I'm working on an experiment in school and I need to create 200 mL of 2 mol/L boric acid. According to a formula I found on wikipedia, boric acid can be created with a reaction with hydrochloric acid.

$$\ce{Na2B4O7·10H2O + 2 HCl -> 4 H3BO3 + 2 NaCl + 5 H2O}$$

So to calculate the amount of moles of boric acid needed, I multiplied the volume with the concentration to get 0.4 mol. After using the molar ratios in the equation, the molar mass of borax decahydrate and the concentration of HCl, I calculated that I would need 38.14 g of borax and 33.56 mL of 5.96 mol/L HCl in order for a complete reaction.

After mixing these together, I ended up with about 40 mL of a really thick solution, which is not the intended 200 mL. What did I do wrong and how do I actually create boric acid of this volume and concentration? Am I missing steps in between?

  • $\begingroup$ Boric acid is not soluble to make a $2~M$ solution at room temperature. $\endgroup$
    – A.K.
    Jan 13 '16 at 4:23
  • $\begingroup$ At what concentration would it be soluble at? I can change the concentrations as long as I'm getting the right volume for each concentration . And is it supposed to look like a thick paste or is it supposed to dissolve? $\endgroup$ Jan 13 '16 at 4:33
  • 1
    $\begingroup$ Refer to the wikipedia page: en.wikipedia.org/wiki/Boric_acid $\endgroup$
    – A.K.
    Jan 13 '16 at 4:35
  • $\begingroup$ I never noticed that table before, thanks. So if i were to dissolve the right amount at room temperature, would ny process be fine? $\endgroup$ Jan 13 '16 at 5:04

Assuming that boric acid is completely soluble in water, you did everything correctly so far, you simply missed the last step: dilution.

First you calculated the amount of substance of boric acid you would need to produce a two molar solution. With that you basically assume, that you dissolve solid boric acid in water.
You then calculated the amount of borax you need for complete reaction with hydrochloric acid. Since this is already hydrogen chloride dissolved in water, you bring some of the water you assumed before. But this is as your calculation says only about 40 mL.
Still assuming that it is completely soluble, you would have produced a solution that is about ten molar. You still need to dilute it to the final volume, to have the desired concentration.

Now let's go one step further. Boric acid is not very soluble in water under standard conditions, see Wikipedia. As A.K. pointed out, solubility is 47.2 g/L at 20 °C. That translates to 0.76 mol/L. As you can see solubility increases with temperature, in boiling water the maximum concentration is about 4.5 mol/L.
So what you are trying to attempt is not possible, at least not in pure water at room temperature. Try heating it up a bit, somewhere around 70 °C should be enough.

Alternatively PubChem has kindly provided us with a little more statistics about solubility.
In methanol you can obtain the desired two molar solution at room temperature, as solubility is 173.9 g/L at 25 °C.

  • $\begingroup$ Just for clarification, do i add 200 mL of water to the solution or 160 mL to the existing 40mL? $\endgroup$ Jan 13 '16 at 15:29
  • $\begingroup$ @BobMarshall you dilute to 200 mL as this is your final volume, hence you only have to add about 160 mL. $\endgroup$ Jan 13 '16 at 18:01

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