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
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.

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  1. (a) 0.5 * (20.00 / 25.0) * (250.0 / 25.0)

    (b) multiply the answer from (a) by the molar mass of HCl

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  2. 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.

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