Acid present in milk, need to check

[Sussan_Potts]

A titration was performed in order to find the amount of lactic acid within milk.
When we did the titration, 5.8mL of 0.1 NaOH was used before the endpoint was found.
There was 20mls of milk.

We’ve tried to do the calculations but were not sure.

Chemical formula of Lactic acid is CHH₆₀₃, mass is 90.049

Here’s our working out:

Calculating the moles of base used
5.8/1000 = 0.0058 (converting to liters)

0.1/1000 = 0.0001 (do I need this step, I don’t know why it is here)

0.0058*0.0001 = 0.00000058

Moles of base = moles of acid

Calculating mass of lactic acid
0.00000058 * 90.049 = 0.00005222842

Finding the percentage
20/1000 (converting to liters)
= 0.02

0.00005222842 / 0.02 = 0.002611421

*100 = 0.26%

Is this calculation correct? It seems kind of small.

[Zedekiah]

You do not need the step where you divide the molarity of the base by 1000.

Multiply the number 0.00000058 by 1000 to get the number of moles.

0.00058 moles.

Yep, moles of acid = moles of base through the miracle of titration .

Hmmm, yep, that is one of doing it.

0.05222842 grams is the new solution when you have the modified molarity.

I think this is another minor error. Now that you have grams of lactic acid, all you have to do now is divide by grams of milk total to find the mass percent of lactic acid that you have. Without the density, however, this is not possible. You could approximate and get the answer by dividing

0.0522 by 20, because it is probably about 1.0 grams per ml.

I got 0.26% as my answer. You divided by 1000 and then multiplied by a 1000 so the errors cancelled.

You're right, it seems too low.

[Zedekiah]

Another way of doing this would be to solve for the molarity of the milk and use the density to find the % concentration.

0.00058 / .02 = 0.029 M Milk has a reported density of 1.028 g / mL.

Since (10cd)/W = M, you can rearrange this equation to find c, the % concentration.

MW/10d = c

(90.049*0.029)/10.28 = c

c = .25%. Heh, close to the other answer.