For the reaction KOH(aq) --> K+(aq) + OH-(aq) , K = .316. If the concentration of KOH is 4.75M at equilibrium, what is the concentration...?

KOH(aq) ⇌ K⁺(aq) + OH⁻(aq) ...... K = 0.316

At equilibrium :
[KOH] = 4.75 M
[K⁺] = [OH⁻] = y M

K = [K⁺][OH⁻] / [KOH]
0.316 = y² / 4.75
y² = 0.316 × 4.75
y = 1.23 M

[OH⁻] at equilibrium = 1.23 M