How can 100ml of sodium hydroxide solution with a ph of 13.00 be converted to a sodium hydroxide solution with a ph of 12.00?

Chemical reaction: NaOH(aq) → Na⁺(aq) + OH⁻(aq).
V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0.1 L.
pOH = -log[OH⁻].
pH + pOH = 14.
pOH₁ = 14 - 13 = 1.
pOH₂ = 14 - 12 = 2.
[OH⁻]₁ = 10∧(-pOH₁) = 10⁻¹ M = 0.1 M.
[OH⁻]₂ = 10(-pOH₂) = 10⁻² M = 0.01 M.
[OH⁻]₁ · V₁(NaOH) = [OH⁻]₂ · V₂(NaOH).
0.1 M · 0.1 L = 0.01 M · V₂(NaOH).
V₂(NaOH) = 1 L, we can add 900 mL of water to convert solution of sodium hydroxide from pH=12 to pH=13, or we can add some strong acid.

C) electrons encircling a nucleus with neutrons and protons