Find the percent acid (eq wt 173.8) if 20.07 mL of 0.1100 N base is required to neutralize 0.721 g of a sample.
mL x N x milliequivalent weight = grams
20.07 x 0.1100 x 0.1738 = 0.3837
% acid = (0.3837/0.721)*100 = ?
These can be combined into one step as follows:
% acid = (mL x N x mew/mass sample)*100 = (20.07 x 0.1100 x 0.1738/0.721)*100 = ?
The International Union of Pure and Applied Chemistry (IUPAC) recommended, about 30 or more years ago, that equivalent weights not be used because the method was redundant. I'm pleased to see someone is using the method for I think it is easier, simpler, more straightforward, and frankly more esoteric. Way to go!