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