請教化學大神有關酸鹼滴定的問題?

1.
It is known that Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴

In pure water : [H⁺] = [OH⁻]
Then [H⁺]² = 1 × 10⁻¹⁴
[H⁺] = √(1 × 10⁻¹⁴) = 1 × 10⁻⁷ M
pH = -log(1 × 10⁻⁷) = 7


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2.
The volume of 0.008 mol of NaOH is negligible.

[OH⁻] = 0.008 M >> (1 × 10⁻⁷ M)
Thus, the self-dissociation of water can be neglected.

pOH = -log(0.008) = 2.1
pH = 14 - 2.1 = 11.9


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3.
Denote the weak acid MOPS as HA, and its conjugate base as A⁻.

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

pH = pKa - log([HA]/[A⁻]
pH = 7.2 - log(61/39)
pH = 7.0