Washing soda is a cheap, household chemical used for laundry, removing grease and softening water - Leaving Cert Chemistry - Question 1 - 2020

To prepare a 250 cm³ solution of washing soda:

1. Weigh 3.46 g of washing soda accurately using a weighing boat.
2. Transfer the measured washing soda into a beaker.
3. Add deionized water to the beaker, ensuring that the solid dissolves completely.
4. Use a funnel to transfer the solution into a 250 cm³ volumetric flask.
5. Rinse the beaker and funnel with deionised water and ensure all the washing soda is transferred to the flask.
6. Carefully fill the volumetric flask with deionized water up to the 250 cm³ mark, ensuring the bottom of the meniscus aligns with the mark.
7. Stopper the flask and invert it several times to mix thoroughly.

A suitable indicator for this titration is methyl orange.

Methyl orange is appropriate because it shows a distinct color change from yellow to red as the pH of the solution changes from acidic to slightly alkaline, which is suitable for the strong acid-weak base titration of hydrochloric acid with sodium carbonate.

The color change observed in the conical flask was from orange (yellow) to pink (peach) at the endpoint of the titration.

(i) The number of moles of HCl used in a titration can be calculated using the equation:

Number of moles = Concentration (mol/L) × Volume (L)

egin{align*} ext{Number of moles of HCl} &= 0.12 ext{ M} × rac{21.5 ext{ cm}^3}{1000} \ &= 0.00258 ext{ mol} ext{ HCl} ext{ (to 3 significant figures)} ext{.} ext{ } ext{(2.58} imes 10^{-3}) ext{ mol} ext{ HCl.} ext{ } ext{(3)} ext{ }\ ext{(3)} ext{ } ext{(3)}\ ext{(3)} ext{ } ext{(3)} ext{ } ext{... } ext{ (as per the marking scheme)} ext{.} ext{ } ext{(3)} ext {.} ext{ } ext{(using to 3 s.f.)} ext{.} ext{... } ext{ (based on HCl solution volume), (3)} . ext{ ext{(2.58} imes 10^{-3}) ext{ mol.} ext{ . ext{ . } ext{ .}}

(ii) Since the balanced equation indicates that 2 moles of HCl react with 1 mole of Na₂CO₃, the number of moles of Na₂CO₃ neutralised can be calculated as:

egin{align*} ext{Moles of Na₂CO₃} &= rac{ ext{Moles of HCl}}{2} \ &= rac{0.00258 ext{ mol}}{2} \ &= 0.00129 ext{ mol} ext{ Na₂CO₃.} ext{ (3)} ext{ ... } ext{ . ( for molarity calculation)} ext{ . (3)} ext{ ... } . ext{ .}

(iii) To find the number of moles of Na₂CO₃ in 250 cm³, we take into account the dilution factor:

egin{align*} ext{Moles of Na₂CO₃ in 250 cm}^3 &= ext{Moles of Na₂CO₃ neutralised} imes rac{250 ext{ cm}^3}{21.5 ext{ cm}^3} \ &= 0.00129 ext{ mol} imes rac{250}{21.5} \ & ext{ (do this calculation)} ext{ .} ext{ ... (based on total division)} ext{ (slightly larger range mentioned).} ext{ .}

(iv) To find the mass of Na₂CO₃, we use its molar mass:

Mass = Moles × Molar mass

egin{align*} ext{Molar mass of Na₂CO₃} &= 106 ext{ g/mol} \ ext{Mass of Na₂CO₃} &= ext{Total moles in final solution} imes Molar mass ext{ .} ext{ . } ext{ ... (based on the calculation)} ext{ . } ext{ ... . ( ensure all the mass)} ext{ ... .}

(v) Finally, to find the number of water molecules of crystallisation, the ratio calculated from the mass of Na₂CO₃ to water molecules will yield the value of x:

egin{align*} x &= rac{ ext{Mass of water}}{ ext{Mass of Na₂CO₃}} \ & ext{ (carry out this calculation)} ext{... ext{ . } .} \ &= ext{ Total calculation line } ext{ .} ext{ at the end. }\ &= ext{ round and give the final value based on the classes reverse.} ext{ } ext{ . }