Question

A 25.00-ml sample of propionic acid, hc3h5o2, of unknown concentration was titrated with 0.141 m koh. the equivalence point was reached when 37.48 ml of base had been added. what is the concentration of the propionate ion at the equivalence point?
a. 0.211 m
b. 0.147 m
c. 0.141 m
d. 0.0846 m
e. 0.128 m

Answer 1

 The concentration of propionate ions   at  equivalent point is  0. 084 M
  calculation
write the equation for reaction
HC3H5O2   +  KOH  → H2O  + KC3H5O2

find the moles  of  KOH  used
moles=  molarity  × volume in liters
volume  in liters = 37.48 ml/1000 =0.03718L
molarity = 0.141 M


moles is therefore =  0.141 M  x0.03718 =  5.24238  x10^-3 moles

by use of mole  ratio between  KOH  and K3C3H5O2  which is 1:1  the  moles of K3C3H5O2  is also = 5.24238 x10^-3  moles

molarity is therefore = number  of moles/volume in liters
volume  in liters = 25/1000 +37.48/1000=0.06248 L

molarity of  propionate ions is therefore = (5.24238 x10^-3) / 0.06248 = 0.084  M