25cm3 of a 0.2 mol dm3 solution of Na2CO3 requires 20 cm3 of a solution of HCl for neutralization. The concentration of the HCl solution is

Answer Details

The question describes a neutralization reaction between a solution of Na2CO3 and HCl. The volume of HCl required for neutralization is given as 20 cm3. From the balanced equation for the reaction, we know that 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, the moles of HCl required for neutralization can be calculated as follows: moles of HCl = (moles of Na2CO3) x 2 moles of Na2CO3 = concentration x volume / 1000 Substituting the given values, we get: moles of Na2CO3 = 0.2 x 25 / 1000 = 0.005 moles of HCl = 0.005 x 2 = 0.01 The concentration of HCl solution can then be calculated by dividing the moles of HCl by the volume of HCl solution used: concentration of HCl = moles of HCl / volume of HCl solution concentration of HCl = 0.01 / 20 / 1000 concentration of HCl = 0.5 mol dm^-3 Therefore, the concentration of the HCl solution is 0.5 mol dm^-3.