Question

I'm working on my final lab. I need to figure the molarity of sodium hydroxide by reaction with oxalic acid. My data is as follows:

0.0119g H2C2O4

Net mL 21.05 NaOH

Here's my calculations:

molar mass of oxalic acid: 126.07 g

molar mass of NaOH: 39.99711

Mass of oxalic acid: 0.0119g

number of moles of H2C2O4 x 2H2O= 0.0119g/126.07g=9.4392x10^5

9.4392x10^5 moles of acid reacts with 2 x 9.4392 x 10^-5

moles of base number of moles of base: 1.88784x10^-4

volume of NaOH = 21.05 mL = 0.02105L

NaOH= 1.88784x10^-4/0.02105 L = 0.0089683

Is this correct?

Answer 1

(COOH)2 + 2NaOH --------------> (COONa)2 + H2O

Here we are observing that 1 mole of oxalic acid requires two mole of NaOH

now given = (COOH)2 = 0.0119g , molar mass = 90.03

now moles of oxalic acid = 0.0119/ 90.03 = 0.0001322 moles

now required moles of NaOH = 2 * 0.0001322 = 0.0002644 MOLES

now volume of Naoh = 21.05 ml or 0.02105 ml

now molarity of NaOH = moles /vol(in L)

= 0.0002644/ 0.02105 = 0.01256057 M

you have taken the wrong molar mass of oxalic acid . but your way towards answer is correct .

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