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How is this formula derived? What does it mean; what is the use of this formula?

I know that at infinite dilution, ionization extent should be very high because of Le Chatlier's principle.

But I'm not seeing how this formula is derived and its significance in actual chemistry. I tried rewriting the formula as:

$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} \cdot 100\%}$

But this isn't elucidating where the formula came from or its significance.

I could also rewrite the formula as this for the hypothetical acid, $\ce{HA}$:

$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} = \frac{\frac{[H_3O^+][A^-]}{[HA]}}{1.0 \cdot 10^{-7}+\frac{[H_3O^+][A^-]}{[HA]}}}$

I think the trouble I'm having is defining maximum ionization extent. For example, I know that in general ionization extent is defined as (where x is the concentration of the derived species - i.e. $\ce{[A^{-}]}$ for the hypothetical acid $\ce{HA}$ and $\ce{M_{i}}$ the initial molarity of the hypothetical acid.

$\mathrm{\frac{x}{M_{i}}}=\text{ionization extent}$

I have a feeling that the $1.0*10^{-7}$ term refers to the concentration of one of the products of the auto-ionization of pure water at 25 degrees Celsius. This makes sense because at infinite dilution the concentration of solute should be negligible (=0).

Upon further thought ...

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  • $\begingroup$ I've been looking on the web for information and came across this: en.wikipedia.org/wiki/Law_of_dilution $\endgroup$ – LDC3 May 26 '14 at 2:45
  • $\begingroup$ Thanks, but I don't see the relationship between that article and my question. Interesting reading though! $\endgroup$ – Dissenter May 26 '14 at 2:50

For every reaction with an incomplete dissociation process in aqueous solution, you have to consider water's autoprotolysis. \begin{aligned}\ce{ (1)&&HA + H2O &<=> H3+O + A^{-}\\ (2)&&2H2O &<=> H3+O + {}^{-}OH \\\hline (3)&&HA + 3H2O &<=> 2H3+O + A^{-} + {}^{-}OH }\end{aligned}

The degree of dissociation is usually defined as \begin{aligned} &&\alpha &= \frac{c(\ce{A^{-}})}{c_0}&\text{with}&&c_0&=c(\ce{A^{-}})+c(\ce{HA})\\ \therefore&&\alpha &= \frac{c(\ce{A^{-}})}{c(\ce{A^{-}})+c(\ce{HA})}&\text{with}&& c(\ce{A^{-}})&=K_a\frac{c(\ce{HA})}{c(\ce{H3+O})}\\ (4)\therefore&&\alpha &= \frac{K_a}{K_a + c(\ce{H3+O})} \end{aligned}

At infinite solution, the concentration of hydronium ions will be governed by the autoprotolysis $(2)$ and can therefore be rewritten as \begin{aligned} &&K_w&=c(\ce{H3+O})c(\ce{{}^{-}OH})\\ (5)\therefore&&c(\ce{H3+O})&=\sqrt{K_w} \end{aligned}

At $25~^\circ{}C$ you will find your given equation: \begin{aligned} \alpha[\%] &= \frac{K_a}{K_a + 1\cdot10^{-7}}\cdot100\% \end{aligned}

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  • $\begingroup$ This is a masterful derivation. Thanks Martin. I just showed my friend this derivation and it made more sense than a book's derivation. $\endgroup$ – Dissenter Jul 7 '14 at 14:33

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