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I have a pretty basic question but the last time I took Chemistry was in 2007. I am studying the Navy's Nuclear study guide for their interviews and one of the question I am faced with is below.

Determine the final pH and temperature when these two solutions are mixed together in a 3L container.

Sol A: 2L, pH = 3, and 80F

Sol B: 1L, pH = 5, and 40F

Would this be as simple as $(2/3) \cdot 3 + (1/3)\cdot5$ for the pH and $(2/3)\cdot80 + (1/3)\cdot40$ for the temperature?

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  • $\begingroup$ I've edited tag,formatting and title let me know if you don't agree with my edit. $\endgroup$ – G M Apr 13 '14 at 11:57
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For the temperature, this method of calculation will give you a rough estimate of the final temperature, but it will not be exactly right. The reason is that mixing two solutions comes at an energy cost/gain due to the interaction of the two solution molecules. For some mixtures this enthalpy of mixing yields a temperature rise (i.e. energy is released from the molecular potential energy), for others it yields a temperature decrease (i.e. energy is taken up to 'make the mixing possible'). For many mixtures this difference is not more than a few degrees, which makes your linear estimate reasonable, but it is good to keep in mind that it will not be exact.

For the pH, the shortcut you are taking is incorrect and will at best give you a very crude estimate. The reason is that the pH scale is logarithmic, it is correlated with the concentration of protons in the solution as $\ce{pH}=-\log_{10}\; [\ce{H+}] $, where $[\ce{H+}]$ is the proton concentration. (I will make the assumption here that you have simple liquids, which are not buffered.) Therefore, you would first need to calculate the proton concentration in both solutions from this equation, yielding $10^{-3}\,\ce{M}$ and $10^{-5}\,\ce{M}$ for sols A and B resp. This means that the amount of $\ce{H+}$ in sol A is $2\cdot10^{-3}\,\ce{mol}$ and that in sol B $10^{-5}\,\ce{mol}$. Mixing the two you get 3 liter of solution with $2.01\cdot10^{-3}\,\ce{mol}$ protons, which is $6.7\cdot10^{-4}\,\ce{M}$ which works out to $\ce{pH=}3.17$. Indeed, very far off from the 3.67 predicted in your linear method.

Short discussion on the pH: because the scale is logarithmic and your 2 $\ce{pH}$ values are actually not that far from each other the linear estimate is still ok-ish. If, for example, you would have used $\ce{pH}$ 0 and 7 you would have found $\ce{pH}=0.17$ vs. $2.33$ from your estimate.

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  • $\begingroup$ Where did 10^{-3} and 10^{-5} come from? Then why did you multiple by 2 on both of them? $\endgroup$ – dustin Apr 13 '14 at 15:04
  • $\begingroup$ $\ce{pH}=-\log_{10}[\ce{H+}]$ gives $10^{-\ce{pH}}=[\ce{H+}]$ which shows how I got the $10^{-3}$ and $10^{-5}$, then I multiplied with 2 ONLY for the $10^{-3}$ because its units are $\ce{mol/l}$ and you indicated that you have 2 liters of sol A, which then makes it $2\cdot10^{-3} \ce{mol}$ $\endgroup$ – Michiel Apr 13 '14 at 16:18
  • $\begingroup$ and the $\log$ is a $\log_{10}$, just to be clear. I edited that in for clarity $\endgroup$ – Michiel Apr 13 '14 at 16:21
  • $\begingroup$ Minor mistake, $2\times 10^{-3} + 10^{-5} = 2.01 \times 10^{-3}$, not $2.05 \times 10^{-3}$. Mixed into 3 liters gives $6.7\times 10^{-4}$ M. Close enough. $\endgroup$ – LDC3 Apr 13 '14 at 16:25
  • $\begingroup$ @LDC3 whoops, you are absolutely right, corrected it! $\endgroup$ – Michiel Apr 13 '14 at 16:28
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The temperature would be correct. And the pH would be correct for this exercise, but not in reality. Since most solutions contain weak acid and base, it is difficult to determine the pH unless you knew which buffers and quantities are in the system.

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  • $\begingroup$ Can you explain what you mean by knowing which buffers and quantities are in the system? $\endgroup$ – dustin Apr 13 '14 at 0:03
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    $\begingroup$ The pH portion of the answer is not correct. Dilution of species is linear in concentrations, but the pH scale is logarithmic in concentration. For example, 1 L of non-buffered pH 3 solution + 9 L of pure water (pH 7 solution) makes 10 L of pH 4 solution. $\endgroup$ – Nicolau Saker Neto Apr 13 '14 at 0:25
  • $\begingroup$ @NicolauSakerNeto can you write an answer explaining how you solve that part so I can see? $\endgroup$ – dustin Apr 13 '14 at 1:48
  • $\begingroup$ Does this answer give you enough of an idea of how to calculate pH after mixing? $\endgroup$ – Nicolau Saker Neto Apr 13 '14 at 1:57
  • $\begingroup$ @NicolauSakerNeto that is too complicated. I need something simpler as an example. $\endgroup$ – dustin Apr 13 '14 at 2:14

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