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### Topic: How would I calculate the concentration of malonic acid through titration?  (Read 9313 times)

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#### a_huynh00

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##### How would I calculate the concentration of malonic acid through titration?
« on: April 11, 2009, 08:42:15 PM »
Malonic acid (H2C3H2O4) is a diprotic acid. In the titration of malonic acid with NaOH, stoichiometric (aka endpoint) points occur at pH = 3.9 and 8.8. A 25.00-mL sample of malonic acid of unknown concentration is titrated with 0.0984M NaOH, requiring 31.50 mL of the NaOH solution to reach the phenolphthalein end point.

Calculate the concentration of malonic acid in the unknown solution.

The answer is 0.0620 M Malonic Acid but I don't know how to arrive to that answer.

This is what I've figured out so far.

Ka1 = 1.5 x 10^-3
Ka2 = 2.0 x 10^-6

H2C3H2O4 + OH- ↔ H2O + HC3H2O4- ==> Ka1=[HC3H2O4-]/[H2C3H2O4][OH-]
HC3H2O4- + OH- ↔ H2O + C3H2O4-- ==> Ka2=[C3H2O4--]/[HC3H2O4-][OH-]

I tried setting up rice tables but I still don't know how to figure out the initial concentration of malonic acid.

#### a_huynh00

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##### Re: How would I calculate the concentration of malonic acid through titration?
« Reply #1 on: April 11, 2009, 09:27:01 PM »
K so I figured

phenolphthalein has a pKa about 9 so the end point represents the loss of 2 protons from the acid

at this eq point the moles of NaOH = .0984M * 0.0315L

= 3.1*10^-3 moles

number of moles of acid at that point = 1.5 * 10^-3 in 0.025L so the mole in a liter = 1.5*10^-3*0.025L

= .0600 M

But is this right?

#### Borek

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##### Re: How would I calculate the concentration of malonic acid through titration?
« Reply #2 on: April 12, 2009, 05:33:28 AM »
= .0600 M

Correct approach, but incorrect result. You should not round down intermediate results. Do the calculations with full precision and you will get 0.0620 (0.061992, rounded down to three significant digits 0.0620).
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