If 5.00 mL of the citric acid, H3C6H5O7, required 47.8 mL of 0.121 M NaOH for neutralization....

Question

What was the molarity of the citric acid in the H3C6H5O7?

The reaction is:

3 NaOH H3C6H5O7 --> 3H2O Na3C6H5O7

the answer is 0.386 M, But please give the equation and steps?

when I tried this one equation the closest answer I got was .394...

): My exam is tomorrow evening, and I'm having a lot of trouble with this, so much so it's hard to focus on the other problems I need to study.

Answer 1

The reaction indicates that citric acid is tribasic.  Since 47.8 x 0.121 = 5.7838 millimoles of NaOH were used, 5.7838/3 = 1.9279 millmoles of citric acid were present.  Multiply this by 1000/5 to bring the volume to 1 liter, and get a molarity of 0.38558.

Answer 2

3NaOH + H3C6H5O7 --> 3H2O + Na3C6H5O7

Molarity = moles / volume

moles = Molarity X volume

moles of NaOH = 0.121 M X 0.0478 L = 0.0058 mols

According to the equation(Stoichiometry):

0.0058mols /3(NaOH) X 1(H3C6H5O7) = 0.0019 mols of Citric Acid.

Molarity  = mols / volume = 0.0019mols / 0.005L = 0.386 M