1.06g of sodium trioxocarbonate(iv) reacted with excess dilute 0.1M hydrochloric acid, after the reaction, the unreacted acid required 24cm3 of 0.1M sodium hydroxide for its complete neutralization. Calculate the original volume of the hydrochloric acid used.

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  1. First you need to know that Na2CO3 is sodium carbonate and not that funny name you have. The reaction with HCl is
    Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
    mols Na2CO3 = grams/molar mass = 1.06/106 = 0.01 mols.]
    It will take twice that number of mols to react EXACTLY with the HCl so that is 0.02 mols HCl needed for the 1.06 g Na2CO3.

    The excess HCl required 24 cc of 0.1 M NaOH or mols NaOH = 0.024 x 0.1 = 0.0024 mols.
    NaOH + HCl ==> NaCl + H2O
    0.0024 mols NaOH must take 0.0024 mols HCl for the excess SO you must have had the 0.02 mols initially plus the 0.0024 excess or a total to start with of 0.0024 + 0.02 = 0.0224 mols. Then M = mols/L. You know mols and M, solve for L and conertr to mL. Post your work if you get stuck.

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