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(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\) , which is the stronger base? (b) Which is the stronger acid, the ammonium or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}.\)

Short Answer

Expert verified
(a) Ammonia is the stronger base because its Kb value is larger (\(1.8 \times 10^{-5}\) compared to \(1.1 \times 10^{-8}\) for hydroxylamine). (b) The hydroxylammonium ion is the stronger acid, as it is the conjugate acid of the weaker base hydroxylamine. (c) The Ka values for ammonium and hydroxylammonium ions are approximately \(5.56 \times 10^{-10}\) and \(9.09 \times 10^{-7}\), respectively.

Step by step solution

01

(a) Compare the Kb values of ammonia and hydroxylamine

As we are given the Kb values for ammonia and hydroxylamine, we can compare them directly: \(K_{b,\,NH_3} = 1.8 \times 10^{-5}\) \(K_{b,\,H_2NOH} = 1.1 \times 10^{-8}\) Since \(K_{b,\,NH_3} > K_{b,\,H_2NOH}\), ammonia is the stronger base.
02

(b) Determine which is the stronger acid: ammonium or hydroxylammonium ion

Stronger bases have weaker conjugate acids. Therefore, since ammonia is the stronger base, its conjugate acid, the ammonium ion (NH4+), will be weaker than the hydroxylammonium ion (H3NOH+). This means that the hydroxylammonium ion is the stronger acid.
03

(c) Calculate Ka values for ammonium and hydroxylammonium ions

To calculate the Ka values for the ammonium and hydroxylammonium ions, we can use the relationship between Ka, Kb, and Kw. The ion-product constant of water, Kw, is given by: \(K_w = K_a \cdot K_b\) \(K_w = 1.0 \times 10^{-14}\) at 25°C For ammonium (NH4+), we have: \(K_{a,\,NH_4^+} = \frac{K_w}{K_{b,\,NH_3}}\) \(K_{a,\,NH_4^+} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}\) \(K_{a,\,NH_4^+} = 5.56 \times 10^{-10}\) For hydroxylammonium (H3NOH+), we have: \(K_{a,\,H_3NOH^+} = \frac{K_w}{K_{b,\,H_2NOH}}\) \(K_{a,\,H_3NOH^+} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}}\) \(K_{a,\,H_3NOH^+} = 9.09 \times 10^{-7}\) So, the Ka values for ammonium and hydroxylammonium ions are approximately \(5.56 \times 10^{-10}\) and \(9.09 \times 10^{-7}\), respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ammonia
Ammonia, which is a compound made up of nitrogen and hydrogen (NH₃), is quite fascinating in the world of chemistry. It acts as a base, and one way to measure its strength as a base is by checking its base dissociation constant, commonly known as the Kb value. For ammonia, this value is given as \(1.8 \times 10^{-5}\).

A higher Kb value indicates a stronger base. Hence, ammonia is relatively strong compared to many other bases. Its ability to accept protons easily makes it useful in various chemical reactions.

In everyday life, you'll often find ammonia used in household cleaners. Its basicity helps in removing dirt and stains effectively, making it a staple item for many cleaning products. Understanding ammonia's behavior as a base is essential for grasping the broader concept of acid-base chemistry.
Hydroxylamine
Hydroxylamine (H₂NOH) is another compound with notable roles in chemical reactions. Similar to ammonia, it can function as a base. However, its base dissociation constant, or Kb, is much smaller than that of ammonia, measured at \(1.1 \times 10^{-8}\).

This lower Kb value signifies that hydroxylamine is a weaker base compared to ammonia. Its structure and the presence of the hydroxyl group impact its ability to accept protons, thereby influencing its basic strength.

Despite being a weaker base, hydroxylamine has practical applications, especially in organic synthesis and as a reducing agent. Understanding the properties and behavior of hydroxylamine is valuable for students and chemists handling organic reactions and processes.
Conjugate Acids and Bases
The concept of conjugate acids and bases is critical in understanding acid-base equilibrium. A conjugate acid is what a base becomes after it accepts a proton, whereas a conjugate base is what an acid becomes after it donates a proton.

In the case of ammonia and hydroxylamine, their respective conjugate acids are ammonium (NH₄⁺) and hydroxylammonium (H₃NOH⁺) ions. Since ammonia is the stronger base, its conjugate acid, ammonium, is a relatively weaker acid. Conversely, hydroxylammonium is a stronger acid compared to ammonium because it is derived from the weaker base, hydroxylamine.

This inverse relationship between the strength of an acid and its conjugate base (and vice versa) is a fundamental principle in acid-base chemistry. It helps to predict the behavior of chemical species in reactions and solutions.
  • Strong base = weak conjugate acid
  • Weak base = strong conjugate acid
Ion-Product Constant of Water
The ion-product constant of water, indicated by \(K_w\), is a crucial parameter in the study of acid-base chemistry. At 25°C, \(K_w\) is generally accepted as \(1.0 \times 10^{-14}\). This value is derived from the fact that water, even though neutral, self-ionizes into hydrogen ions (H⁺) and hydroxide ions (OH⁻).

Knowing \(K_w\) is particularly useful when you're calculating the dissociation constants for acids (Ka) and bases (Kb), as it provides a relationship between them: \(K_w = K_a \cdot K_b\). This equation allows chemists to convert between Ka and Kb values, facilitating comparisons between the strengths of acids and bases.

In practical scenarios, understanding \(K_w\) helps in predicting the pH of solutions and the extent of ionization, thus playing a vital role in both analytical and theoretical chemistry disciplines. Remember, a clear grasp of \(K_w\) sets the foundation for exploring more complex chemical equilibria involving water and aqueous solutions.

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Most popular questions from this chapter

codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. \(\mathrm{A} 5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of \(9.95 .\) Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{pK}_{b}\) for this base?

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Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2}\) (b) 2.250 \(\mathrm{g}\) of LiOH in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 1.00\) mL of 0.175 M NaOH diluted to \(2.00 \mathrm{L},\) (d) a solution formed by adding 5.00 \(\mathrm{mL}\) of 0.105 \(\mathrm{M} \mathrm{KOH}\) to 15.0 \(\mathrm{mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}.\)

(a) Which of the following is the stronger Bronsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2} ?\) (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{HS}^{-}\) or \(\mathrm{HSO}_{4}^{-}\) ?

(a) Given that \(K_{a}\) for acetic acid is \(1.8 \times 10^{-5}\) and that for hypochlorous acid is \(3.0 \times 10^{-8}\) , which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypochlorite ion? (c) Calculate \(K_{b}\) values for \(C H_{3} C O O^{-}\) and \(C 1 O^{-}.\)

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