A rain sample of pH 4.00 contains nitric acid at $4.0 \times 1 0^{- 5} mol L^{- 1}$ and sulphuric acid, with no other acid or base. Assume complete dissociation of both acids (both protons of $H X_{2} SO X_{4}$ ) and neglect water ionisation. The concentration of $H X_{2} SO X_{4}$ is

Answer: A

[H X^{+}] = 1 0^{- 4.00} = 1.0 \times 1 0^{- 4}

M. From

HNO X_{3}

the contribution is

4.0 \times 1 0^{- 5}

M, so

H X_{2} SO X_{4}

must give

6.0 \times 1 0^{- 5}

M of

H X^{+}

. Each

H X_{2} SO X_{4}

gives 2

H X^{+}

, hence

[H X_{2} SO X_{4}] = 3.0 \times 1 0^{- 5}

M. Using 6.0e-5 directly forgets the factor of 2.