What's the volume ratio of two KOH solutions with conentrations of 10^-2M and 10^-4M to obtain a solution with a pH=11? I care more about how it's solved

We know, pH + pOH = 14

pOH = -log[OH⁻]

11 = -log[OH⁻]

log[OH⁻] = -11

[OH⁻] = 10⁻¹¹ M

For 10⁻⁴ M of solution, volume required is :

10⁻⁴ × 1 = 10⁻¹¹ × V₁

V₁ = 10⁷ units.

Similarly for 10⁻² M :

10⁻²  × 1 = 10⁻¹¹ × V₂

V₂ = 10⁹ units.

Therefore, the ratio of volume is 10⁷/10⁹  = 10⁻² units.