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The flowchart shown outlines the sequence of steps used to determine the concentration of an unknown hydrochloric acid solution - HSC - SSCE Chemistry - Question 28 - 2010 - Paper 1
Question 28
The flowchart shown outlines the sequence of steps used to determine the concentration of an unknown hydrochloric acid solution. Preparation of 500 mL 0.100 mol L⁻¹... show full transcript
Worked Solution & Example Answer:The flowchart shown outlines the sequence of steps used to determine the concentration of an unknown hydrochloric acid solution - HSC - SSCE Chemistry - Question 28 - 2010 - Paper 1
Step 1
Preparation of 500 mL 0.100 mol L⁻¹ sodium carbonate standard solution
Answer
To prepare a 0.100 mol L⁻¹ sodium carbonate solution, first, calculate the required mass of sodium carbonate (Na₂CO₃). The molar mass of Na₂CO₃ is approximately 105.99 g/mol.
Using the formula:
we determine the mass:
Weigh out 5.30 g of sodium carbonate using a balance and dissolve it in distilled water to make a total volume of 500 mL in a volumetric flask.
Step 2
Titration
Answer
For the titration, use a burette to dispense the hydrochloric acid solution into a conical flask containing the sodium carbonate solution. Ensure to use a suitable indicator, such as phenolphthalein, which changes color at the endpoint of the titration. As you add the hydrochloric acid, swirl the conical flask to mix the solution and add acid dropwise as you approach the endpoint, which is indicated by a color change in the solution. Record the volume of hydrochloric acid used to reach the endpoint, which is noted to be 21.4 mL on average.
Step 3
Concentration of hydrochloric acid solution
Answer
To calculate the concentration of the hydrochloric acid (HCl), we apply the neutralization reaction:
From the balanced equation, 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid. The moles of sodium carbonate in 25.0 mL of the 0.100 mol L⁻¹ solution can be calculated:
Thus, the moles of hydrochloric acid required will be:
To find the concentration of HCl solution:
ext{Concentration of HCl} = rac{ ext{Moles}}{ ext{Volume}} = rac{0.0050 ext{ mol}}{0.0214 ext{ L}} \\ ext{Concentration of HCl} ext{ (mol L}^{-1} ext{)} = 0.233 ext{ mol L}^{-1}