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Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$ .

$\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})$ reacts with $\mathrm{OH}^{-}(\mathrm{aq})$ .

\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})

Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction.

Table 2.1

The experiment is repeated using a large excess of $\mathrm{OH}^{-}(\mathrm{aq})$ .

Under these conditions, the rate equation is:

\begin{array}{ll} & \text { rate }=k_{1}\left[\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})\right] \\ k_{1}=8.25 \times 10^{-5} \mathrm{~s}^{-1} & \end{array}

Calculate the value of the half-life, $t_{\frac{1}{2}}$ , of the reaction.

\begin{aligned} & t_{\frac{1}{2}}= \\ & \text { s [1] } \end{aligned}

$t_{1 / 2}=0.693 / 8.25 \times 10^{-5}=\mathbf{8 4 0 0}(\mathrm{s})$ OR $t_{1 / 2}=\ln 2 / 8.25 \times 10^{-5}=\mathbf{8 4 0 1 . 8}(\mathrm{s})$

CAIE A-Level Chemistry A2 26.1.3 Show Understanding That the Half Life Topic Practice