Calculate the pH at 298 K of $1.0 \times 1 0^{- 8} M$ $HCl$ , taking the ionisation of water into account ( $K_{w} = 1.0 \times 1 0^{- 14}$ ). Give the answer to two decimals.

Numerical answer type, tolerance 0.01.

Answer: 6.98

Total

[H X^{+}] = C + [OH X^{-}]

with

[OH X^{-}] = K_{w} / [H X^{+}]

, giving

[H X^{+}]^{2} - 1 0^{- 8} [H X^{+}] - 1 0^{- 14} = 0

. Hence

[H X^{+}] = 1.05 \times 1 0^{- 7} M

and

pH = 6.978 \approx 6.98