**Problem Statement:**Determine the pH of a 1.2 M solution of ammonia, NH₃, with \( K_b = 1.8 \times 10^{-5} \).---For educational purposes, this would involve calculating the hydroxide ion concentration ([OH⁻]) from the base equilibrium equation and using it to find the pOH and subsequently the pH. ### Steps for Solution:1. **Write the balanced equation for ammonia in water:** \[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]2. **Write the expression for the base dissociation constant (\( K_b \)):** \[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]3. **Make assumptions for initial concentrations and changes:** - Initial \([\text{NH}_3] = 1.2 \, \text{M}\) - Changes in concentration are symbolized as \(-x\) for \([\text{NH}_3]\) and \(+x\) for \([\text{NH}_4^+]\) and \([\text{OH}^-]\).4. **Solve for \(x\):** - Substitute into \(K_b\) expression: \[ 1.8 \times 10^{-5} = \frac{x \cdot x}{1.2 - x} \] - Assume \(1.2 - x \approx 1.2\) due to small value of \(K_b\): \[ 1.8 \times 10^{-5} = \frac{x^2}{1.2} \] \[ x^2 = 1.8 \times 10^{-5} \times 1.2 \] \[ x^2 = 2.16 \times 10^{-5} \] \[ x = \sqrt{2.16 \times 10^{-5}} \approx 0.00464 \]5. **Calculate pOH and pH:** - pOH = \(-

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Chemistry

Determine the pH of a 1.2 M solution of ammonia, NH3, with K 1.8 x 105

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

**Problem Statement:**

Determine the pH of a 1.2 M solution of ammonia, NH₃, with \( K_b = 1.8 \times 10^{-5} \).

---

For educational purposes, this would involve calculating the hydroxide ion concentration ([OH⁻]) from the base equilibrium equation and using it to find the pOH and subsequently the pH.

### Steps for Solution:

1. **Write the balanced equation for ammonia in water:**

\[
\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-
\]

2. **Write the expression for the base dissociation constant (\( K_b \)):**

\[
K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}
\]

3. **Make assumptions for initial concentrations and changes:**

- Initial \([\text{NH}_3] = 1.2 \, \text{M}\)
- Changes in concentration are symbolized as \(-x\) for \([\text{NH}_3]\) and \(+x\) for \([\text{NH}_4^+]\) and \([\text{OH}^-]\).

4. **Solve for \(x\):**

- Substitute into \(K_b\) expression:

\[
1.8 \times 10^{-5} = \frac{x \cdot x}{1.2 - x}
\]

- Assume \(1.2 - x \approx 1.2\) due to small value of \(K_b\):

\[
1.8 \times 10^{-5} = \frac{x^2}{1.2}
\]

\[
x^2 = 1.8 \times 10^{-5} \times 1.2
\]

\[
x^2 = 2.16 \times 10^{-5}
\]

\[
x = \sqrt{2.16 \times 10^{-5}} \approx 0.00464
\]

5. **Calculate pOH and pH:**

- pOH = \(-

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