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I weighted my hydroiodic acid sample: 25 ml weight 34.1 g, so the density is about 1.364 g/ml.

I want to know its concentration based on density. Does anyone have the corresponding density chart?

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    $\begingroup$ You should just titrate it. $\endgroup$ – Mithoron Jul 1 at 19:45
  • $\begingroup$ Your right, but i just run out of indicator, have to resupply soon. $\endgroup$ – THF Jul 2 at 5:14
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In 1981, Nishikata et al. published Viscosities of Aqueous Hydrochloric Acid Solutions, and Densities and Viscosities of Aqueous Hydroiodic Acid Solutions in the Journal of Chemical Engineering Data.

Table III from this paper shows what you are after.

HI_density_concentration_chart

I don't know what temperature you used to make your density measurement, but assuming it was 20 °C, then your $\ce{HI}$ looks to have a concentration of about 39 weight percent.

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Here is the density/concentration chart for aqueous $\ce{HI}$ at $\pu{20 ^\circ C}$:

$$ \begin{array}{ccc} \hline \text{Density, }\pu{kg/L}(\text{@ }\pu{20 ^\circ C}) & \text{Concentration, %}(w/w) & \text{Concentration, }(\pu{mol/L}) \\ \hline 1.0342 & 5.2 & 0.4204 \\ 1.0812 & 10.8 & 0.9129\\ 1.1226 & 16.4 & 1.4393 \\ 1.1765 & 22.4 & 2.0603 \\ 1.2333 & 27.2 & 2.6226 \\ 1.2918 & 33.1 & 3.3429 \\ 1.3605 & 38.7 & 4.1163\\ 1.4208 & 42.9 & 4.7653 \\ 1.5072 & 48.7 & 5.7385 \\ 1.5913 & 53.0 & 6.5936 \\ 1.6933 & 57.0 & 7.5458 \\ \hline \end{array} $$

HI Densities

Evidently, the density and molarity of aqueous solutions of $\ce{HI}$ is directly proportional at $\pu{20 ^\circ C}$ (LHS Chart) with the equation of straight line is $y = 10.965x - 10.87$ where $x$ is density in $\pu{kg/L}$ and $y$ is the molarity ($\pu{mol/L}$). Thus,

Molarity of your $\ce{HI}$ solution is: $10.965\times 1.364 - 10.87 = \pu{4.086 mol/L}$.

I also make a plot to see how percentage of the solution behave with density. Evidently (and also agree with literature finding), the density and $w/w\%$ of aqueous solutions of $\ce{HI}$ have second order relationship at $\pu{20 ^\circ C}$ (RHS Chart) with the equation of $y = -72.044x^2 + 274.62x - 201.55$ where $x$ is density in $\pu{kg/L}$ and $y$ is the percent concentration ($w/w$). Thus,

Percent concentration of your $\ce{HI}$ solution is: $-72.044\times 1.364^2 + 274.62\times 1.364 - 201.55 = 38.99\% (w/w)$.

Reference for densities and percentages:

E. Nishikata, T. Ishii, T. Ohta, "Viscosities of aqueous hydrochloric acid solutions, and densities and viscosities of aqueous hydroiodic acid solutions," J. Chem. Eng. Data 1981, 26(3), 254-256 (https://doi.org/10.1021/je00025a008).

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