Question

Question #33da5

Answer 1

An aqueous solution for which `[HO^-]=4.2xx10^-8*mol*L^-1` is slightly acidic.

Explanation:

For neutrality under standard conditions `[H_3O^+]=[HO^-]=10^-7*mol*L^-1`.
Alternatively since, `pH+pOH =14`, where `pH=-log_10[H_3O^+]`, if `pH<7`, then the solution is acidic.

Here `pOH=7.38`, thus `pH=6.62`.