99.9224 % of the acid is not ionized.
____HA + H₂O ⇌ A⁻ + H₃O⁺
I: ___<em>c</em> ________0 ____0
C: _-α<em>c</em> _______+α<em>c</em> __+α<em>c
</em>
E: <em>c</em>(1-α) _______α<em>c</em> ___α<em>c
</em>
pH = 4.110
[H₃O⁺] = α<em>c</em> = 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 %
Answer:
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