Asked by Ellen
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.
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.
Answers
Answered by
DrBob222
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.
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.
Answered by
Erika
24.4%
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