Acid and Base Worksheet featuring questions on Brønsted-Lowry theory, conjugate pairs, and base definitions.

Problem Analysis:

The worksheet focuses on acid-base chemistry, specifically using the Bronsted-Lowry theory and other acid-base theories (Arrhenius and Lewis). Let's solve each part step by step.

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1. Using the Bronsted-Lowry Theory

#### a) HNO₃ + OH⁻ →
- HNO₃ is a strong acid, and OH⁻ is a strong base.
- According to the Bronsted-Lowry theory:
- An acid donates a proton (H⁺).
- A base accepts a proton (H⁺).

The reaction is:
\[
\text{HNO}_3 + \text{OH}^- \rightarrow \text{NO}_3^- + \text{H}_2\text{O}
\]

- Conjugate acid-base pairs:
- HNO₃ (acid) → NO₃⁻ (conjugate base)
- OH⁻ (base) → H₂O (conjugate acid)

#### b) CH₃NH₂ + H₂O →
- CH₃NH₂ is methylamine, a weak base.
- H₂O can act as a weak acid in this context.

The reaction is:
\[
\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{OH}^-
\]

- Conjugate acid-base pairs:
- CH₃NH₂ (base) → CH₃NH₃⁺ (conjugate acid)
- H₂O (acid) → OH⁻ (conjugate base)

#### c) OH⁻ + HPO₄²⁻ →
- OH⁻ is a strong base.
- HPO₄²⁻ is the hydrogen phosphate ion, which can act as an acid by donating a proton.

The reaction is:
\[
\text{OH}^- + \text{HPO}_4^{2-} \rightarrow \text{PO}_4^{3-} + \text{H}_2\text{O}
\]

- Conjugate acid-base pairs:
- HPO₄²⁻ (acid) → PO₄³⁻ (conjugate base)
- OH⁻ (base) → H₂O (conjugate acid)

---

2. Describing NaOH as a Base Using Three Theories

#### a) Arrhenius Base Theory
- According to the Arrhenius theory, a base is a substance that produces hydroxide ions (OH⁻) in water.
- NaOH dissociates in water as follows:
\[
\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-
\]
- Therefore, NaOH is an Arrhenius base because it produces OH⁻ ions in solution.

#### b) Bronsted-Lowry Base Theory
- According to the Bronsted-Lowry theory, a base is a proton acceptor.
- NaOH can accept a proton (H⁺) from another species. For example:
\[
\text{OH}^- + \text{H}_3\text{O}^+ \rightarrow 2\text{H}_2\text{O}
\]
- Here, OH⁻ accepts a proton from H₃O⁺, making NaOH a Bronsted-Lowry base.

#### c) Lewis Base Theory
- According to the Lewis theory, a base is an electron pair donor.
- OH⁻ has a lone pair of electrons on the oxygen atom, which it can donate to an electron-deficient species (Lewis acid).
- For example:
\[
\text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O}
\]
- Here, OH⁻ donates its lone pair to H⁺, making NaOH a Lewis base.

---

3. Reaction of Hydrogen Chloride with Ammonia

#### Reaction Equation
When hydrogen chloride (HCl) reacts with ammonia (NH₃), ammonium chloride (NH₄Cl) is formed. The reaction is:
\[
\text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4^+ + \text{Cl}^-
\]

#### Identifying Lewis Acid and Base
- Lewis Acid: A species that accepts an electron pair.
- Lewis Base: A species that donates an electron pair.

In this reaction:
- HCl acts as a Lewis acid because it accepts a lone pair of electrons from NH₃.
- NH₃ acts as a Lewis base because it donates a lone pair of electrons to HCl.

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Final Answers

1. Reactions and Conjugate Pairs:
- a) \(\boxed{\text{HNO}_3 + \text{OH}^- \rightarrow \text{NO}_3^- + \text{H}_2\text{O}}\)
- Conjugate pairs: HNO₃/NO₃⁻, OH⁻/H₂O
- b) \(\boxed{\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{OH}^-}\)
- Conjugate pairs: CH₃NH₂/CH₃NH₃⁺, H₂O/OH⁻
- c) \(\boxed{\text{OH}^- + \text{HPO}_4^{2-} \rightarrow \text{PO}_4^{3-} + \text{H}_2\text{O}}\)
- Conjugate pairs: HPO₄²⁻/PO₄³⁻, OH⁻/H₂O

2. Describing NaOH as a Base:
- Arrhenius: Produces OH⁻ in water.
- Bronsted-Lowry: Accepts a proton (H⁺).
- Lewis: Donates a lone pair of electrons.

3. Reaction and Identification:
- Equation: \(\boxed{\text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4^+ + \text{Cl}^-}\)
- Lewis Acid: HCl
- Lewis Base: NH₃

---

Final Boxed Answers:
\[
\boxed{
\begin{aligned}
&1. \text{a) } \text{HNO}_3 + \text{OH}^- \rightarrow \text{NO}_3^- + \text{H}_2\text{O} \\
&\quad \text{b) } \text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{OH}^- \\
&\quad \text{c) } \text{OH}^- + \text{HPO}_4^{2-} \rightarrow \text{PO}_4^{3-} + \text{H}_2\text{O} \\
&2. \text{Arrhenius: Produces OH}^-\text{, Bronsted-Lowry: Accepts H}^+\text{, Lewis: Donates lone pair} \\
&3. \text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4^+ + \text{Cl}^- \quad \text{(Lewis Acid: HCl, Lewis Base: NH}_3\text{)}
\end{aligned}
}
\]