Question 3.9.5: Analyzing a Titration Curve Find the value of x at which the...
Analyzing a Titration Curve
Find the value of x at which the rate of change of pH is the smallest. Identify the corresponding point on the titration curve in Figure 3.100.
The pH is given by the function p(x)=c+\ln \frac{x}{1-x}. The rate of change of pH is then given by the derivative p′(x). To make this computation easier, we write p(x)=c+\ln x-\ln (1-x) The derivative is then
p^{\prime}(x)=\frac{1}{x}-\frac{1}{1-x}(-1)=\frac{1}{x(1-x)}=\frac{1}{x-x^2} .The problem then is to minimize the function g(x)=\frac{1}{x-x^2}=\left(x-x^2\right)^{-1}, with 0 < x < 1. Critical points come from the derivative
g^{\prime}(x)=-\left(x-x^2\right)^{-2}(1-2 x)=\frac{2 x-1}{\left(x-x^2\right)^2} .Notice that g^{\prime}(x) \text { does not exist if } x-x^2=0 \text {, } which occurs when x = 0 or x = 1, neither of which is in the domain 0 < x < 1. Further, g^{\prime}(x)=0 \text { if } x=\frac{1}{2}, which is in the domain. You should check that which is in
the domain. You should check that g^{\prime}(x)<0 \text { if } 0<x<\frac{1}{2} \text { and } g^{\prime}(x)>0 \text { if } \frac{1}{2}<x<1 \text {, } which proves that the minimum of g(x) \text { occurs at } x=\frac{1}{2}. Although the horizontal axis in Figure 3.100 is not labeled, observe that we can still locate this point on the graph. We found the solution of g^{\prime}(x)=0 \text {. Since } g(x)=p^{\prime}(x) \text {, we have } p^{\prime \prime}(x)<0 \text { for } 0<x<\frac{1}{2} and p^{\prime \prime}(x)>0 \text { for } \frac{1}{2}<x<1, so that the point of minimum change is an inflection point of the original graph.
