The hydrolysis of an ester in dilute acid is first order in the ester. The titre (mL of NaOH) is $V_{0} = 20$ at $t = 0$ , $V_{t} = 44$ at $t = 30 min$ and $V_{\infty} = 80$ at completion. Use $k = \frac{1}{t} ln \frac{V _{\infty} - V _{0}}{V _{\infty} - V _{t}}$ with $ln 5 = 1.609$ and $ln 3 = 1.099$ . Find $k$ in units of $1 0^{- 2} mi n^{- 1}$ (two decimals).

Numerical answer type, answer in 10^-2 min^-1, tolerance 0.02.

Answer: 1.7 10^-2 min^-1

The ester left is proportional to

V_{\infty} - V_{t}

. Here

V_{\infty} - V_{0} = 60

and

V_{\infty} - V_{t} = 36

, a ratio of

60/36 = 5/3

. So

k = (ln 5 - ln 3) /30 = (1.609 - 1.099) /30 = 0.0170 mi n^{- 1}

, i.e.

1.70 \times 1 0^{- 2}