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title: "A student prepares two aqueous solutions of ammonia, \\(\\text{NH}_3\\text{(aq)}\\), at \\(25^\\circ\\text{C}\\). Solution 1 has an initial concentration of \\(0.010\\text{ M}\\), and Solution 2 has an initial concentration of \\(0.10\\text{ M}\\). The equilibrium reaction is represented by the equation below.  \\[ \\text{NH}_3\\text{(aq)} + \\text{H}_2\\text{O(l)} \\rightleftharpoons \\text{NH}_4^+\\text{(aq)} + \\text{OH}^-\\text{(aq)} \\]  The student determines that Solution 2 has a higher \\(\\text{pH}\\) than Solution 1, even though the percent ionization of \\(\\text{NH}_3\\) is lower in Solution 2. Which of the following statements correctly explains why Solution 2 has a higher \\(\\text{pH}\\)?"
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date_modified: "2026-09-28T12:30:23+00:00"
---

# A student prepares two aqueous solutions of ammonia, \(\text{NH}_3\text{(aq)}\), at \(25^\circ\text{C}\). Solution 1 has an initial concentration of \(0.010\text{ M}\), and Solution 2 has an initial concentration of \(0.10\text{ M}\). The equilibrium reaction is represented by the equation below.

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

The student determines that Solution 2 has a higher \(\text{pH}\) than Solution 1, even though the percent ionization of \(\text{NH}_3\) is lower in Solution 2. Which of the following statements correctly explains why Solution 2 has a higher \(\text{pH}\)?

A student prepares two aqueous solutions of ammonia, \(\text{NH}_3\text{(aq)}\), at \(25^\circ\text{C}\). Solution 1 has an initial concentration of \(0.010\text{ M}\), and Solution 2 has an initial concentration of \(0.10\text{ M}\). The equilibrium reaction is represented by the equation below.

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

The student determines that Solution 2 has a higher \(\text{pH}\) than Solution 1, even though the percent ionization of \(\text{NH}_3\) is lower in Solution 2. Which of the following statements correctly explains why Solution 2 has a higher \(\text{pH}\)?

- **A.** The \(\text{pH}\) is higher in Solution 2 because the equilibrium constant, \(K_b\), is larger at higher solute concentrations, resulting in a higher equilibrium \([\text{OH}^-]\).
- **B.** The \(\text{pH}\) is higher in Solution 2 because the greater initial concentration of \(\text{NH}_3\) yields a higher equilibrium \([\text{OH}^-]\), even though a smaller fraction of the base molecules ionize.
- **C.** The \(\text{pH}\) is higher in Solution 2 because having a lower percent ionization results in a higher equilibrium \([\text{H}_3\text{O}^+]\), which directly raises the \(\text{pH}\).
- **D.** The \(\text{pH}\) is higher in Solution 2 because higher solute concentrations shift the equilibrium further to the right, increasing the percentage of \(\text{NH}_3\) that is converted to products.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123809/*