1. A student determined the concentration of a hydrochloric acid solution by titration with 25.0 cm³ portions of a 0.05 M primary standard solution of anhydrous sodium carbonate - Leaving Cert Chemistry - Question 1 - 2012

To prepare 500 cm³ of a 0.05 M sodium carbonate solution:

1. Calculate the mass of sodium carbonate needed using the formula:

   $ext{mass} = ext{molarity} imes ext{volume} imes ext{molar mass}$

   The molar mass of Na₂CO₃ is approximately 106 g/mol. Thus,

   $ext{mass} = 0.05 imes 0.500 imes 106 = 2.65 ext{ g}$

2. Weigh 2.65 g of anhydrous sodium carbonate on a balance.

3. Dissolve the solid in a small volume of distilled water in a beaker.

4. Transfer the solution to a 500 cm³ volumetric flask and fill to the mark with distilled water, ensuring thorough mixing.

The burette was filled above the zero mark and then the tap was opened slightly to allow some solution to flow out. The liquid was adjusted until the bottom of the meniscus was exactly on the zero mark.

The pipette was filled to the stop line to ensure that a precise volume of the sodium carbonate solution was delivered for the titration. This accuracy is essential for reliable and reproducible results.

A suitable indicator for this titration is methyl orange. The colour changes from yellow in the alkaline solution to red in the acidic solution, indicating the end point of the titration.

Using the titration data:

1. From the balanced equation: 2 moles of HCl react with 1 mole of Na₂CO₃.

2. The molarity of Na₂CO₃ is 0.05 M and the volume used is 0.025 L, thus moles of Na₂CO₃:

   $ext{moles of Na₂CO₃} = 0.05 imes 0.025 = 0.00125$

3. Therefore, moles of HCl required:

   $ext{moles of HCl} = 2 imes 0.00125 = 0.0025$

4. Mean titre in L: 20.8 cm³ = 0.0208 L. Thus,

   ext{Concentration of HCl} = rac{ ext{moles}}{ ext{volume}} = rac{0.0025}{0.0208} = 0.120 ext{ M}

   The correct concentration is therefore 0.12 M.

To find the concentration in grams per litre:

1. The concentration of HCl is 0.12 M, which implies:

   $ext{mass (g)} = ext{molarity (M)} imes ext{molar mass (g/mol)} imes ext{volume (L)}$

   The molar mass of HCl is approximately 36.5 g/mol, thus:

   $ext{mass} = 0.12 imes 36.5 = 4.38 ext{ g/L}$

   Therefore the concentration is 4.38 g/L.