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HClO4= 100.5, d HClO4=1.8, (ACO)2O=102, d (ACO)2O=1.1, H2O=18, d H2O=1

A) Preparation of 100 ml perchloric acid 0.1 Molar aqueous solution. below is my answer, but what I'm not sure is that the density the problem has given is for 100% pure acid or 70%. below I calculated the density for 70% assuming 1.8 is for 100%.

d_70%= 100/(70 x 1/1.8+30 x 1/1)=1.452
M=10αd/MW=10x70x1.452/100.5=10.11 M

M_1 V_1= M_2 V_2→10.11 x V_1=0.1 x 100→ V_1≅1ml

B) Preparation of 100 ml perchloric acid 0.1 Molar non-aqueous solution in acetic acid. Okay about this one. So all I know is that I need to remove all water in the acid by adding acetic anhydride, and the ratio is 1/1: enter image description here

I thought about how to calculate the volumes of both acid and acetic anhydride, and maybe I can use the density, or tried to create some equation, but none actually reached a final answer. So I tried this, I'm not sure this actually makes sense:

70/100 x 1/1.8=0.39 ml

M_pure= ( 0.7/100.5 x1000)/0.39=17.86 M

M_1 V_1= M_2 V_2→17.86 V_1=0.1 x 100→V_(pure acid)=0.56ml

1-0.56=0.44 ml water

0.44ml x 1g/1ml x 1mol/18g x (1 mol (ACO)_2 O)/(1 mol water) x 102g/1mol x 1ml/1.1g=2.27ml acetic anhydrade

So 1 ml perchloric acid and 2.27 ml acetic anhydride, up to 100 with acetic acid

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  • $\begingroup$ Your question is not very clear. Would you please add some details? For example, are you given 70% perchloric acid? $\endgroup$ Commented Apr 29, 2020 at 16:03
  • $\begingroup$ Yes, I have perchloric acid 70% $\endgroup$ Commented Apr 29, 2020 at 16:14
  • $\begingroup$ Your conversion for the density of perchloric acid is wrong. You generally can't convert densities by doing a linear extrapolation. Assume density of 1.8 is for the 70% perchloric acid by weight. $\endgroup$
    – MaxW
    Commented Apr 29, 2020 at 16:51
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    $\begingroup$ Actual density of 70% perchloric acid is 1.67 g/mL. $\endgroup$ Commented Apr 29, 2020 at 17:13
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    $\begingroup$ Crimson-Ruby - @MathewMahindaratne's point is absolutely correct. I absolutely believe that problems should be realistic to teach some chemistry too. Why just make up a number when you can use the real values? $\endgroup$
    – MaxW
    Commented Apr 29, 2020 at 17:43

3 Answers 3

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Given:

HClO4= 100.5, d HClO4=1.8
(ACO)2O=102, d (ACO)2O=1.1
H2O=18, d H2O=1

A) Preparation of 100 ml perchloric acid 0.1 Molar aqueous solution.

I had to do some poking around to understand the density number. Without a temperature the density of liquids is nebulous. Wikipedia lists a density of 1.768 g/ml which seems to be for the 72.5% azeotrope.

So just blundering with the information given assume density is for 70% by weight perchloric acid.

Molarity $\ce{HClO4}$

$$\dfrac{0.70\times\pu{1.8 g/ml} \times \pu{1000 ml/L}}{\pu{100.5 g/mol}} = 12.5 M$$

Volume of conc $\ce{HClO4}$ for 100 mls of 0.1 molar

$$\dfrac{\pu{0.1 L}\times \pu{0.1 M}}{\pu{12.5 M}}= \pu{0.0008 L} = \pu{0.8 ml}$$

B) Preparation of 100 ml perchloric acid 0.1 Molar non-aqueous solution in acetic acid.

This is a bit gnarly.

Volumes are not typically additive. So typically an excess of the acetic acid would be made and this would be used to dilute the 0.8 ml of perchloric acid to 100 ml. However in this case the 0.8 ml of perchloric acid contains water too. Also no density is provided for glacial acid acid, nor for the 0.1 Molar perchloric acid in glacial acetic acid. So the problem isn't really solvable as is.

The other wrinkle here is that mixing exact stoichiometric proportions is impossible. If an anhydrous solution was wanted, then an excess of acetic anhydrous would be used or some other means to dry the solution would be necessary.

Never the less, let's blunder through some sort of solution, as poor as the problem is...

Assume that the volumes are additive.

So we'll get the right proportions of the three components but only somewhere near 100 ml of solution, not 100 ml exactly.

$$V_{solution} = V_{anhydride} + V_{water} + V_{\pu{conc acid}}$$

$$V_{\pu{conc acid}} = \pu{0.8 ml}$$

$$\therefore V_{anhydride} + V_{water} = \pu{99.2 ml}$$

However as mentioned before, the concentrated perchloric acid contains 30% water by weight.

$$\pu{g water in acid} = 0.3 \times \pu{0.8 ml}\times \pu{1.8 g/ml} =\pu{0.4 g}$$

Since water has a density of 1 g/ml that is also 0.4 ml.

