A sewage cleanser contains concentrated hydrochloric acid (HCl) as the active ingredient. A titration
experiment was carried out to determine the concentration of hydrochloric acid in the domestic
cleanser.
25.0 mL of the cleanser were diluted to 250.0 mL with distilled water. 25.0 mL of the diluted cleanser
were then titrated against 0.5 mol L-1 sodium hydroxide solution. 20.00 mL of sodium hydroxide
solution were required for complete neutralization.
(a) Calculate the molarity of hydrochloric acid in the cleanser. (7 marks)
(b) Calculate the concentration, in g dm–3, of hydrochloric acid in the cleanser.
(a) 0.5 * (20.00 / 25.0) * (250.0 / 25.0)
(b) multiply the answer from (a) by the molar mass of HCl
The titration is
HCl + NaOH ==> NaCl + H2O
mols NaOH used = M x L = approx 0.01 but you need to confirm that.Since the ratio in the titration is 1 mol NaOH to 1 mol HCl, that means mols HCl in that 25 mL aliquot is the same as mols NaOH used to titrate it. Since this was a diluted sample of 25 to 250 and you took 25, that means mols in the original 25.0 mL = 10x that or 0.1.
M HCl = mols/L (= mols/dm^3)= 0.1/0.025 = ?
To convert mol/dm^3 to g/dm^3, convert mols to grams. grams HCl = mols HCl x molar mass HCl.
To calculate the molarity of hydrochloric acid in the cleanser, we need to use the information given in the problem.
(a) Calculate the molarity of hydrochloric acid in the cleanser:
1. First, calculate the moles of sodium hydroxide (NaOH) used in the titration.
Moles of NaOH = Molarity of NaOH x Volume of NaOH solution used (in liters)
Given: Molarity of NaOH = 0.5 mol L-1
Volume of NaOH solution used = 20.00 mL = 0.02000 L
Moles of NaOH = 0.5 mol L-1 x 0.02000 L = 0.0100 mol
2. The balanced chemical equation for the neutralization reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH.
3. Since the moles of NaOH used in the titration are equal to the moles of HCl in the cleanser, the molarity of HCl can be calculated as:
Molarity of HCl = Moles of HCl / Volume of HCl solution used (in liters)
Given: Volume of HCl solution used = 0.0250 L (diluted cleanser volume)
Molarity of HCl = 0.0100 mol / 0.0250 L = 0.400 mol L-1
Therefore, the molarity of hydrochloric acid in the cleanser is 0.400 mol L-1.
(b) Calculate the concentration, in g dm–3, of hydrochloric acid in the cleanser:
1. Calculate the moles of HCl in the cleanser using the molarity and volume of HCl solution.
Moles of HCl = Molarity of HCl x Volume of HCl solution used (in liters)
Moles of HCl = 0.400 mol L-1 x 0.0250 L = 0.0100 mol
2. Calculate the mass of HCl in grams using the molar mass of HCl.
Molar mass of HCl = 1.007 g/mol (hydrogen) + 35.453 g/mol (chlorine) = 36.460 g/mol
Mass of HCl = Moles of HCl x Molar mass of HCl
Mass of HCl = 0.0100 mol x 36.460 g/mol = 0.3646 g
3. Calculate the concentration of HCl in grams per decimeter cubed (g dm-3) using the diluted cleanser volume.
Concentration = (Mass of HCl / Volume of HCl solution used) x 1000
Concentration = (0.3646 g / 0.0250 L) x 1000 = 14.58 g dm-3
Therefore, the concentration of hydrochloric acid in the cleanser is 14.58 g dm-3.