The first and second dissociation constants of an acid $$H_2A$$ are $$1.0 \times 10^{-5}$$ and $$5.0 \times 10^{-10}$$ respectively. The overall dissociation constant of the acid will be

The diprotic acid $$H_2A$$ ionises in two successive steps:

Step-1 $$H_2A \rightleftharpoons H^+ + HA^- \qquad K_{a1} = 1.0 \times 10^{-5}$$

Step-2 $$HA^- \rightleftharpoons H^+ + A^{2-} \qquad K_{a2} = 5.0 \times 10^{-10}$$

The overall (net) dissociation process is

$$H_2A \rightleftharpoons 2H^+ + A^{2-}$$

For consecutive equilibria, the overall equilibrium constant is the product of the individual step constants: $$K_\text{overall} = K_{a1}\,K_{a2}$$.

Substitute the given values:
$$K_\text{overall} = \left(1.0 \times 10^{-5}\right)\left(5.0 \times 10^{-10}\right) = 5.0 \times 10^{-15}$$

Hence the overall dissociation constant of the acid is $$5.0 \times 10^{-15}$$.

Option C which is: $$5.0 \times 10^{-15}$$