The pH of a drop of water containing ammonia

Presented with the following equations: What would the $\mathrm{pH}$ of a drop of water be, while in equilibrium with $\ce{NH3(g)}$ at a concentration of $\pu{150 ppb}$?


$$\ce{NH3\_{(g)} + H2O\_{(l)} <=> NH3\_{(aq)} + H2O} \qquad K\_\mathrm{H} = \pu{1.76 \cdot 10^{-2} M/atm}$$


$$\ce{NH4+\_{(aq)} <=> NH3\_{(aq)} + H+} \qquad K\_\mathrm{a} = 5.62 \cdot 10^{-10}$$


So my issue is that I don't know how to express the concentration of $\ce{[NH4+]}$.


Calculating the concentration of $\ce{NH3\_{(aq)}}$ is simple: $K\_\mathrm{H} \cdot P\_\ce{NH3}$, while $P\_\ce{NH3} = C \cdot P\_\mathrm{air} = 150 \cdot 10^{-9} \cdot \pu{1atm}$, yields $\ce{NH3\_{(aq)}} = \pu{2.64 \cdot 10^{-9} M}$.


If we are to tally up all contributions of $\ce{H+}$ then we get:
$\ce{[H+]} = K\_\mathrm{w}/\ce{[H+]}$ (from water dissociation) + $K\_\mathrm{a} \ce{[NH4+]}/\ce{[H+]}$.


Then $\ce{[H+]} = (K\_\mathrm{w} + K\_\mathrm{a}\ce{[NH4]})^{0.5}$ and we'd get the $\mathrm{pH}$.


I'm stuck unsure how to express $\ce{[NH4]}$.


