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Chapter 13: Problem 13

If the \(K_{\mathrm{a}}\) for \(\mathrm{HNO}_{2}\) is \(5.6 \times 10^{-4}\), what is the \(K_{\mathrm{b}}\) for \(\mathrm{NO}_{2}^{-}(\mathrm{aq})\) ?

Short Answer

Expert verified
The \(K_b\) for \(\mathrm{NO}_2^-\) is approximately \(1.79 \times 10^{-11}\).

Step by step solution

01

Understanding the Relationship between Ka and Kb

To find the relationship between the acid dissociation constant \(K_a\) for \(\mathrm{HNO}_2\) and the base dissociation constant \(K_b\) for its conjugate base \(\mathrm{NO}_2^-\), we must use the equation: \(K_w = K_a \cdot K_b\), where \(K_w\) is the ion-product constant for water, typically \(1.0 \times 10^{-14}\) at 25 掳C.
02

Rearranging the Formula

We need to solve for \(K_b\). Rearrange the equation \(K_w = K_a \cdot K_b\) to find: \(K_b = \frac{K_w}{K_a}\).
03

Substitute the Known Values

Now, substitute the given values into the equation. Use \(K_w = 1.0 \times 10^{-14}\) and \(K_a = 5.6 \times 10^{-4}\). This gives us \(K_b = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-4}}\).
04

Calculate Kb

Perform the division: \(K_b = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-4}}\), which equals approximately \(1.79 \times 10^{-11}\).
05

Verify the Units

Verify that the dissociation constants retain the correct units. Both \(K_a\) and \(K_b\) are dimensionless, maintaining consistency in our calculations.

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Key Concepts

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

Acid Dissociation Constant
The acid dissociation constant, commonly referred to as "Ka," is a fundamental concept in chemistry that measures the strength of an acid in a solution. It is specifically the equilibrium constant for the dissociation of an acid into its conjugate base and a proton. A higher Ka value indicates a stronger acid, which means the acid dissociates more completely in water. To understand the importance of Ka, consider its role in equilibrium reactions:
  • When an acid like \(\mathrm{HNO}_2\) dissociates in water, it forms \(\mathrm{NO}_2^-\) (its conjugate base) and \(\mathrm{H}^+\) ions.
  • The reaction reaches equilibrium, and the extent to which the acid dissociates is quantified by the Ka.
Knowing the Ka helps predict how an acid will behave in different chemical environments. It鈥檚 an essential parameter when calculating pH and understanding acid-base equilibria in aqueous solutions.
Conjugate Base
A conjugate base is what remains of an acid molecule after it donates a proton in a chemical reaction. In an acid-base reaction, an acid donates its hydrogen ion (proton) to another molecule, resulting in the formation of its conjugate base.For example, when \(\mathrm{HNO}_2\) (nitrous acid) donates a proton, it becomes \(\mathrm{NO}_2^-\), which is its conjugate base. This transformation highlights several key points:
  • Acids and their conjugate bases exist in equilibrium in aqueous solutions.
  • The strength of a conjugate base is inversely related to the strength of its acid. Strong acids have weak conjugate bases and vice versa.
  • Understanding conjugate pairs is vital in predicting the outcome of acid-base reactions.
The relationship between a conjugate acid-base pair is essential in calculating the pH of solutions and in various applications such as buffer solutions.
Ion-Product Constant for Water
The ion-product constant for water, denoted as \(K_w\), is a crucial concept in understanding acid-base chemistry. \(K_w\) is the product of the concentrations of hydrogen ions \(\mathrm{[H^+]}\) and hydroxide ions \(\mathrm{[OH^-]}\) in pure water. At room temperature (25掳C), \(K_w\) is typically \(1.0 \times 10^{-14}\). Here are a few important points about \(K_w\):
  • It represents the extent of water dissociation into \(\mathrm{H^+}\) and \(\mathrm{OH^-}\).
  • The value of \(K_w\) remains constant for dilute aqueous solutions at a given temperature.
  • In the context of acid-base theories, the ion-product constant relates to other constants such as \(K_a\) and \(K_b\).
In calculations like determining the base dissociation constant, \(K_b\), for a conjugate base, the constant \(K_w\) is used alongside \(K_a\) to find the desired equilibrium constant, showing the interconnectedness of these equilibrium properties.

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Most popular questions from this chapter

The following reaction is at equilibrium: $$ \mathrm{CH}_{4}+2 \mathrm{Cl}_{2} \rightleftarrows \mathrm{CH}_{2} \mathrm{Cl}_{2}+2 \mathrm{HCl} $$ If \(\left[\mathrm{CH}_{4}\right]\) is \(0.250 \mathrm{M},\left[\mathrm{Cl}_{2}\right]\) is \(0.150 \mathrm{M},\) and \(\left[\mathrm{CH}_{2} \mathrm{Cl}_{2}\right]\) is \(0.175 \mathrm{M}\) at equilibrium, what is \([\mathrm{HCl}] \mathrm{at}\) equilibrium if the \(K_{\text {eq }}\) is \(2.30 ?\)

What is the relationship between the equilibrium row in an ICE chart and the other two rows?

Explain the difference between the \(K_{\mathrm{eq}}\) and the \(K_{\mathrm{sp}}\).

For a salt that has the general formula MX, an ICE chart shows that the \(K_{\mathrm{sp}}\) is equal to \(x^{2}\), where \(x\) is the concentration of the cation. What is the appropriate formula for the \(K_{\text {sp }}\) of a salt that has a general formula of \(\mathrm{MX}_{2}\) ?

Explain what is meant when it is said that chemical equilibrium is dynamic.
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