Saponification

We denote the concentrations as $x = [\text{Ester}] \, , \quad y = [\text{OH}^-]$ with the initial concentrations $x_0 = y_0$ . Due to particle number conservation, $x = y$ holds, so that the reaction equation simplifies: $r = -\frac{d x}{dt} = k_2 x y = k_2 x^2$ This differential equation can be solved by separating the variables $\begin{aligned} & & \frac{dx}{x^2} &= -k_2 dt \\ \Rightarrow & & \int_{x_0}^x \frac{ dx}{x^2} &= - k_2 \int_0^t dt \\ \Rightarrow & & \frac{1}{x_0} - \frac{1}{x} &= - k_2 t \\ \Rightarrow & & \frac{1}{x} &= \frac{1}{x_0} + k_2 t \\ \Rightarrow & & x(t) &= \frac{x_0}{1 + \kappa t} \end{aligned}$ where $\kappa = x_0 k_2$ . For $t = 2 \,\text{min}$ the concentration $x(t) = 0.8 \cdot x_0$ is observed. Therefore, $\kappa = \left(\frac{x_0}{x} - 1\right) \frac{1}{t} = \left(\frac{1}{0.8} - 1\right) \cdot \frac{1}{2}\, \frac{1}{\text{min}} = \frac{1}{8} \,\frac{1}{\text{min}}$ For $x = 0.2\cdot x_0$ the time results to $t = \left(\frac{x_0}{x} - 1\right) \frac{1}{\kappa} = \left(\frac{1}{0.2} - 1\right) \cdot 8 \,\text{min} = 32 \,\text{min}$