How to find pH of an aqueous ammonia solution at high pressures?

Here's the simplified question: I start with $15~\%$ of $\ce{NH\_3}$ and I want to find the $\mathrm{pH}$ of the following reaction at $25~\mathrm{^\circ C}$ and $5\ \mathrm{kbar}$:
$$\ce{NH\_3 + H\_2O -> NH\_4^+ + OH^-}$$


Now I believe I understand the basic procedure. I was able to find the pH at STP, but there are a few things that confuse me about finding the pH at high pressures. Here's the procedure I followed for STP:


$$[\ce{NH3}]\_\mathrm{eq} = 0.15-x\tag{1}$$


$$[\ce{NH4+}]\_\mathrm{eq} = x\tag{2}$$


$$[\ce{OH-}]\_\mathrm{eq} = x\tag{3}$$


$$K\_\mathrm{b} = \frac{x^2}{0.15-x}\tag{4}$$


where $x = [\ce{OH-}], \mathrm{pOH} = -\log[\ce{OH-}],$ and $\mathrm{pH} = 14 - \mathrm{pOH} = 11.2$


Now for my questions about applying this to $5~\mathrm{kbar}$: 


1. Unfortunately I'm not entirely clear about what they mean by $15~\%$—whether it's a mass fraction or in g/ml or what. 


But I did manage to get the right $\mathrm{pH}$ using $0.15$ as my initial $[\ce{NH3}]$ … though maybe that was just a coincidence. Actually if I assume that it's a mol fraction and then convert it to molarity, I do get something close to the right answer but not quite … if anyone has any ideas about this please let me know.


Also if I am missing some conversion from $15~\%$, is there any reason $[\ce{NH\_3}]\_\mathrm{initial}$ would be different at STP vs $5\ \mathrm{kbar}$? If both start with $15~\%\ \ce{NH3}$?
2. From what I've read, the value for $K\_\mathrm{b}$ shouldn't have any pressure dependence at all. And for this specific reaction, $K\_\mathrm{b}$ should be equal to $K\_\mathrm{p}$ since there are two molecules on the left and two molecules on the right sides of the equation. 


However when I calculate $K$ using $\Delta H$ values at $5~\mathrm{kbar}$, I do get a different value (at STP, $K = 10^{-4.754}$ and at $5~\mathrm{kbar}$ $K = 10^{-2.81}$). So does $K\_\mathrm{b}$ really not change with pressure or is this just an approximation at low pressures? Or am I dealing with two different $K$ values? Does the $K$ that I calculated from $\Delta H$ only equal $K\_\mathrm{b}$ at STP?


Actually now that I think about it, the value for $K\_\mathrm{w}$ definitely changes at high pressures. At STP $K\_\mathrm{w} = 10^{-14}$ and at $5~\mathrm{kbar}$ $K\_\mathrm{w} = 10^{-12.559}$


So if $K\_\mathrm{w}$ changes at high pressures, shouldn't that mean that $K\_\mathrm{b}$ does as well?
3. Is there some sort of step I'm missing at the end? Because I tried solving for $\mathrm{pH}$ using both K values (the value at STP and the value I found at $5~\mathrm{kbar}$) but I still don't get the right value for $\mathrm{pH}$. 


If I had to guess, I'd say that $[\ce{OH-}] = a\_\ce{OH-}$ at STP which is why $-\log a\_\ce{OH-}$ gives me the right $\mathrm{pOH}$ and $\mathrm{pH}$ at STP. But at $5~\mathrm{kbar}$ $[\ce{OH-}] \neq a\_\ce{OH-}$ so I'd need the activity coefficient? But if activity coefficients were needed, their source should have been cited which it wasn't … Unless there's a straightforward way to solve for the activity coefficient?


Let me know if any clarifications are needed.


Edit: I've just found that if I take the ratio of the vapor pressure of Ammonia (~10.1 bars) to the total pressure (5000 bars) and multiply that by .15, I do get very close to reproducing the 5 kbar results. Any ideas why that might be? 


