Question

if 24.83cm3 of modm-3 NAOH is titrated to it's end point with 39.45cm3 of HCL, what is the molar concentration of the HCL.

Answer 1

Volume of $NaOH(V_1)=24.83 \ cm^3$

Concentration of $NaOH(M_1)=mol/dm^3=10^{-3} mol/cm^3$

Volume of $HCl(V_2)=39.45 \ cm^3$

Using molarity equation,$M_1V_1=M_2V_2$

$10^{-3}\times 24.83=39.45\times M_2$

$M_2=0.6294\times 10^{-3} mol/cm^3=0.6294 mol/dm^{3}=0.6294 mol/L$