posted by jhon .
If a 2.6 g of weak diprotic acid were dissolved in 100 mL of water and a 10 mL aliquot of this solution required 21.6 mL of 0.1 M NaOH to reach the first endpoint, what are the equivalent and formula weights of H2A?
I think I first need to find the moles of NaOH. Then, by using the formula.
H2A + 2NaOH ----> Na2A + 2 H20
I am stuck from here. Do I find the equivalence point of each and add them?
The easy way to do this is
mL x N x millieq wt = grams.
You had 2.6g in 100 and took a 10 mL aliquot to titrate; therefore, the mass of the sample titrated was 0.26g
So 21.6 x 0.1 x mew = 0.26.
Solve for mew. Multiply by 1000 to find equivalent weight and multiply by 2 to find the formula mass. I am so sorry that texts/IUPAC and others have done away with normality. One step solved this problem. It would have taken half a page and multiple steps to do it the "new" way.
This topic's pretty old, but I figured it would be worth it to have the new way to solve the problem posted as well. If you're like me you're a young whippersnapper who's never even heard of normality.
It's actually not as hard as DrBob says.
Molarity = moles/liter, so convert your NaOH added to liters then plug everything in like so:
0.1 M = moles/0.0216 L
This will give you the moles of NaOH in the solution. From the balanced equation you know there are 2 moles NaOH per 1 mol of H2A, so dividing moles of NaOH by 2 will give you moles of H2A.
The formula weight is simply the grams of H2A in the solution (0.26g) divided by the moles of H2A you just calculated. Divide this answer by 2 to find the equivalent weight (because H2A has 2 equivalents, or H+ ions).