What is the correct method for converting a concentration given in ppm to molarity?

Question

I need to convert a concentration of dissolved boric acid $(\ce{H_3BO_3})$ in water (at high temperature and pressure) from ppm to mol/L.

Concentration of boric acid: $c_{ppm}=1500\ \ce{ppm}$

Molar mass of boric acid: $M_i=61.83\ \ce{g/mol}$

Density of water at $25\ ^{\circ}\ce{C}$ : $\rho_{25}=1.0\ \ce{kg/L}$

Density of water at $300\ ^{\circ}\ce{C}$ : $\rho_{300}=0.7125\ \ce{kg/L}$

Note that the water density is low due to high temperature ($300\ ^{\circ}\ce{C}$) and pressures.

Method 1: At $300\ ^{\circ}\ce{C}$

$$ c_i = \frac{c_{ppm}\rho_{300}}{1000M_i} = \frac{1500\times0.7125}{1000\times61.83}=0.0173 \ [\ce{mol/L}] $$

Method 1: At $25\ ^{\circ}\ce{C}$

$$ c_i = \frac{c_{ppm}\rho_{25}}{1000M_i} = \frac{1500\times1}{1000\times61.83}=0.0243 \ [\ce{mol/L}] $$

Method 2:

I have seen the following method that takes the temperature to be 25C, despite modelling the same high temperature/pressure body of water:

https://i.stack.imgur.com/mS2hV.jpg

Which yields a final concentration of $8.33\times10^{-8}\ [\ce{mol/L}]$

I am convinced that method 2 given in the attached image is incorrect (as the molarity calculated is far lower than expected), but am unsure where the fault lies.

Edit: I have improved formatting and readability.

Answer 1

This question is confusing to me, and the calculation that you show doesn't look right to me.

You state that the question has to do with a solution of boric acid in water. Fair enough, until you report that the temperature of the solution is 300 °C. This would imply a pressure of at least about 85 atmospheres; any less and the solution would boil, assuming that the solution behaves as water would. This is a fair assumption, as the concentration of boric acid (I'll get to that below) is much too low to affect the physical properties of water all that much.

This isn't wrong. I'm just trying to orient myself to the problem.

The calculation that you reference by link to imgur.com is suspect. Notice that it has conversion factors that undo each other (the molar mass of boric acid). In addition, it uses a density conversion that is valid under standard pressure (1 atm), and not at a minimum of about 85 atm, where the density of water at 300 °C is, as you correctly state, 0.7125 kg/L.

Let's start fresh:

The concentration of the boric acid solution is given as 1500 parts per million. Parts per million is defined as mg of solute/liter of solution. Molarity is defined as moles of solute/liter of solution. The denominators match, so the problem is just one of converting the numerator from milligrams to moles. Which I do, thusly:

(1500 mg H3BO3 / L sol'n) * (1 g / 1000 mg) * (1 mole H3BO3 / 61.83 g H3BO3)

= 0.0243 M H3BO3

Instead of calling it a day, let me consider the calculation on imgur.com. It starts with the expression:

1500 ppm H3BO3 = 1500 E -6 (molecules H3BO3 / molecules H2O)

This seems to imply that "parts per million" refers to a molecule fraction of solute in solvent, which is not used in chemistry, at least not as far as I know. But if we accept this understanding of "parts per million", the logic and calculation becomes this:

By the definition of "mole":

1500 molecules H3BO3 / 1 E 6 molecules H2O is equivalent to:

1500 moles H3BO3 / 1 E 6 moles H2O

So:

(1500 moles H3BO3 / 1 E 6 moles H2O) * (1 mole H2O / 18.01 g H2O) * (1000 g / kg) * (0.7125 kg / L)

= 0.0593 M H3BO3

...which matches the calculation by the formula you stated at the top of your post:

c_i = (1500 * 0.7125) / (1000 * 18.01) = 0.0593 M H3BO3

The omission of units of measure is unsettling to me, but we're all friends here, right?

I should point out that the conversion in the imgur.com file: 1 cm^3 = 1000 dm^3 is wrong. The correct relationship is: 1000 cm^3 = 1 dm^3. The difference between the calculated answer in the imgur.com file and the answer above can be traced to this improper conversion of cm^3 to dm^3, and the density of water at a temperature and pressure (1.0 kg/L at ~25 °C and 1 atm) that does not apply to the conditions (0.7125 kg/L at 300 °C and at least ~85 atm).

I hope this helps, or at least, that I haven't added to your confusion.