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Shouldn't we also add the volume of the HI to the volume of the water (500mL) when we calculate the molarity? My book simply divided the number of mol by 500 mL.

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    $\begingroup$ Hydrogen iodide is a gas that is highly soluble in water, so any volume change caused by its dissolution was likely considered too small to bother with. The problem is just an exercise in computing pH. $\endgroup$ – Ed V May 18 at 18:50
  • $\begingroup$ oh ok thanks! but when calculating the pH, shouldnt we also consider the pH of water by itself? $\endgroup$ – 12345bird May 18 at 19:00
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    $\begingroup$ Only unless you have a molar concentration down around 100 nM or so! An aqueous solution of HI is like an aqueous solution of HCl, i.e., they are strong acids. $\endgroup$ – Ed V May 18 at 19:05
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Simple questions are often interesting.

(1) Most importantly, volumes are rarely additive. So the notion of adding the volume of the HI and the volume of water just doesn't work.

(2) In an answer to a question about hydroiodic acid, Curt F. showed a chart of concentration and density. For the $\pu{20 ^\circ C}$ values the liters of water in a liter of the acid solution can easily be calculated. So for the OP to get an error of 1% or less, the molarity would have to be about 0.2 molar or less.

\begin{array}{|c|c|c|c|} \hline \text{Concentration, %}(w/w) & \text{Density, }\pu{kg/L}(\text{@ }\pu{20 ^\circ C}) & \text{Molarity} & \pu{L(water)/L(acid)} \\ \hline 5.2 & 1.0342 & 0.420 & 0.980 \\ \hline 10.8 & 1.0812 & 0.922 & 0.964\\ \hline 16.4 & 1.1226 & 1.44 & 0.938 \\ \hline 22.4 & 1.1765 & 2.06 & 0.913 \\ \hline 27.2 & 1.2333 & 2.62 & 0.898 \\ \hline 33.1 & 1.2918 & 3.34 & 0.864 \\ \hline 38.7 & 1.3605 & 4.12 & 0.834\\ \hline 42.9 & 1.4208 & 4.77 & 0.811 \\ \hline 48.7 & 1.5072 & 5.74 & 0.773 \\ \hline 53.0 & 1.5913 & 6.59 & 0.748 \\ \hline 57.0 & 1.6933 & 7.55 & 0.728 \\ \hline \hline \end{array}

You can see from the red line that the data is not linear.

enter image description here

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  • $\begingroup$ hi, thanks for the detailed reply! do you think i should just do the simplified version and disregard that the water actually has some H3O by itself, too, in the test? $\endgroup$ – 12345bird May 19 at 7:19
  • $\begingroup$ oh wow this is getting confusing since the H3O isnt additive... $\endgroup$ – 12345bird May 19 at 7:20
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    $\begingroup$ @12345bird - Yes, just assume that the volume doesn't change. I did the calculation as a matter of curiosity. $\endgroup$ – MaxW May 19 at 8:29
  • $\begingroup$ thanks! sad to know that what we learn is just simplification of reality :/ $\endgroup$ – 12345bird May 19 at 8:42
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    $\begingroup$ @12345bird - Chemistry uses good enough rather than perfect extensively. Mathematicians can calculate $\pi$ to million of places. Chemistry just doesn't work like that. $\endgroup$ – MaxW May 19 at 8:46

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