Rain water contains sulphuric acid at a concentration of $2.5 \times 1 0^{- 5} mol L^{- 1}$ and no other acid or base. Assume $H X_{2} SO X_{4}$ is fully dissociated in both steps and neglect the ionisation of water. Taking $lo g 5 = 0.699$ , calculate the pH of the rain water. Give the answer to two decimal places.

Numerical answer type, tolerance 0.01.

Answer: 4.3

Full dissociation of

H X_{2} SO X_{4}

gives

[H X^{+}] = 2 \times 2.5 \times 1 0^{- 5} = 5.0 \times 1 0^{- 5} M

. pH =

- lo g (5.0 \times 1 0^{- 5}) = 5 - 0.699 = 4.301

. The pH is 4.30, below 5.6, so this is acid rain.