Answer
$$pH =13.842$$
Work Step by Step
$$3.85 \% \space KOH : \frac{3.85 \space g \space KOH}{100 \space g \space Solution}$$ Molar mass :
$K: 39.10g $
$O: 16.00g $
$H: 1.008g$
39.10g + 16.00g + 1.008g = 56.11g: $\frac{56.11 \space g \space KOH}{1 \space mole \space KOH}$
$$\frac{1.01 \space g \space Solution}{1 \space mL \space Solution} \times \frac{3.85 \space g \space KOH}{100 \space g \space Solution} \times \frac{1 \space mole \space KOH}{56.11 \space g \space KOH} \times \frac{1000 \space mL}{1 \space L} = 0.693 \space M \space KOH$$
$$[OH^-] = [KOH] = 0.693 \space M$$ $$[H_3O^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.693} = 1.44 \times 10^{-14}$$ $$pH = -log(1.44 \times 10^{-14}) = 13.842$$
