In an ester hydrolysis followed by titration, the NaOH volumes are 10.0 mL (t = 0), 26.0 mL (t = 45 min) and 50.0 mL (infinite time). Find the first-order rate constant in units of $1 0^{- 3} mi n^{- 1}$ (two decimal places).

Numerical answer type, answer in 10^-3 min^-1, tolerance 0.05.

Answer: 11.35 10^-3 min^-1

k = \frac{1}{t} ln \frac{V _{\infty} - V _{0}}{V _{\infty} - V _{t}} = \frac{1}{45} ln \frac{40}{24} = 11.35 \times 1 0^{- 3} mi n^{- 1}