Ionic Equilibrium (Class 11): pH, Buffers and Solubility Product Solved

The relations everything else is built on

The qualifier at 25 °C matters. K_w rises with temperature, so neutral water at 60 °C has a pH below 7 while remaining neutral — a favourite conceptual question.

Buffers and the Henderson equation

A buffer resists pH change because it holds a reservoir of both a weak acid and its conjugate base. Add acid and the base component mops it up; add alkali and the acid component neutralises it. The ratio in the logarithm changes only slightly, so the pH barely moves.

Two consequences worth remembering: a buffer is most effective when [salt] = [acid], because then pH = pK_a and the log term vanishes; and buffer capacity is finite — once one component is consumed, the pH moves sharply.

Solubility product and the common ion effect

For a sparingly soluble salt A_xB_y with solubility s:

So for AgCl, K_sp = s², while for PbCl₂ it is 4s³. Getting the stoichiometric coefficients into the exponents correctly is where most numerical marks are lost.

Adding a common ion pushes the equilibrium back toward the solid, so solubility falls. This is why AgCl is markedly less soluble in dilute HCl than in pure water.

FAQs

Why is pH + pOH equal to 14?

Because Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25 °C. Taking negative logarithms of both sides gives pH + pOH = 14.

Can pH be negative?

Yes. For a very concentrated strong acid, [H⁺] can exceed 1 mol/L, which makes the logarithm positive and the pH negative.

When is a buffer most effective?

When the salt and acid concentrations are equal, so pH equals pKa and the buffer resists change in both directions equally.

What is the common ion effect?

The suppression of ionisation or solubility caused by adding an ion already present in the equilibrium, shifting it backward.