TeXes Science Practice Test 2026 - Free Science Exam Practice Questions and Study Guide

Question: 1 / 400

Given the dissociation equation of nitrous acid, what is Ka for a solution with a pH of 3?

1.25 x 10^-4

6.25 x 10^-4

To determine the acid dissociation constant (Ka) for nitrous acid from the given pH, it’s important to first understand the relationship between pH, hydrogen ion concentration, and Ka.

When a weak acid like nitrous acid (HNO2) dissociates in solution, it can be represented by the following equation:

HNO2 ⇌ H⁺ + NO2⁻

The pH of the solution is given as 3, which means the concentration of hydrogen ions, [H⁺], can be calculated using the formula:

\[ [H⁺] = 10^{-pH} = 10^{-3} \]

This results in a hydrogen ion concentration of 0.001 M.

In a weak acid equilibrium, we can express the dissociation in terms of concentrations. Let the initial concentration of nitrous acid be \( C \). For a weak acid at equilibrium, since it only partially dissociates, we can denote the concentration of the dissociated species as \( x \), which in this case is the concentration of H⁺ ions produced from the dissociation. Thus, at equilibrium:

- [H⁺] = x = 0.001 M

- [NO2

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2.5 x 10^-3

4.00 x 10^-5

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