Answer to Question #235514 in General Chemistry for Lily

Question #235514

A) What mass of H2C2O4 is present in a 25cm3 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.

B) What is the mass concentration of H2C2O4 in the sample?


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Expert's answer
2021-09-10T08:00:12-0400

Solution.

V(NaOH)=18.09mL=18.09103L;V(NaOH)=18.09mL=18.09\sdot10^{-3}L;

Concentration of NaOH =0.2235mol/L;0.2235 mol/L;

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

Amount of NaOH=0.2235mol/L18.09103L=4.043103mol;0.2235mol/L\sdot18.09\sdot10^{-3}L=4.043\sdot10^{-3}mol;

x mol 4.043103mol4.043\sdot10^{-3}mol

H2C2O4+2NaOHNa2C2O4+2H2O;H_2C_2O_4+2NaOH→Na_2C_2O_4+2H_2O;

1 2

Amount of H2C2O4=14.043103mol2=2.0215103mol;H_2C_2O_4=\dfrac{1\sdot4.043\sdot10^{-3}mol}{2}=2.0215\sdot10^{-3}mol;

ν(H2C2O4)=m(H2C2O4)M(H2C2O4)    \nu(H_2C_2O_4)=\dfrac{m(H_2C_2O_4)}{M(H_2C_2O_4)}\implies

m(H2C2O4)=ν(H2C2O4)M(H2C2O4);m(H_2C_2O_4)=\nu(H_2C_2O_4)M(H_2C_2O_4);

M(H2C2O4)=90g/mol;M(H_2C_2O_4)=90g/mol;

m(H2C2O4)=2.0215103mol90g/mol=0.182g;m(H_2C_2O_4)=2.0215\sdot10^{-3}mol\sdot90g/mol=0.182g;

Answer: m(H2C2O4)=0.182g.m(H_2C_2O_4)=0.182g.

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