Question

Question #db107

Answer 1

Here's how you could do that.

Explanation:

Some theoretical aspects first.

Borax, `"Na"_2"B"_4"O"""_7 * 10"H"_2"O"`, will dissolve in water to produce sodium cations, `"Na"^(+)`, and tetraborate anions, `"B"_4"O"_7^(2-)`.

The tetraborate anion will then react with water to produce boric acid, `"H"_3"BO"_3`, and hydroxide anions, `"OH"^(-)`.

`"B"_4"O"_text(7(aq])^(2-) + 7"H"_2"O"_text((l]) -> 7"H"_3"BO"_text(3(aq]) + 2"OH"_text((aq])^(-)`

Notice that this reaction produces two moles of hydroxide anions for every mole of borate anions, which is equivalent to the number of moles of borax, that reacts.

In order to neutralize the produced hydroxide anions, you would need to match the number of moles of added `"HCl"` with half the number of moles of borax.

This means that you need 2 moles of hydrochloric acid for every mole of borax.

The balanced chemical equation for the overall reaction looks like this

`"Na"_2"B"_4"O"_7 * 10"H"_2"O"_text((aq]) + color(red)(2)"HCl"_text((aq]) -> 2"NaCl"_text((aq]) + 4"H"_3"BO"_text(3(aq]) + 5"H"_2"O"_text((l])`

So, you start by weighing 4.0 g of borax. Use the compound's molar mass to determine how many moles you would get

`4.0color(red)(cancel(color(black)("g"))) * "1 mole"/(381.38color(red)(cancel(color(black)("g")))) = "0.0105 moles borax"`

You dissolve the borax in 250 mL of water, which means that the resulting solution will have a molarity of

`C = n/V`

`C = "0.0105 moles"/(250 * 10^(-3)"L") = "0.0420 M"`

Now, take 25 mL of this solution and add a drop or two of bromocresol green indicator. This volume of solution would contain 10 times fewer moles of borax than the parent 250-mL solution

`25color(red)(cancel(color(black)("mL"))) * "0.0105 moles"/(250color(red)(cancel(color(black)("mL")))) = "0.00105 moles borax"`

Start titrating this 25-mL sample with hydrochloric acid. Bromocresol green is blue at pH values greater than 5.8 and yellow at pH values smaller than 3.8.

The pH of the solution at the end-point will be acidic because of the presence of the boric acid, which is a weak acid. So, when the solution starts going from blue to yellow, record the volume of hydrochloric acid used.

You might want to repeat the procedure for other 25 mL samples as well.

Once you're satisfied with the values you got, use the recorded volume and the mole ratio that exists between borax and hydrochloric acid to find the molarity of the acid.

`n_(HCl) = color(red)(2) * n_"borax"`

`C_(HCl) = n_(HCl)/V_(HCl)`