Aweak acid (ha) has a pka of 4.001. if a solution of this acid has a ph of 4.110, what percentage of the acid is not ionized? (assume all h in solution came from the ionization of ha.)

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 %

i dont know

explanation: