$10 mL$ of $HCl$ of pH $6$ is diluted to $100 mL$ at 298 K. Taking water ionisation into account, the pH of the dilute solution is

Answer: D

Acid after dilution:

C = 1 0^{- 7} M

. Then

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

gives

[H X^{+}]^{2} - 1 0^{- 7} [H X^{+}] - 1 0^{- 14} = 0

, so

[H X^{+}] = 1.62 \times 1 0^{- 7} M

and

pH = 6.79

. Simple addition (

2 \times 1 0^{- 7}

) would give

6.70

, which is wrong.