Buffer Sensitivity

Use the Henderson-Hasselbalch equation: $\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right).$

Initially, $[\text{A}^-] = [\text{HA}] = 0.2$ , and the logarithm is equal to 0. One way of finding the answer is simply by considering the effect of a small amount of acid, say 0.0001 mol/L. This would lower the concentration of the base and increase the concentration of the acid accordingly, and $\text{pH}_\text{new} = \text{p}K_a + \log\left(\frac{0.1999}{0.2001}\right) \approx \text{p}K_a - 0.0004343,$ so that the rate of change of pH is $\frac{0.0004343}{0.0001\:\text{mol/L}} = \boxed{4.3434}.$

A more sophisticated approach tells us to consider the function $\text{pH} = \text{p}K_a + \log\left(\frac{b-x}{a+x}\right),$ and consider its derivative. This derivative is $\frac{d\text{pH}}{dx} = -\frac 1{\ln 10}\left(\frac 1{b-x}+\frac 1{a+x}\right),$ so at $x = 0$ it becomes $\frac{d\text{pH}}{dx} = -\frac 1{\ln 10}\left(\frac 1 b +\frac 1 a \right).$ Substituting $a = b = 0.2$ we find $\frac{d\text{pH}}{dx} = -\frac {10}{\ln 10} \approx 4.343.$