Question

An organic acid concentration in a hard sample was determined by diluting a 15.00 mL sample of 200.00 mL solution. A 20.00 mL portion of the diluted sample reacted with 40.0 mL of 0.05417 M NaOH completed. The excess OH- was back titrated with 5.21 mL of 0.02154 M HCL. The concentration of the organic acid in the orginal sample is:

Answer is 1.370 M

Answer 1

we know that

moles = molarity x volume (ml) / 1000

so

moles of HCl added = 0.02154 x 5.21 / 1000

moles of HCl added = 1.122234 x 10-4

now

first consider the reaction of OH- and HCl

the reaction is

H+ + OH- ---> H20

we can see that

moles of H+ added = moles of OH-

so

excess moles of OH- = 1.122234 x 10-4

now

moles of NaOH taken = 0.05417 x 40 / 1000 = 2.1668 x 10-3

so

moles of OH- reacted with acid = moles of NaOH taken - excess moles of OH-

moles of OH- reacted with acid = 2.1668 x 10-3 - ( 1.122234 x 10-4)

moles of OH- reacted with acid = 2.0545766 x 10-3

we know that

moles of acid present = moles of OH- reacted

so

moles of acid in diluted sample = 2.0545766 x 10-3

volume of diluted sample taken = 20 ml

so

conc of diluted sample = moles x 1000 / volume (ml)

conc of diluted sample = 2.0545766 x 10-3 x 1000 / 20

conc of diluted sample = 0.10272883

now

we know that

for dilution

M1 x V1 = M2 x V2

so

M1 x 15 = 0.10272883 x 200

M1 = 1.3697

so

the concentration of the original sample is 1.370 M