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) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}:\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.34 \times 10^{-3} \mathrm{M}$ $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]=1.34 \times 10^{-3} \mathrm{M}$ $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=9.866 \times 10^{-2} \mathrm{M}$ (b) $\mathrm{ClO}^{-}:\left[\mathrm{OH}^{-}\right]=4.0 \times 10^{-4} \mathrm{M}$ $[\mathrm{HClO}]=2.38 \times 10^{-4} \mathrm{M}$ $\left[\mathrm{ClO}^{-}\right]=0.273 \mathrm{M}$ (c) $\mathrm{HCO}_{2} \mathrm{H}:\left[\mathrm{HCO}_{2} \mathrm{H}\right]=0.524 \mathrm{M}$ $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.8 \times 10^{-3} \mathrm{M}$ $\left[\mathrm{HCO}_{2}^{-}\right]=9.8 \times 10^{-3} \mathrm{M}$ (d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}: \quad\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right]=0.233 \mathrm{M}$ $\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=2.3 \times 10^{-3} \mathrm{M}$ $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.3 \times 10^{-3} \mathrm{M}$

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) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}:\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.34 \times 10^{-3} \mathrm{M}$ $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]=1.34 \times 10^{-3} \mathrm{M}$
$\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=9.866 \times 10^{-2} \mathrm{M}$
(b) $\mathrm{ClO}^{-}:\left[\mathrm{OH}^{-}\right]=4.0 \times 10^{-4} \mathrm{M}$
$[\mathrm{HClO}]=2.38 \times 10^{-4} \mathrm{M}$
$\left[\mathrm{ClO}^{-}\right]=0.273 \mathrm{M}$
(c) $\mathrm{HCO}_{2} \mathrm{H}:\left[\mathrm{HCO}_{2} \mathrm{H}\right]=0.524 \mathrm{M}$
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.8 \times 10^{-3} \mathrm{M}$
$\left[\mathrm{HCO}_{2}^{-}\right]=9.8 \times 10^{-3} \mathrm{M}$
(d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}: \quad\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right]=0.233 \mathrm{M}$
$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=2.3 \times 10^{-3} \mathrm{M}$
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.3 \times 10^{-3} \mathrm{M}$

Key Concepts

Acid Dissociation Constant (Ka)

Ka is a quantitative measure of the strength of a weak acid in solution. It is defined by the equilibrium concentrations of the products and reactants in the acid dissociation reaction, typically given by Ka = [H3O+][A–] / [HA]. Evaluating Ka using the measured equilibrium concentrations helps in understanding the degree of ionization of the weak acid.

Base Dissociation Constant (Kb)

Kb similarly quantifies the strength of a weak base by measuring the extent to which it reacts with water to form its conjugate acid and hydroxide ions. The equilibrium constant for a base dissociation reaction is expressed as Kb = [OH–][BH+] / [B]. Calculating Kb gives insight into the basicity and the tendency of the base to accept protons.

Equilibrium Constant Expression

The equilibrium constant expression relates the concentrations of all reactants and products at equilibrium according to the balanced chemical equation. For acid-base reactions, it shows how the formation of ions from parent molecules is balanced. This expression is fundamental for computing values such as Ka and Kb from given equilibrium data.

Weak Acid and Base Equilibria

In weak acid and base systems, the acid or base does not fully dissociate in solution, leading to an important equilibrium between the undissociated form and the ions produced. Understanding these equilibria is crucial for predicting the pH of the solution and for calculating the respective dissociation constants using the equilibrium concentration values.

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