I can do this step by step if you wish but I remember one neat formula that wraps all of it up together.
mL x N x mew = grams.
24mL x 0.1M x mew = 0.125
Solve for mew. 0.05208 so eq. wt. will be 52.08 (but that's too many significant figures). I looked up the molar mass and found it to be 104.06 so the eq wt would be 52.03
A student attempted to synthesize malonic acid which has a molecular weight of 104
g/mol by the same method used to synthesize aspirin. A titration was performed to find
the equivalent weight. 0.125 g of malonic acid product was used for a titration that
required 24 mL of 0.10 M NaOH. Calculate the equivalent weight from this experiment.
What does this value indicate about the acid?
4 answers
Thank you very much! The formula is very useful,but if it's no trouble for you could you please explain the concept?
On re-reading the problem I note that the problem lists NaOH as 0.1M (and I jumped to the conclusion that was 0.1N). It certainly IS 0.1N but the point is that I assumed the prof was working in normality and not molarity. Therefore, this may not have been a normality problem in the first place; however, since you wanted the concepts here they are. Most schools are phasing out the concept of normality. I'm not in favor of that, because it has such useful tools, but the world is not going that way.
The idea is that 1 equivalent weight of A = 1 equivalent weight B = 1 equivalent weight C = 1 equivalent weight of anything else we can have. (Most of the time we work in mols and we write an equation and convert mols of one thing to mols of another in the equation by using the coefficients in the balanced equation. When using equivalents we don't need to do that; i.e., the equivalent weight takes care of all of that "nonsense".)
Just as mols = M x L (mols/L x L), when we work with equivalents it is equivalents = N x L = normality x liters = equivalents/L x L.
So we have the normality of NaOH and that x volume gives # equivalents or 0.025L x 0.1N = 0.0025 equivalents NaOH = 0.0025 equivalents of malonic acid.
Then just as mols = g/molar mass, it is equivalents = g/equivalent weight(mass).
So equivalent weight = g/equivalents = 0.125/0.0025 = 50
Hope this helps.
The idea is that 1 equivalent weight of A = 1 equivalent weight B = 1 equivalent weight C = 1 equivalent weight of anything else we can have. (Most of the time we work in mols and we write an equation and convert mols of one thing to mols of another in the equation by using the coefficients in the balanced equation. When using equivalents we don't need to do that; i.e., the equivalent weight takes care of all of that "nonsense".)
Just as mols = M x L (mols/L x L), when we work with equivalents it is equivalents = N x L = normality x liters = equivalents/L x L.
So we have the normality of NaOH and that x volume gives # equivalents or 0.025L x 0.1N = 0.0025 equivalents NaOH = 0.0025 equivalents of malonic acid.
Then just as mols = g/molar mass, it is equivalents = g/equivalent weight(mass).
So equivalent weight = g/equivalents = 0.125/0.0025 = 50
Hope this helps.
Thank you so much! This makes much more sense now. Indeed a great tool to know