From the equilibrium concentrations given, calculate \(K_{\mathrm{a}}\) for each of the weak acids and \(K_{\mathrm{b}}\) for each of the weak bases. (a) \(\mathrm{NH}_{3}:\left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{NH}_{4}+\right]=3.1 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{NH}_{3}\right]=0.533 \mathrm{M}\). (b) \(\mathrm{HNO}_{2}:\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.011 \mathrm{M}\) \(\left[\mathrm{NO}_{2}^{-}\right]=0.0438 \mathrm{M}\) \(\left[\mathrm{HNO}_{2}\right]=1.07 \mathrm{M}\). (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}:\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right]=0.25 \mathrm{M}\) \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]=4.3 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{OH}^{-}\right]=3.7 \times 10^{-3} \mathrm{M}\). (d) \(\mathrm{NH}_{4}^{+}: \quad\left[\mathrm{NH}_{4}^{+}\right]=0.100 \mathrm{M}\) \(\left[\mathrm{NH}_{3}\right]=7.5 \times 10^{-6} \mathrm{M}\) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.5 \times 10^{-6} \mathrm{M}\).

Step by step solution

01

Determine the Equilibrium Expression for NH3(aq)

The equilibrium expression for the reaction of NH3 with water to produce NH4+ and OH- is given by: \[ K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]} \] Plug in the equilibrium concentrations to solve for the base dissociation constant (Kb).

02

Calculate Kb for NH3(aq)

Substitute the given concentrations into the equilibrium expression: \[ K_{\mathrm{b}}=\frac{(3.1 \times 10^{-3})(3.1 \times 10^{-3})}{0.533} \]. Perform the multiplication and division to get the value of Kb.

03

Determine the Equilibrium Expression for HNO2(aq)

The equilibrium expression for the reaction of HNO2 with water to produce H3O+ and NO2- is given by: \[ K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right]}{\left[\mathrm{HNO}_{2}\right]} \] Plug in the equilibrium concentrations to solve for the acid dissociation constant (Ka).

04

Calculate Ka for HNO2(aq)

Substitute the given concentrations into the equilibrium expression: \[ K_{\mathrm{a}}=\frac{(0.011)(0.0438)}{1.07} \]. Perform the multiplication and division to get the value of Ka.

05

Determine the Equilibrium Expression for (CH3)3N(aq)

The equilibrium expression for the reaction of (CH3)3N with water to produce (CH3)3NH+ and OH- is given by: \[ K_{\mathrm{b}}=\frac{\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right]} \] Plug in the equilibrium concentrations to solve for the base dissociation constant (Kb).

06

Calculate Kb for (CH3)3N(aq)

Substitute the given concentrations into the equilibrium expression: \[K_{\mathrm{b}}=\frac{(4.3 \times 10^{-3})(3.7 \times 10^{-3})}{0.25} \]. Perform the multiplication and division to get the value of Kb.

07

Determine the Equilibrium Expression for NH4+(aq)

The equilibrium expression for the reaction of NH4+ with water to produce NH3 and H3O+ is given by: \[ K_{\mathrm{a}}=\frac{\left[\mathrm{NH}_{3}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \] Plug in the equilibrium concentrations to solve for the acid dissociation constant (Ka).

08

Calculate Ka for NH4+(aq)

Substitute the given concentrations into the equilibrium expression: \[ K_{\mathrm{a}}=\frac{(7.5 \times 10^{-6})(7.5 \times 10^{-6})}{0.100} \]. Perform the multiplication and division to get the value of Ka.

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Key Concepts

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

Weak Acid Dissociation Constant

The weak acid dissociation constant (\(K_a\)) is a measure of the strength of an acid in solution. It quantifies the acid's tendency to donate a proton to water, forming its conjugate base and hydronium ions \(H_3O^+\). For a generic weak acid \(HA\), the dissociation in water can be represented as \(HA + H_2O \rightleftharpoons A^- + H_3O^+\) . The equilibrium expression for this reaction is \(K_a = \frac{[A^-][H_3O^+]}{[HA]}\). A higher value of \(K_a\) indicates a stronger acid, meaning it more readily donates protons. In the case of \(HNO_2\), for instance, the given concentrations lead to the calculation of its \(K_a\) as shown in the solution steps.

Weak Base Dissociation Constant

The weak base dissociation constant (\(K_b\)) is similar to \(K_a\) but applies to bases. It represents the extent to which a base forms its conjugate acid and hydroxide ions \(OH^-\) when dissolved in water. For a generic weak base \(B\) reacting with water, the reaction is \(B + H_2O \rightleftharpoons BH^+ + OH^-\) and the equilibrium expression is \(K_b = \frac{[BH^+][OH^-]}{[B]}\). A higher \(K_b\) value indicates a more capable base of accepting protons. As shown in the given exercise, for substances like \(NH_3\) and \(\left(CH_3\right)_3N\) the \(K_b\) can be determined using equilibrium concentrations.

Chemical Equilibrium

Chemical equilibrium occurs in a reversible chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. Equilibrium does not mean the reactants and products are present in equal amounts; rather, their ratios do not change. The position of equilibrium is described by the equilibrium constant \(K\), which is derived from the concentrations of the reactants and products at equilibrium. For example, in the solutions to the exercise, equilibrium is assumed to find the values of \(K_a\) and \(K_b\).

Equilibrium Expressions

Equilibrium expressions mathematically describe the state of a chemical equilibrium. For a generic reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium expression is \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\) where \(K\) is the equilibrium constant, and the brackets denote the concentrations of the substances. This expression assumes that the reaction occurs in a closed system at a constant temperature. When dealing with weak acids and bases, the equilibrium constants, \(K_a\) and \(K_b\) respectively, are specific instances of such expressions as illustrated through the solution to the given exercise.