Question

A 250.0-mg sample of an organic weak acid is dissolved in an appropriate solvent and titrated with 0.0556 M NaOH, requiring 32.58 mL to reach the end point. Determine the compound’s equivalent weight.

Answer 1

$\star$ Moles of titrant

The moles of NaOH needed to reach the end point are:

$M_b\times V_b= 0.0556M\times 0.03258L$

$=0.00181$ moles of NaOH

$\star$ Wt. of analyte=250 X10^-3 g

This gives an equivalent weight of analyte:

E .W. = (g analyte/mole titrant)

=$250\times 10^{-3} \over 0.00181$

=$138.1$ g/mol