Question

50mL of 0.010M NaOH were titrated with 0.50M HCl, added with a dropper pipet. If each drop has a volume of 0.040mL, calculate how many drops of HCl are needed to reach the end-point of the titration.

Answer 1

Number of moles of NaOH , n = Molarity x volume in L

                                           = 0.010 M x 0.050 L

                                           = 0.0005 mol

The reaction between NaOH & HCl is

     NaOH + HCl   NaCl + H2O

According to the balanced equation ,

1 mole of NaOH reacts with 1 mole HCl

0.0005 mol of NaOH reacts with 0.0005 mol HCl

So Volume of HCl required is , V = number of moles / Molarity

                                               = 0.0005 mol / 0.50 M

                                               = 0.001 L

                                              = 1.0 mL

So 1.0 mL of HCl is enough for complete neutralization.

Given 1 drop has volume = 0.040 mL

So number of drops in 1.0 mL are ( 1.0 / 0.040) = 25 drops

Therefore the number of drops of HCl needed to reach the end-point are 25 drops