In aqueous solutions H₂SO₄ ionises as :
$\begin{array}{l}{\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_2{\mathrm{SO}}_4^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+};{\mathrm{Ka}}_1\\ {\mathrm{H}}_2{\mathrm{SO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{SO}}_4^{2-}+{\mathrm{H}}_3{\mathrm{O}}^{+};{\mathrm{Ka}}_2\end{array}$
The relation between Ka₁ and Ka₂ is

The given equation shows that the sulfuric acid (H₂SO₄) ionizes in aqueous solution to form$\mathrm{HSO}{ }_4^{-}$ion and H₃O⁺  ion. This is the first ionization reaction, and its equilibrium constant is denoted by Ka₁. The $\mathrm{HSO}{ }_4^{-}$ ion can also undergo further ionization to form ${\mathrm{SO}}_4^{-2}$ ion and H₃O⁺ ion. This is the second ionization reaction, and its equilibrium constant is denoted by Ka₂.

The chemical equations for the two ionization reactions are:

H₂SO₄+ H₂O ⇌ $\mathrm{HSO}{ }_4^{-}$ +H₃O⁺ (Ka₁)

$\mathrm{HSO}{ }_4^{-}$ +H₂O ⇌ ${\mathrm{SO}}_4^{-2}$ + H₃O⁺ (Ka₂)

The value of Ka₁ is greater than Ka₂ for sulfuric acid because the first ionization reaction (Ka₁) is more favorable than the second ionization reaction (Ka₂). This is because the first ionization reaction involves the breaking of a strong O-H bond, while the second ionization reaction involves the breaking of a weaker O-S bond.

As a result, the first ionization reaction is more exothermic and releases more energy, making it more favorable than the second ionization reaction. This means that sulfuric acid will dissociate more readily in water to produce $\mathrm{HSO}{ }_4^{-}$ and H₃O⁺ ions than it will dissociate further to produce ${\mathrm{SO}}_4^{-2}$ and H₃O⁺ ions. Hence, Ka₁ is greater than Ka₂ for sulfuric acid.