= Solution
A <humidity> assumption is necessary: warming does not by itself specify how much <water> is available. At fixed <relative humidity>, the <Clausius-Clapeyron equation> gives the first-order <vapor pressure> response
$$
\Delta\ln e_s\simeq\frac{45050}{8.314(285)^2}(4)=0.2668.
$$
If <water> abundance is approximated as proportional to $e_s$, the usual first-order estimate is \b[a $26.7\%$ increase in <hydroxyl radical> production]. Integrating the constant-latent-heat expression instead of linearizing gives $e_s(289)/e_s(285)=\exp[(45050/8.314)(1/285-1/289)]=1.3010$.
There is a small but explicit <concentration> correction when the square brackets are literal <molecules> per volume. Since $[\mathrm H_2\mathrm O]=\phi e_s/(k_BT)$ and $P_{\rm OH}=2k[\mathrm O({}^1D)][\mathrm H_2\mathrm O]$,
$$
\boxed{\frac{\Delta P_{\rm OH}}{P_{\rm OH}}
\simeq\left(\frac{\Delta H_{\rm vap}}{RT^2}-\frac1T\right)\Delta T
=0.2528\quad\text{at fixed relative humidity}.}
$$
The corresponding finite-change <concentration> result is $(285/289)(1.3010)-1=0.2830$. The $25.3\%$ linear <concentration> estimate and the $26.7\%$ <vapor pressure> approximation express different stated approximations; neither follows without a <humidity> condition.
Extra <water> increases the fraction of <oxygen singlet D atoms> reacting to form <hydroxyl radicals> rather than being collisionally quenched and recycled into <ozone>. In clean, low-<nitrogen oxides> air this tends to increase <ozone> loss. In <nitrogen oxides>-rich air, increased <hydroxyl radical> can accelerate precursor <oxidation> and <ozone> production, while <radical termination> also changes. \b[Greater <hydroxyl radical> production therefore does not imply a universally positive or negative <ozone> response.]
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