What is [OH-] in a 0.0035 M HCl solution?

We are given with concentration of H⁺, Starting by convertiong this concentration into pH as follow,

pH = - log [H⁺]

pH = - log [0.0035] ∴ conc. of HCl

pH = 2.45

Also,

pH + pOH = 14

pOH = 14 - 2.45

pOH = 11.55

Changing pOH into [OH⁻]

[OH⁻] = 10⁻¹¹°⁵⁵ ∴ Taking Antilog

[OH⁻] = 2.81 * 10⁻¹²