Asked by Ellen
A 25.00mL sample of a household cleaning solution was diluted to 250.0mL in a volumetric flask. A 50.00mL aliquot of this solution required 40.38mL of 0.2506 M HCl to reach a bromocresol green end point. Calculate the weight/volume percentage of NH3 in the sample. (Assume that all the alkalinity results from the ammonia.)
Can someone please explain how to get the volume when calculating for the weight/volume percentage. I know how to get the weight but don't know how to get the volume for this question. Please explain it throughly. Thanks.
Can someone please explain how to get the volume when calculating for the weight/volume percentage. I know how to get the weight but don't know how to get the volume for this question. Please explain it throughly. Thanks.
Answers
Answered by
DrBob222
You took a 25.00 mL sample and diluted t 250.0 mL and titrated a 50.00 mL aliquot. You have the weight in the 50.00 mL sample. The weight in the original 25.00 mL sample is weight in titrated sample x (250/50) = ? Now you have the weight in the original 35.00 mL sample and that is the volume you use.
Answered by
Lance
333
1mmolNH17.031gNH0.1943mmolHCl× 41.27mLHCl × ×mL1mmolHCl1000mmol × 100%50.00mL × 25.00mL250.0mL = 2.731% (w/v)NH
Answered by
Lance
NH3 + HCl = NH4Cl
[(0.1943mmol/mL HCl)× 41.27mL HCl × (1mmol NH3/ 1mmol HCl) x 17.031g NH3 /1000mmol NH3)] / [50.00mL × 25.00mL250.0mL ] x100% = 2.731% (w/v)NH3
[(0.1943mmol/mL HCl)× 41.27mL HCl × (1mmol NH3/ 1mmol HCl) x 17.031g NH3 /1000mmol NH3)] / [50.00mL × 25.00mL250.0mL ] x100% = 2.731% (w/v)NH3
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