Question

What mass of H2C2O4 is present in a sample if it is titrated to its equivalence point with
18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows:
H2C2O4+2NaOH→Na2C2O4+2H2O

Answer 1

Solution.

$V(NaOH)=18.09mL=18.09\sdot10^{-3}L;$

Concentration of NaOH =$0.2235 mol/L;$

Amount of NaOH (mol) = Concentration of NaOH (mol/L) • Volume of NaOH consumed (L);

Amount of NaOH=$0.2235mol/L\sdot18.09\sdot10^{-3}L=4.043\sdot10^{-3}mol;$

x mol $4.043\sdot10^{-3}mol$

$H_2C_2O_4+2NaOH→Na_2C_2O_4+2H_2O;$

1 2

Amount of $H_2C_2O_4=\dfrac{1\sdot4.043\sdot10^{-3}mol}{2}=2.0215\sdot10^{-3}mol;$

$\nu(H_2C_2O_4)=\dfrac{m(H_2C_2O_4)}{M(H_2C_2O_4)}\implies$

$m(H_2C_2O_4)=\nu(H_2C_2O_4)M(H_2C_2O_4);$

$M(H_2C_2O_4)=90g/mol;$

$m(H_2C_2O_4)=2.0215\sdot10^{-3}mol\sdot90g/mol=0.182g;$

Answer: $m(H_2C_2O_4)=0.182g.$