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

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Answer 1 · 2017-07-31T03:54:13

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

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 $10xx10^-3*Lxx0.430*mol*L^-1xx90.03*g*mol^-1=$

$~=0.4*g$ $""$

A 10*mL10*mL volume of oxalic acid was titrated with a 34.40*mL34.40*mL volume of NaOH(aq)NaOH(aq) whose concentration was 0.250*mol*L^-10.250*mol*L^-1. What was the mass of oxalic acid present in the initial 10*mL10*mL volume?