A solution of sodium carbonate of concentration 0.100moldm-3 is used to standardize a solution of hydrochloric acid. 25cm3 of the standard solution of sodium carbonate require 35cm3 of the acid for neutralization. Calculate the concentration of the acid.
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols Na2CO3 used = M x L = 0.100 x 0.025 L = ?
mols HCl needed = 2 x mols Na2CO3 (from the coefficients in the balanced equation.
M HCl = mols HCl/0.035 L= ?
To calculate the concentration of the hydrochloric acid solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl).
The balanced chemical equation for this reaction is:
Na2CO3 + 2 HCl -> 2 NaCl + CO2 + H2O
First, let's determine the number of moles of sodium carbonate used:
Number of moles of sodium carbonate = concentration (in moldm^-3) × volume (in dm^3)
Number of moles of sodium carbonate = 0.100 moldm^-3 × (25/1000) dm^3
Number of moles of sodium carbonate = 0.0025 moles
Now, using the stoichiometry of the balanced equation, we can determine the number of moles of hydrochloric acid used in the reaction:
From the balanced equation, we see that it takes 2 moles of HCl to react with 1 mole of Na2CO3.
Therefore, the number of moles of HCl = 0.0025 moles × (2/1) = 0.005 moles
Next, we need to calculate the concentration of the hydrochloric acid solution:
Concentration of hydrochloric acid = number of moles of HCl / volume of HCl solution (in dm^3)
Concentration of hydrochloric acid = 0.005 moles / (35/1000) dm^3
Concentration of hydrochloric acid = 0.1429 moldm^-3
Therefore, the concentration of the hydrochloric acid solution is 0.1429 moldm^-3.