We were given that (ACO)2O=102, d (ACO)2O=1.1 so for the anhydride the moles per ml is:

$$\dfrac{\pu{1.1 g/ml}}{\pu{102 g/mol}} = \pu{0.0108 mol/ml}$$

We were given that H2O=18, d H2O=1 so for the water the moles per ml is:

$$\dfrac{\pu{1 g/ml}}{\pu{18 g/mol}} = \pu{0.0556 mol/ml}$$

Now since the perchloric acid contained 0.4 ml of water we need to figure out how many ml of acetic anhydride we need to neutralize that water.

$$\dfrac{\pu{0.0556 mol/ml}\times\pu{0.4 ml}}{\pu{0.0108 mol/ml}} = \pu{2.1 ml}$$

So we now need 99.2 - 2.1 = 97.1 ml of water/acetic anhydride mixture in a 1:1 molar ratio. For 1 ml of acetic anhydride the ml of water needed is

$$\dfrac{\pu{0.0108 mol/ml}}{\pu{0.0556 mol/ml}}= \pu{0.194 ml water}$$

Let $x$ be the multiplicative factor

$$x = \dfrac{97.1}{1.194}= 81.3$$

So: 0.8 ml conc perchloric acid
83.4 ml acetic anhydride
15.8 ml water


I loathe that the problem is inferring that volumes are additive.

Looking up some data...

\begin{array}{|c|c|c|c|} \hline & Mol. Wt. & Density\ (g/ml) & mol/ml \\ \hline \pu{Water} & 18.015 & \pu{0.9950 @ 25 °C} & 0.05523\\ \hline \pu{Acetic anhydride} & 102.09 & \pu{1.082 @ 20 °C} & 0.01052 \\ \hline \pu{Acetic acid} & 60.05 & \pu{1.0446 @ 25 °C} & 0.01740 \\ \hline \end{array}

The density data is out of Pubchem. I looked for a while and couldn't find density values at same temperature, but the values are close enough to make the point...

assume 1.0000 ml of acetic anhydride and appropriate water to give 1:1 molar ratio

\begin{array}{|c|c|c|} \hline & ml & g \\ \hline \pu{Water} & 0.1904 & 0.1894\\ \hline \pu{Acetic anhydride} & 1.0000 & \pu{1.082 g} \\ \hline \pu{Totals} & 1.1904 & ‭1.271‬ \\ \hline \end{array}

So density of acetic acid by combining volumes is $$\dfrac{1.271}{1.1904} = 1.067$$ which gives an error of $$\dfrac{1.067-1.0446}{1.0446}\times 100\% = +2.1\% $$

Thus the total solution as calculated above would be closer to 98 ml than 100.

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This is not an attempt to answer the question(s). However, it is seemingly clear that OP is doing a lab practical, and I'd like to clear his assumption about density of $\ce{HClO4}$ acid.

The series of density ($d$) of $\ce{HClO4}$ acid by percentage at $\pu{25 ^\circ C}$ have been published in 1941 (Ref.1):

$$ \begin{array}{cccc} \% \ \ce{HClO4} \ (w/w) & d\text{ at } \pu{25 ^\circ C} \ (\pu{gcm-3}) & d\text{ at } \pu{30 ^\circ C} \ (\pu{gcm-3}) & \text{Conc. at } \pu{25 ^\circ C} \ (\pu{molL-1}) \\ \hline 0 & (0.99707) & (0.99568) & - \\ 2 & 1.00828 & - & 0.2007 \\ 4 & 1.01975 & - & 0.4059\\ 6 & 1.03154 & - & 0.6158\\ 8 & 1.04355 & - & 0.8307\\ 10 & 1.05591 & 1.05388 & 1.0507\\ 20 & 1.1228 & 1.11999 & 2.2344\\ 30 & 1.20002 & 1.19651 & 3.5821\\ 40 & 1.29073 & 1.28658 & 5.1372\\ 45 & 1.34252 & - & 6.0113\\ 50 & 1.39937 & 1.39435 & 6.9620\\ 55 & 1.46134 & - & 7.9974\\ 60 & 1.52766 & 1.52177 & 9.1203\\ 65 & 1.59628 & - & 10.3242\\ 70^a & 1.6644 & 1.658 & 11.5928\\ \hline \end{array} $$ $^a$ The values are from Ref.2.

Some values in the publication was avoided for convenience (total of 27 density values for 27 percentages not including 0% have been reported). The authors have given best fitting empirical equation as:

$$d = 4.9593 \times 10^{-7}x^3 + 2.30045 \times 10^{-5}x^2 + 5.62796 \times 10^{-3}x + 0.99707 \tag{1}$$

where $d$ is density at $\pu{25 ^\circ C}$ and $x$ is percentage of $\ce{HClO4}$. When plotted using an excel spread sheet, the same data set gives me a parabolic relationship with $R^2 = 0.9996$ agreement. The relevant equation is:

$$d = 7.0 \times 10^{-5}x^2 + 4.8 \times 10^{-3}x + 0.9971 \tag{2}$$

% HClO4  vs density

This scientific data shows two facts:

