Question

At what $pH$ does a buffer solution exhibit maximum buffer capacity?

Answer 1

let’s first see what a buffer is. A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its $pH$ changes very little when a small amount of strong acid or base is added to it. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications.

Complete answer:
The buffer solution is most effective when it does not let the addition of a small amount of base change the value of $pH$ . This happens when the buffer capacity is maximum.
According to Henderson-Hasselbalch equation,
$\Rightarrow pH = pKa + \log \dfrac{{[salt]}}{{[acid]}}$ for a buffer solution.
Let [salt] $= a$ , [acid] = $b$
Hence, $pH = pKa + \log \left( {\dfrac{a}{b}} \right)$
When, $x$ molar base is added to the solution, the equation becomes,
$\Rightarrow pH = pKa + \log \left( {\dfrac{{a + x}}{{b - x}}} \right)$ (consider this as equation $1$ )
Now, Buffer capacity is defined as,
$\Rightarrow \beta = \dfrac{x}{{\Delta (pH)}}$
Therefore, it can be written as,
$\Rightarrow \beta = \dfrac{{dx}}{{d(pH)}}$
Using value of pH from equation $1$ ,
$\Rightarrow \beta = \dfrac{{dx}}{{d(pKa + \log \dfrac{{(a + x)}}{{(b - x)}})}}$
i.e. $\beta = \dfrac{1}{{\left( {\dfrac{{d(pKa + \log \dfrac{{(a + x)}}{{(b - x)}}}}{{dx}}} \right)}}$
On differentiating, we get,
$\Rightarrow \beta = \dfrac{{2.303}}{{\left( {\dfrac{{(b - x)((b - x) + (a + x))}}{{(a + x){{(b - x)}^2}}}} \right)}}$
On simplifying,
$\Rightarrow \beta = \dfrac{{2.303(a + x)(b - x)}}{{(a + b)}}$
i.e. $\beta = \dfrac{{2.303( - {x^2} + (b - a)x + ab}}{{(a + b)}}$
Now, for maximum buffer capacity, we need to find the maxima of $\beta$ .
For that, $\dfrac{{d\beta }}{{dx}} = 0$
i.e. $\dfrac{{d\beta }}{{dx}} = \dfrac{{2.303( - 2x + (b - a))}}{{(a + b)}} = 0$
Thus, $- 2x + (b - a) = 0$
Which gives us, $x = \dfrac{{(b - a)}}{2}$ (consider it as equation $2$ )
For checking when the buffer is most effective without adding base, put $x = 0$ in equation $2$ ,
We get, $a = ba = b$
This tells us that when no base is added to the solution, buffer is most effective i.e. buffer capacity is maximum when $a = ab = b$ or [salt] = [acid]
Putting [salt] = [acid] in the Henderson-Hasselbalch equation,
$\Rightarrow pH = pKa + \log 1$
Hence, $pH = pKa$ .

Note:
Buffer solutions resist $pH$ change because of an equilibrium between the weak acid $HA$ and its conjugate base ${A^ - }$ . When some strong acid is added to an equilibrium mixture of the weak acid and its conjugate base, hydrogen ions are added, and the equilibrium is shifted to the left, in accordance with Le Chatelier's principle. Because of this, the hydrogen ion concentration increases by less than the amount expected for the quantity of strong acid added.