Question

During a titration, 0.200 M HCl is added to a NaOH solution of unknown concentration. What is the concentration of the NaOH solution if 20.0mL of it is neutralized by 30.7mL of the standard solution? Show your work.

Answer 1

The amount of mole at the neutralization point is equal for both components of the reaction:

$n_1=n_2$

$c_{NaOH}V_{NaOH}=c_{HCl}V_{HCl}$

$c_{NaOH} = \frac{c_{HCl}V_{HCl}}{V_{NaOH}} = \frac{0.200 mol/L\times 30.7 ml}{20.0 ml} = 0.307 mol/L$