  1. The relationship between percent $\ce{HClO4}$ and its density is not linear. It is most probably parabolic.
  2. The density of $70\% \ \ce{HClO4}$ solution is: $$d_{70\%} = 7.0 \times 10^{-5} \times 70^2 + 4.8 \times 10^{-5} \times 70 + 0.9971 \approx 1.676$$

According to the empirical formula of Ref.1, the density of $70\% \ \ce{HClO4}$ solution can also be calculated:

$$d = 4.9593 \times 10^{-7} \times 70^3 + 2.30045 \times 10^{-5} \times 70^2 + 5.62796 \times 10^{-3} \times 70 + 0.99707 \\ = 1.674$$

The both calculated values have good agreement with the experimental value of $\pu{1.67 gcm-3}$ (Ref.2 has given a little deviated value of $\pu{1.6644 gcm-3}$ for the $70\%$ solution at $\pu{25 ^\circ C}$).

If you have done for the similar calculations for 100% solution, you'd have ended up with the value, $\pu{2.286 gcm-3}$ from equation $(1)$ and $\pu{2.177 gcm-3}$ from equation $(2)$ for $d_{100\%}$. Either value is much bigger than the given value of $\pu{1.8 gcm-3}$. However, as MaxW pointed out in his comment, extrapolation is not a great idea for this kind of data sets. Based on MaxW's graph in one of his answers, either values of $\pu{2.286 gcm-3}$ or $\pu{2.177 gcm-3}$ are not acceptable. Hence, it is safe to say that it would be acceptable to consider $d=\pu{1.8 gcm-3}$ for $d_{100\%}$ as OP originally did.

Note: I have also calculated the concentration of each solution in terms of $\pu{mol/L}$ using density values. Based on that values, it seems $1\%$ solution is exactly equal to $\pu{0.10 M}$ (at least for 2 significant figures).

References:

  1. Aaron E. Markham, “Density of Perchloric Acid Solutions,” J. Am. Chem. Soc. 1941, 63(3), 874-875 (https://doi.org/10.1021/ja01848a509).
  2. Langhorne H. Brickwedde, “Properties of Aqueous Solutions of Perchloric Acid,” Journal of Research of the National Bureau of Standards 1949, 42, 309-329 (PDF).
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    $\begingroup$ Great data showing that volumes are not additive! // It is a big dangerous to extrapolate a fitted curve beyond the bounds of the data. I'd agree that the 70% number is good to 2 decimal places. Extrapolating to 72.5% gives 1.713 vs 1.768 with equation 2. However extrapolating to pure perchloric acid is too much. // I looked for the density of the pure acid and couldn't find it... $\endgroup$
    – MaxW
    Commented Apr 29, 2020 at 20:49
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    $\begingroup$ density of 70% acid is 1.6644 @ 25 C which agrees with an earlier study // "Properties of Aqueous Solutions of Perchloric Acid"; Langhorne H . Brickwedde; Journal of Research of the National Bureau of Standards Volume 4 2, March 1949, page 309 $\endgroup$
    – MaxW
    Commented Apr 29, 2020 at 22:15
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    $\begingroup$ 75.0% acid = 1.731 @ 25 C // Pub Chem - Other Experimental Properities // data from 10th edition Merck Index $\endgroup$
    – MaxW
    Commented Apr 29, 2020 at 22:23
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    $\begingroup$ "Perchloric acid; authentic, scientific and practical considerations relative to its use as an important research and routine analytical chemical reagent with special reference to rapidity, accuracy and economy"; Data compiled by G. Frederick Smith; Columbus, G. Frederick Smith Chemical Co., 1940, 4th ed., page 60 // table for 25 C for 75% to 65% by 0.25% to 5 decimal places. // 75% =1.73175; 74%=1.71824; 73%=1.70743; 72%=1.69122; 71%=1.67771; 70%=1.66420; 65%=1.59665 $\endgroup$
    – MaxW
    Commented Apr 29, 2020 at 22:53
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    $\begingroup$ 100% Perchloric acid 1.7676 @ 20 C // "PERCHLORIC ACID AND PERCHLORATE" by ALFRED A. SCHILT, G. FREDERICK SMITH CHEMICAL COMPANY, 867 McKinley Avenue Columbus, Ohio 43223 LCCN 79-63068 © COPYRIGHT 1979, table 2, page 14 which references A. A. Zinov'ev and N. A. Shchirova, ZNK, 5, 1299 (1960). $\endgroup$
    – MaxW
    Commented Apr 29, 2020 at 23:28
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This isn't an answer but a plot of the density data from:

100% Perchloric acid 1.7676 @ 20 C // "PERCHLORIC ACID AND PERCHLORATE" by ALFRED A. SCHILT, G. FREDERICK SMITH CHEMICAL COMPANY, 867 McKinley Avenue Columbus, Ohio 43223 LCCN 79-63068 © COPYRIGHT 1979, table 2, page 14 which references A. A. Zinov'ev and N. A. Shchirova, ZNK, 5, 1299 (1960).

The density curves exhibits a maximum at about 80% perchloric acid. Also note the danger of extrapolating curves beyond the fitted data range.

enter image description here

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