Question

How much 12M HCl is needed to quench a setup consisting of .250mg benzoic acid (FW=122.12) with 10mL 3M NaOH? Please show all work and express the answer in mL.

Answer 1

C6H5COOH + NaOH   --------------------> C6H5COONa +   H2O

    1                         1                                          1

mass of benzoic acid = 0.250mg = 2.5 x 10^-4 g

molar mass = 122.12 g/mol

moles = 2.5 x 10^-4 / 122.12

          = 2.047 x 10^-6

moles of NaOH = 10 x 3 / 1000 = 0.03

moles of NaOH remained = 0.02999

moles of HCl = moles of NaOH remained

12 x V = 0.02999

V = 2.5 x 10^-3 L

volume of HCl = 2.5 mL