To determine the original volume of the hydrochloric acid used, we need to consider the balanced chemical equation for the reaction between sodium trioxocarbonate(IV) (Na2CO3) and hydrochloric acid (HCl):
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
From the balanced equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, we need to calculate the number of moles of Na2CO3 that reacted.
Given that the mass of Na2CO3 is 1.06g, we can calculate the number of moles of Na2CO3 using its molar mass:
Molar mass of Na2CO3 = 2(23.0) + 12.0 + 3(16.0) = 106.0 g/mol
Number of moles of Na2CO3 = mass / molar mass = 1.06g / 106.0 g/mol = 0.01 mol
Since Na2CO3 and HCl react in a 1:2 ratio, we have twice the number of moles of HCl compared to Na2CO3. Therefore, the number of moles of HCl is 2 times the number of moles of Na2CO3.
Number of moles of HCl = 2 * 0.01 mol = 0.02 mol
Now, we can use the given information about the unreacted acid requiring 24cm3 of 0.1M sodium hydroxide for complete neutralization.
From the balanced equation, we can see that 2 moles of HCl react with 2 moles of NaOH. Therefore, the number of moles of HCl is equal to the number of moles of NaOH used to neutralize the unreacted HCl.
Number of moles of NaOH = concentration * volume
Number of moles of NaOH = 0.1 mol/L * 0.024 L = 0.0024 mol
Since the number of moles of HCl is equal to the number of moles of NaOH, we have:
Number of moles of HCl = 0.0024 mol
Now, we can calculate the original volume of the hydrochloric acid used:
Original volume of HCl = number of moles / concentration
Original volume of HCl = 0.0024 mol / 0.1 mol/L = 0.024 L = 24 cm3
Therefore, the original volume of the hydrochloric acid used is 24 cm3.