration of 0.20 M, acting as a weak acid. First, obtain \(K_a\) using the relationship \(K_w = K_b \, K_a\), where \(K_w = 1 \times 10^{-14}\) at 25掳C. Look up \(K_b = 1.1 \times 10^{-8}\) for NH鈧侽H to find \(K_a = \frac{1 \times 10^{-14}}{1.1 \times 10^{-8}} \approx 9.09 \times 10^{-7}\).
05
Use \\ K_a to find [H鈦篯
Using the relationship \( [\text{H}^+]^2 = K_a \times [\text{NH}_2\text{OH}_2^+] \) and solving for \([\text{H}^+]\), we have \([\text{H}^+] = \sqrt{9.09 \times 10^{-7} \times 0.20} \approx 1.35 \times 10^{-4} \text{ M}\).
06
Calculate \\ \\mathrm{pH}
Finally, the \(\mathrm{pH}\) of the solution is calculated by \(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.35 \times 10^{-4}) \approx 3.87 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acid-Base Reactions
In chemistry, acid-base reactions are fundamental processes where an acid reacts with a base to form a salt and often water. These reactions are crucial for understanding various chemical behaviors and are essential for calculating the pH of solutions. In the given problem, hydroxylamine (NH鈧侽H), a weak base, reacts with hydrochloric acid (HCl), a strong acid, to form hydroxylammonium ion (NH鈧侽H鈧傗伜) and chloride ion (Cl鈦�).
- The base (NH鈧侽H) accepts protons from the acid (HCl), transforming into its conjugate acid (NH鈧侽H鈧傗伜).
- This process illustrates a typical acid-base reaction where the base gains a hydrogen ion.
- The strong acid, HCl, fully dissociates into its ions in solution, completing the reaction efficiently.
Equilibrium Constants
Equilibrium constants are vital for understanding the extent of reactions, particularly in reversible reactions. For acid-base reactions involving weak acids or bases, equilibrium constants play a significant role in determining the concentrations of ions in solution. In this exercise, the equilibrium constant for hydroxylamine is initially given as a base, and its strength is expressed through its base ionization constant, \(K_b\).
- The autoprotolysis constant of water (\(K_w\)) at 25掳C is \(1 \times 10^{-14}\), a crucial tool for relating \(K_a\) and \(K_b\).
- The relationship \(K_w = K_b \times K_a\) allows conversion of the base equilibrium constant \(K_b\) of hydroxylamine to its corresponding acid equilibrium constant \(K_a\).
- Through this calculation, using \(K_b = 1.1 \times 10^{-8}\), we can determine \(K_a\) for the NH鈧侽H鈧傗伜 ions as approximately \(9.09 \times 10^{-7}\).
Hydroxylamine
Hydroxylamine (NH鈧侽H) is a chemical compound notable for its use as a reducing agent and as a weak base in various chemical reactions. In this particular exercise, understanding hydroxylamine's behavior as a weak base is crucial for calculating the pH of solutions it is part of.
- As a weak base, hydroxylamine partially accepts protons from strong acids, like HCl.
- Its reaction with HCl results in the formation of its conjugate acid, NH鈧侽H鈧傗伜, changing the pH of the solution to more acidic.
- The formation of NH鈧侽H鈧傗伜 ions dictates how the solution will behave in terms of pH, a parameter seen through its base ionization constant \(K_b = 1.1 \times 10^{-8}\).
