[Ellie]

Presented with the following equations, what would the pH of a drop of water be, while in equilibrium with NH_3(g) at a concentration of 150ppb?

NH_3(g)+H₂O_(l)⟷NH_3(aq)+H₂O K_H=1.76⋅10⁻²M/atm

NH₄⁺_(aq)⟷NH_3(aq)+H⁺ Ka=5.62⋅10⁻¹⁰

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

Calculating the concentration of NH_3(aq) is simple: K_H⋅P_NH3, while P_NH3=C⋅P_air=150⋅10⁻⁹⋅1atm, yields NH_3(aq)=2.64⋅10⁻⁹M.

If we are to tally up all contributions of H+ then we get: [H+]=Kw/[H⁺] (from water dissociation) + Ka[NH₄⁺]/[H⁺].

Then [H⁺]=(Kw+Ka[NH₄⁺])0.5 and we'd get the pH.

I'm stuck unsure how to express [NH₄⁺].

Thanks!

[Borek]

It is not about dissociating NH₄⁺. What you have is a very diluted solution of a weak base. The real question is: is it diluted enough to be negligible?

Compare https://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution

[Ellie]

I am not sure. Could you provide a hint of what to do? Thanks,