What is total concentration of buffer used in derivation of buffer capacity?

[ChemBuddy — The buffer capacity](http://www.chembuddy.com/?left=pH-calculation&right=pH-buffer-capacity) provides the following:


$$\ce{HA <=> H+ + A-}$$



> 
> Total concentration of the buffer $c\_\mathrm{buf}$ is given by
> 
> 
> $$c\_\mathrm{buf} = [\ce{HA}] + [\ce{A-}]\tag{19.3}$$
> 
> 
> From dissociation constant definition we have
> 
> 
> $$[\ce{HA}] = \frac{[\ce{H+}][\ce{A-}]}{K\_\mathrm a}\tag{19.4}$$
> 
> 
> 


But shouldn't


$$c\_\mathrm{buf} = [\ce{HA}] + [\ce{A-}] + [\ce{H+}]?\tag{1}$$


The total concentration at equilibrium is


$$c(1 - \alpha) + 2c\alpha = c(1 + \alpha)\neq c,\tag{2}$$


hence the initial concentration of buffer is different from that at equilibrium.




Initially in the solution the following equilibria exist


$$\ce{HA <=> H+ +A-}$$  

and  

$$\ce{H2O <=> H+ +OH-}$$


Let initial concentration of $\ce{HA}$ be $a\_1$ and that of $\ce{A-}$ be $a\_2$. Now to a litre of this buffer solution, let's say you have added $\mathrm dn$ moles of strong acid to it. The acid dissociation constant of $\ce{HA}$ be $K\_a$ and the ionic product of water, $K\_w$. By adding acid, you have disturbed the equilibrium, and from Le Chatelier's principle, both the reactions go backwards.


$$\ce{HA <<=> H+ + A- }\tag{1} \label{eqn:1}$$  

$K\_{eq}=K\_a$
$$\ce{H2O <<=> H+ +OH-} \tag{2} \label{eqn:2}$$  

$K\_{eq}=K\_w$


Let the new equilibrium concentrations of $\ce{HA},~\ce{A-},~\ce{H+},~\ce{OH-}$ be $a\_1+\mathrm dx,~a\_2-\mathrm dx,~\frac{K\_a\cdot a\_1}{a\_2}+\mathrm dn-\mathrm dx-\mathrm dy,~\frac{K\_w\cdot a\_2}{K\_a.a\_1} - \mathrm dy$ respectively.


At equilibrium, for $\ref{eqn:1}$,


$$K\_a = \frac{(\frac{K\_a\cdot a\_1}{a\_2} +\mathrm dn-\mathrm dx-\mathrm dy)\cdot (a\_2 - \mathrm dx)}{a\_1+\mathrm dx} \tag{3} \label{eqn:3}$$


and for $\ref{eqn:2}$,


$$K\_w = (\frac{K\_a\cdot a\_1}{a\_2}+\mathrm dn-\mathrm dx-\mathrm dy)\cdot (\frac{K\_w\cdot a\_2}{K\_a.a\_1}-\mathrm dy) \tag{4} \label{eqn:4}$$


Dividing $\ref{eqn:3}$ by $\ref{eqn:4}$, we get


$$(K\_a)\cdot (\frac{K\_w\cdot a\_2}{K\_a\cdot a\_1}-\mathrm dy)=(K\_w)\cdot (\frac{a\_2-\mathrm dx}{a\_1+\mathrm dx})\\
K\_w\cdot a\_2+\frac{K\_w\cdot a\_2}{a\_1}\cdot \mathrm dx-K\_aa\_1\mathrm dy-K\_a\mathrm dx\mathrm dy=K\_wa\_2-K\_w\mathrm dx$$


Neglecting the term $K\_a\mathrm dx\mathrm dy$, we get


$$dx= \frac{K\_a\cdot {a\_1}^2}{K\_w(a\_1+a\_2)}\cdot \mathrm dy$$


and also from $\ref{eqn:3}$


$$K\_a\cdot (a\_1+\mathrm dx)=(\frac{K\_a\cdot a\_1}{a\_2}+\mathrm dn-\mathrm dx-\mathrm dy)\cdot (a\_2-\mathrm dx)$$


Which on simplifying and neglecting terms of the type $\mathrm da\cdot \mathrm db$ will give,


$$\mathrm dn= (\frac{{K\_a}^2\cdot {a\_1}^2}{K\_w\cdot {a\_2}^2}+\frac{K\_a\cdot {a\_1}^2}{K\_w\cdot (a\_1+a\_2)}+1)\cdot \mathrm dy \tag{5} \label{eqn:5}$$


By definition,


$\beta = \frac{\mathrm dn}{\mathrm{dpH}}$ and $\mathrm{pH}=-\log\_{10} {[\ce{H+}]}$.


