From the equilibrium concentrations given, calculate Ka for...

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) $\mathrm{NH}_{4}+:\left[\mathrm{NH}_{4}+\right]=0.100 \mathrm{M}$ $\left[\mathrm{NH}_{3}\right]=7.5 \times 10^{-6} \mathrm{M}$ $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.5 \times 10^{-6} M$

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) $\mathrm{NH}_{4}+:\left[\mathrm{NH}_{4}+\right]=0.100 \mathrm{M}$
$\left[\mathrm{NH}_{3}\right]=7.5 \times 10^{-6} \mathrm{M}$
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.5 \times 10^{-6} M$

Key Concepts

Equilibrium Constant (K_eq)

The equilibrium constant is a ratio that describes the concentrations of products and reactants at equilibrium for a chemical reaction. In the context of acid-base reactions, the constants K_a and K_b are specific forms that quantify the extent of acid or base dissociation in water, providing insight into the strength of the acid or base.

Acid Dissociation Constant (K_a)

K_a is the equilibrium constant for the dissociation of a weak acid into its conjugate base and a proton (or hydronium ion) in aqueous solution. It is calculated by taking the ratio of the concentrations of the ionized products to the concentration of the undissociated acid, with each concentration raised to the power corresponding to its stoichiometric coefficient.

Base Dissociation Constant (K_b)

K_b is the equilibrium constant for the reaction in which a weak base interacts with water to form its conjugate acid and hydroxide ion. Like K_a, it is calculated by the ratio of the concentrations of the products to the concentration of the reactant, providing a measure of the base's strength by indicating its ability to accept a proton.

Conjugate Acid-Base Pairs

In acid-base chemistry, a conjugate acid-base pair consists of two species related by the transfer of a proton. The acid donates a proton to become its conjugate base, while the base accepts a proton to become its conjugate acid. The strengths of these conjugate pairs are interconnected, with the product of the acid and base dissociation constants (K_a and K_b) equaling the ionization constant of water (K_w) in aqueous solutions.

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