99.9224 % of the acid is not ionized.
____HA + H₂O ⇌ A⁻ + H₃O⁺
I: ___c ________0 ____0
C: _-αc _______+αc __+αc
E: c(1-α) _______αc ___αc
pH = 4.110
[H₃O⁺] = αc = 10^(-4.110) mol/L = 7.76 × 10⁻⁵ mol/L
α = 7.76 × 10⁻⁵
1 – α = 1 - 7.76 × 10⁻⁵ = 0.999 224 = 99.9224 %