5% what? w/w or w/v. I'll assume w/v
5% HCOOH is 5 g HCOOH/100 mL.
mols in 5g is 5/48 = about 0.1 mol and that in 0.1 L is 1 M.
5% CH3COOH is 5g/100 mL.
mols in 5g CH3COOH is 5/60 = about 0.08 and that in 0.1L is 0.8 M
So the reach the eq point you will have
mols HCOOH = M x L = ?
mols CH3COOH = M x L = ?
I assume you can take it from here.
you have 5% formic acid solution and a 5% acetic acid solution. You titrate 5mL each with .1868M NaOH.
A. Which will need more NaOH to reach the end point?
B. How much more?
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