posted by Ellen on .
The formaldehyde content of a pesticide preparation was determined by weighing 0.3124g of the liquid sample into a flask containing 50.0mL of 0.0996M NaOH and 50mL of 3% H2O2. Upon heating, the following reaction took place:
OH- + HCHO + H2O2 -> HCOO- + 2H2O
After cooling, the excess base was titrated with 23.3mL of 0.05250M H2SO4. Calculate the percentage of HCHO (30.026g/mol) in the sample.
Please explain throughly step by step on how to do this question. Thanks.
50.00 mL x 0.0996M NaOH = millimoles NaOH initially present(call that #1).
Then the formaldehyde used part of it.
You titrated the excess base with
23.3 mL x 0.05250M H2SO4 = ? millimoles H2SO4. Remembering that 1 mole H2SO4 x 2 = mols NaOH (call that #2).
mols NaOH initially (#1)-moles excess NaOH(#2) = moles used up by the formaldehyde reaction. Use the equation to convert mols NaOH to mols formaldehyde and go from there.