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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?

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  • $\begingroup$ Hi, I’ve cleaned up the MathJax in your post. You should really use \ce{...} (i.e. mhchem) for chemical formulae. It’s much easier to type and it also produces upright (non-italic) letters. Find out more on Chemistry Meta. $\endgroup$ – Jan Aug 22 '16 at 14:53
  • $\begingroup$ A possibility may be that you're expected to use $P_{\text{NH}_3}$ rather than $[\ce{NH3}]$. $\endgroup$ – a-cyclohexane-molecule Aug 23 '16 at 14:47
  • $\begingroup$ I thought about that but aren't partial pressures only used for gases? At these pressures ammonia would be a liquid. Though I suppose if the pressure is high enough, it might be compressible... Thanks for the suggestion! $\endgroup$ – Rika Aug 24 '16 at 15:44
  • $\begingroup$ Although I suppose ammonia is a gas at 1 bar pressures so I'd probably have to take that into consideration for that calculation... Hmmmm $\endgroup$ – Rika Aug 24 '16 at 19:43

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