Question

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Answer 1

The concentration of [OH-]  -hydroxide ion : 1.2 x 10⁺⁴ M

Further explanation

H₂O ionization reaction

2 H₂O ⇌ OH− + H₃O⁺

The water equilibrium constant (Kw) is the product of concentration

the ions:

Kw = [H₃O +] [OH-]

Kw value at 25° C = 10⁻¹⁴

It is known  [H₃O +]=8.4 x 10⁻¹¹ M

then the concentration of [OH-]  -hydroxide ion :

$\tt 10^{-14}=8.4\times 10^{-11}\times [OH^-]$

$\tt [OH^-]=\dfrac{10^{-14}}{8.4\times 10^{-11}}=1.19\times 10^{-4}\approx 1.2\times 10^{-4}$