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We've a solution of $\ce{HCOOH}$ and Water $\ce{H_2O}$,
I want to prove that the extent of this reaction is equivalent to : $$\xi=\frac{K_a}{K_a+10^{-\mathrm{p}H}}$$ Could you help try to figure that out ?
the reaction equation is : $$\ce{HCOOH + H_2O <=> HCOO^- + H_3O^+}$$
Prove this through K value inspection. Note that ${1.0~x~10^{-pH}} = [\ce{H3O^+}]$.
Then note that ionization extent or the extent of reaction is defined as the equilibrium concentration of the reaction product over initial molarity of the solute.
Ionization extent for this reaction:
$$\ce{HA + H_2O <=> H3O^+ + A^- }$$
Is defined as (where $M_i$ is the initial molarity of the solute, before any reaction):
$$\frac{[\ce{A^-}]}{M_{i}}$$
This, through the law of conservation of mass, is equal to the following expression (which only uses equilibrium quantities):
$$\frac{[\ce{A^-}]}{[\ce{A^-}]+[\ce{HA}]}$$
So now we have two working definitions of reaction extent or ionization extent. Let's take your formula, substitute in the K expressions and volia!
However, it is crucial to note that this formula gives you ionization extent of a weak acid at infinite dilution or maximum ionization extent. I'm not sure if this is the case with your question; can you provide more information?