On differentiating this, we get


$$\mathrm{dpH} = \frac{-1}{[\ce{H+}]}\cdot \frac{1}{\log\_e10}\cdot \mathrm d[\ce{H+}]$$


Substituting this in the definition of buffer capacity,


$$\beta = \frac{-\log\_e10\cdot [\ce{H+}]\cdot \mathrm dn}{\mathrm d[\ce{H+}]} \tag{6} \label{eqn:6}$$


Here,


$$\mathrm d[\ce{H+}] = \mathrm dx+\mathrm dy-\mathrm dn = -\frac{{K\_a}^2\cdot {a\_1}^2}{{a\_2^2}\cdot K\_w}\cdot \mathrm dy$$


Substituting this expression of $\mathrm d[\ce{H+}]$ and $\mathrm dn$ from $\ref{eqn:5}$ in $\ref{eqn:6}$, we get,


$$\beta = \log\_e10\cdot [\ce{H+}]\cdot \frac{(\frac{{K\_a}^2\cdot {a\_1}^2}{{a\_2}^2\cdot K\_w}+\frac{K\_a\cdot {a\_1^2}}{K\_w\cdot (a\_1 + a\_2)} +1)}{\frac{{K\_a}^2\cdot {a\_1}^2}{{a\_2}^2\cdot K\_w}}\\
\beta = \log\_e(10)[\ce{H+}](1 + \frac{K\_w}{{[\ce{H+}]}^2} + \frac{{a\_2}^2}{(a\_1+a\_2)\cdot K\_a})\\
\beta = \log\_e10( [\ce{H+}] + \frac{K\_w}{[\ce{H+}]} + \frac{{a\_2}^2\cdot [\ce{H+}]}{(a\_1+a\_2)\cdot K\_a})$$


Substituting $[\ce{H+}]= \frac{K\_a\cdot a\_1}{a\_2}$ in the third term, we get


$$\beta = \log\_e10( [\ce{H+}] + \frac{K\_w}{[\ce{H+}]} + \frac{a\_1\cdot a\_2}{(a\_1+a\_2)})$$


Multiplying numerator and denominator of the third term with $K\_a$,


$$\beta = \log\_e10( [\ce{H+}] + \frac{K\_w}{[\ce{H+}]} + \frac{K\_a\cdot a\_1\cdot a\_2}{K\_a\cdot (a\_1+a\_2)})$$


Taking $a\_2$ common from the denominator and cancelling the $a\_2$ in the numerator and substitute $\frac{k\_a\cdot a\_1}{a\_2}=[\ce{H+}]$ and multiplying numerator and denominator with $[\ce{H+}]+K\_a$, we get


$$\beta = \log\_e10([\ce{H+}] + \frac{K\_w}{[\ce{H+}]} + \frac{(\frac{K\_a\cdot a\_1}{a\_2})\cdot (K\_aa\_1+K\_aa\_2)}{{([\ce{H+}]+{K\_a})}^2}$$


Taking $K\_a$ out from the numerator and substituting $[\ce{H+}]=\frac{K\_a\cdot a\_1}{a\_2}$, we get


$$\beta= \log\_e10([\ce{H+}]+[\ce{OH-}]+ \frac{[\ce{H+}]K\_a(a\_1+a\_2)}{([\ce{H+}]+K\_a)^2})$$


So, in your expression, of course, $C\_{\text{buf}}=a\_1+a\_2=[\ce{HA}]+[\ce{A-}]$ and not $[\ce{HA}]+[\ce{A-}]+[\ce{H+}]$


$$\beta= \log\_e10([\ce{H+}]+[\ce{OH-}]+ \frac{[\ce{H+}]K\_aC\_{\text{buf}}}{([\ce{H+}]+K\_a)^2})$$


You will get the same result if a small amount of base is added.


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