A batch of washing soda crystals (hydrated sodium carbonate, Na2CO3 - Leaving Cert Chemistry - Question 1 - 2014

To dissolve the washing soda crystals, the chemist should first weigh out 2.50 g of the crystals using an analytical balance for accuracy. The weighed sample should be placed into a clean beaker.

Next, a small volume of deionised water (approximately 50-100 cm³) should be added to the beaker. The chemist needs to use a stirring rod to ensure that the crystals dissolve completely, adding more deionised water as necessary.

Once the crystals are fully dissolved, the solution should be carefully transferred to a 250 cm³ volumetric flask. While transferring, deionised water should be rinsed from the beaker into the flask to ensure all sodium carbonate is included.

Finally, the flask should be filled with deionised water up to the 250 cm³ mark. The chemist must ensure that the bottom of the meniscus is at the line for an accurate measurement.

A suitable indicator for this titration is methyl orange, as it shows a distinct color change from red in acidic solutions to yellow in neutral to basic solutions, which is appropriate for the titration of a weak base (Na2CO3) with a strong acid (HCl).

The end point of the titration can be precisely detected by observing the color change of the methyl orange indicator. The solution will change from yellow to a faint pink color at the endpoint, indicating that the acid has completely reacted with the base. The observational method must be consistent, and the solution should ideally be swirled continuously to ensure homogeneity.

The amount of HCl used in titration is averaged from several trials. For this case, the mean volume of HCl is 21.6 cm³, which equals 0.0216 L. The reaction shows that 2 moles of HCl react with 1 mole of Na2CO3.

First, find the moles of HCl used:

$n_{HCl} = M_{HCl} imes V_{HCl} = 0.10 ext{ M} imes 0.0216 ext{ L} = 0.00216 ext{ moles}$

Using the stoichiometric ratio from the balanced equation:

n_{Na2CO3} = rac{1}{2} imes n_{HCl} = rac{1}{2} imes 0.00216 ext{ moles} = 0.00108 ext{ moles}

Now, calculate the concentration of Na2CO3:

C_{Na2CO3} = rac{n_{Na2CO3}}{V_{solution}} = rac{0.00108 ext{ moles}}{0.250 ext{ L}} = 0.00432 ext{ M}.

First, calculate the mass of Na2CO3 in the 2.50 g sample:

The formula mass of Na2CO3 is approx. 106 g/mol.

For sodium carbonate:

$ext{Mass}_{Na2CO3} = 0.00432 ext{ moles} imes 106 ext{ g/mol} = 0.45792 ext{ g}$

Next, calculate the mass of water associated with the crystallisation:

Total mass of crystals = 2.50 g, Mass of water = 2.50 g - 0.45792 g = 2.04208 g.

To find the % water of crystallisation:

ext{Percentage water} = rac{ ext{mass of water}}{ ext{total mass}} imes 100 = rac{2.04208 g}{2.50 g} imes 100 ext{ percent} = 81.68 ext{ percent}

To determine x, we know the molar mass of water is approximately 18 g/mol:

x = rac{2.04208 g / 18 g/mol}{0.45792 g / 106 g/mol} = ext{average } x ext{ is approx. } 10.