A 15.9 mL solution of 0.312 mol L-1 HCOOH is titrated using 0.376 mol L-1 NaOH. What volume of NaOH (in mL) is needed to reach the equivalence point in this experiment?
Let us write the equation
HCOOH + NaOH ====> HCOONa + H2O
One mole of formic acid and sodium hydroxide reacts to give one mole of sodium formate and one mole of water .
Volume of HCOOH = 15.9 ml
Concentration of HCOOH = 0.312 Mole L-1
Moles of HCOOH reacted = 15.9 ml x 0.312 Mole/ 1000 ml = 0.0049608 Moles
Moles of NaOH needed = 0.0049608 Moles
Volume of NaOH = 0.0049608 Moles x 1000 ml / 0.376 Moles = 13.19 ml
Hence 13.19 ml of 0.376 M NaOH is needed to reach the equivalence point in this experiment
A 15.9 mL solution of 0.312 mol L-1 HCOOH is titrated using 0.376 mol L-1 NaOH....