# Question #41c9a

Oct 1, 2015

The HCl solution has a concentration of $\textcolor{b l u e}{\text{0.1041 mol/L}}$.

#### Explanation:

Borax is a primary standard for acid-base titrations.

The equation for the reaction with $\text{HCl}$ is

$\text{Na"_2"B"_4"O"_7"·10H"_2"O" + "2HCl" → "4H"_3"BO"_3 + "2NaCl" + "5H"_2"O}$

Let's start by calculating the molarity of the borax solution

$\text{Moles of borax" = 4.750 color(red)(cancel(color(black)("g borax"))) × "1 mol borax"/(381.37 color(red)(cancel(color(black)("g borax")))) = "0.012 455 mol borax}$

$\text{Molarity" = "moles"/"litres" = "0.012 455 mol"/"0.250 L" = "0.04982 mol/L}$

Now we can calculate the moles of $\text{HCl}$ required to neutralize a 25.0 mL aliquot of the borax solution.

$\text{Moles of HCl" = 0.0250 color(red)(cancel(color(black)("L borax"))) × (0.04982 color(red)(cancel(color(black)("mol borax"))))/(1color(red)(cancel(color(black)("L borax")))) × "2 mol HCl"/(1 color(red)(cancel(color(black)("mol borax")))) = "0.002 491 mol HCl}$

And finally we can calculate the molarity of the $\text{HCl}$.

$\text{Molarity" = "moles"/"litres" = "0.002 491 mol"/"0.023 93 L" = "0.1041 mol/L}$