A #10*mL# volume of oxalic acid was titrated with a #34.40*mL# volume of #NaOH(aq)# whose concentration was #0.250*mol*L^-1#. What was the mass of oxalic acid present in the initial #10*mL# volume?

1 Answer
Jul 31, 2017

Answer:

#[C_2O_4H_2]~=0.5*mol*L^-1#.

Explanation:

We need (i) a stoichiometric equation.....

#"HO(O=)CC(=O)OH + 2NaOH"rarrC_2O_4^(2-)Na_2^+ + 2H_2O#

And (ii) equivalent quantities of sodium hydroxide......

#"Moles of NaOH"=34.40*mLxx10^-3*L*mL^-1xx0.250*mol*L^-1=8.60xx10^-3*mol#.

And HALF this molar quantity was equivalent to the oxalic acid solution contained in #10.00*mL# solution.

#"Concentration of oxalic acid"=(1/2xx8.60xx10^-3*mol)/(10.00xx10^-3*L^-1)#

#=0.430*mol*L^-1#

And thus in a #10*mL# volume, there were #10xx10^-3*Lxx0.430*mol*L^-1xx90.03*g*mol^-1=#

#~=0.4*g# #""#