Oxalic acid is a solid organic acid that can donate two protons and has the molecular formula H2C2O4. How many milliliters of 0.050 M NaOH are needed to neutralize both acidic protons in 0.20 g of oxalic acid? You must show all your calculations to receive credit. 

 

Oxalic acid is a solid organic acid that can donate two protons and has the molecular formula H2C2O4. How many milliliters of 0.050 M NaOH are needed to neutralize both acidic protons in 0.20 g of oxalic acid? You must show all your calculations to receive credit. 

ANSWER

Moles of oxalic acid ued = mass used/molar mass

Molar mass of oxalic acid= 90.03 g/mol

Moles of oxalic acid ued = 0.20 g ÷ 90.03 g/mol

=0.0022moles

Equation for the reaction is;

H₂C₂O_4(aq) + 2 NaOH_(aq) → Na₂C₂O_4(aq) + 2 H₂O_(l)                

The equation shows that the reacting mole ratio of H₂C₂O₄ : NaOH is 1:2

The moles of the acid used is;

H₂C₂O₄ : NaOH

    1     : 2

0.0022 : X

X=(2×0.0022)÷1

= 0.0044 moles

Volume of NaOH used in the reaction is;

0.050 moles are present in 1000 mL

0.0044 moles are present in X mL

X =(0.0044 moles × 1000 mL)÷ 0.050 moles

= 88 mL

Step-by-step explanation

Oxalic acid is a solid organic acid that can donate two protons and has the molecular formula H2C2O4. How many milliliters of 0.050 M NaOH are needed to neutralize both acidic protons in 0.20 g of oxalic acid? You must show all your calculations to receive credit. 

ANSWER

Moles of oxalic acid ued = mass used/molar mass

Molar mass of oxalic acid= 90.03 g/mol

Moles of oxalic acid ued = 0.20 g ÷ 90.03 g/mol

=0.0022moles

Equation for the reaction between oxalic acid and sodium hydroxide is used to calculate moles of acid used. The equation is;

H₂C₂O_4(aq) + 2 NaOH_(aq) → Na₂C₂O_4(aq) + 2 H₂O_(l)                

The equation shows that the reacting mole ratio of H₂C₂O₄ : NaOH is 1:2

The moles of the acid used is;

H₂C₂O₄ : NaOH

    1     : 2

0.0022 : X

X=(2×0.0022)÷1

= 0.0044 moles

Volume of NaOH used in the reaction is;

0.050 moles are present in 1000 mL

0.0044 moles are present in X mL

X =(0.0044 moles × 1000 mL)÷ 0.050 moles

= 88 mL