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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 the following two solutions are mixed together in a $\pu{3L}$ container.
Sol A: $\pu{2L}$, pH = 3, and $\pu{80 ^\circ F}$
Sol B: $\pu{1L}$, pH = 5, and $\pu{40 ^\circ F}$

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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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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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, 2014 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. $\endgroup$ Apr 13, 2014 at 0:25
  • $\begingroup$ Does this answer give you enough of an idea of how to calculate pH after mixing? $\endgroup$ Apr 13, 2014 at 1:57
  • $\begingroup$ @NicolauSakerNeto that is too complicated. I need something simpler as an example. $\endgroup$
    – dustin
    Apr 13, 2014 at 2:14
  • $\begingroup$ @LDC3 - your answer for the temperature would actually have to be the same as that for pH: correct in this exercise, but not in reality, because in reality you have to account for the enthalpy of mixing which could change the temperature of the final mixture by some degrees $\endgroup$
    – Michiel
    Apr 13, 2014 at 5:49

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