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TEA [102]
4 years ago
8

A weak 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.)
Chemistry
1 answer:
OverLord2011 [107]4 years ago
7 0

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 %

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<span>
Now,
Ka=[H=][A-]/[HA] </span>

<span>So,
(2.1 x 10^-3)^2/0.0136 - (2.1 x 10^-3) </span>
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