chem standardization

At STP (22.4dm), 560cm^3 of CO2 at STP were dissolved in 0.1 dm^3 of NaOH solution. The resulting solution required 50cm^3 of a molar solution of HCl for neutralization. Find the molarity of the alkali.


Consider the reaction between CO2 and NaOH.
CO2 + 2NaOH → Na2CO3 + H2O
Mole ratio CO2 : NaOH = 1 : 2
No. of moles of CO2 = 560/22400 = 0.025 mol
No. of moles of NaOH reacted with CO2 = 0.025 x 2 = 0.05 mol

Consider the reaction between HCl and excess NaOH.
HCl + NaOH → NaCl + H2O
Mole ratio HCl : NaOH = 1 : 1
No. of moles of HCl = 1 x (50/1000) = 0.05 mol
No. of moles of NaOH reacted with HCl = 0.05 mol

Total no. of moles of NaOH = 0.05 + 0.05 = 0.1 mol
Volume of NaOH solution = 0.1 dm³
Molarity of HCl = 0.1/0.1 = 1 M