Weak bases create basic solutions by reacting incompletely with water, establishing an equilibrium that produces OH−. AP Chemistry problems focus on using $K_b$ or $pK_b$ with initial concentration to determine pH.

Weak bases and their equilibrium with water

A weak base reacts with water but only partially forms products, so appreciable amounts of both reactants and products are present at equilibrium.

A common generic reaction is:

• $\mathrm{B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)}$

• $\mathrm{B}$ is the base; $\mathrm{BH^+}$ is its conjugate acid

• Water is a reactant but is treated as a constant in the equilibrium expression

The base ionisation constant, $K_b$

The equilibrium position is described by $K_b$, which is specific to the base and temperature.

Larger $K_b$ means the base produces more $\mathrm{OH^-}$ (lies further to products) and therefore gives a higher pH for the same initial concentration.

Using $pK_b$

Chemists often use $pK_b$ to simplify comparisons and calculations involving logarithms.

Finding pH of a weak base (problem setup)

To find pH, you determine the equilibrium $[\mathrm{OH^-}]$ produced by the weak base, then convert to pH.

Core method (ICE + equilibrium)

• Write the weak-base equilibrium: $\mathrm{B + H_2O \rightleftharpoons BH^+ + OH^-}$

• Set up an ICE table (Initial, Change, Equilibrium) in $\mathrm{mol,L^{-1}}$

• Initial: $[\mathrm{B}]_0$ known; typically $[\mathrm{BH^+}]_0 \approx 0$, $[\mathrm{OH^-}]_0 \approx 0$

• Change: let $x$ be the amount that reacts; then $[\mathrm{BH^+}] = x$, $[\mathrm{OH^-}] = x$, $[\mathrm{B}] = [\mathrm{B}]_0 - x$

• Substitute equilibrium concentrations into the $K_b$ expression and solve for $x = [\mathrm{OH^-}]$

• Decide whether the small-$x$ approximation is valid:

  • If $x \ll [\mathrm{B}]_0$, then $[\mathrm{B}]_0 - x \approx [\mathrm{B}]_0$

  • If not valid, solve the resulting quadratic (or use a calculator)

Converting $[\mathrm{OH^-}]$ to pH

Once $[\mathrm{OH^-}]$ is known, compute pOH and then pH (at $25^\circ\mathrm{C}$).

pH–pOH scale chart showing the inverse relationship between $[\mathrm{H_3O^+}]$ and $[\mathrm{OH^-}]$ across many orders of magnitude. It visually reinforces that each 1-unit change in pH corresponds to a tenfold change in concentration, and that at $25^\circ\mathrm{C}$, pH and pOH are linked by $\mathrm{pH}+\mathrm{pOH}=14.00$. Source

What “pH depends on initial base concentration and $pK_b$” means

• Higher initial $[\mathrm{B}]_0$ generally increases equilibrium $[\mathrm{OH^-}]$, raising pH

• Smaller $pK_b$ (larger $K_b$) shifts equilibrium toward $\mathrm{OH^-}$, raising pH

• Because ionisation is partial, $[\mathrm{OH^-}]$ is typically much smaller than $[\mathrm{B}]_0$