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.
3 answers
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.
333
1mmolNH17.031gNH0.1943mmolHCl× 41.27mLHCl × ×mL1mmolHCl1000mmol × 100%50.00mL × 25.00mL250.0mL = 2.731% (w/v)NH
1mmolNH17.031gNH0.1943mmolHCl× 41.27mLHCl × ×mL1mmolHCl1000mmol × 100%50.00mL × 25.00mL250.0mL = 2.731% (w/v)NH
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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