QUESTIONS 1. A0.821 gram sample of pure NH,F was treated with 25.0 mL of 1.00 M...

Transcribed Image Text:

QUESTIONS 1. A0.821 gram sample of pure NH,F was treated with 25.0 mL of 1.00 M NAOH and heated to drive off the NH,. How many mols of NaOH are needed to react with the NH,F? Molar Mass of NH4F = 37.037 grams a. 0.82) %3D 0.02216 ニ molar wass 37.037 0:0 222 mols| b. Нow many milliliters of 0.500 M HCI are needed for thc back titration? l,00mol 25.0 mL > 0.025 mol 1000 m/L し 0.00281 mol 0.500 mol /L 0.025 - 0.0222 = 0,00562L 0.00562L x 1000 mL/m 2. How is the calculated molarity of the NaOH affected (higher, lower, or unaffected) if the KHP used for standardization is contaminated with dirt (i.e., is not 100% pure)? Explain. If the KHP is contaminated, the actual mass of kHP useol Will be less then the meacured oot mass , So ty moles of 5.62mL %3D KHP will be higher resulting in a NAOH Moles.'Se the molaiity of NaOH will be higher. higher nomber of 3. Assume that 0.160 g of an unknown ammonium salt, NH X, is treated with 23.00 mL of 0.200 M NaOH, and 30.00 mỉ of 0.0500 M H,C,O, are necded for the back titration. Calculate the molar mass of the ammonium salt. 23 x 0.2 = 4.6= Torial Aa OH Token %3D 30 ml x 0.05M 2 = 3.0 4.6 -3=1,6 NG OH reacteal with ammouium salF 0.160g/larmass = l,l6 x 103M) Molar Mass =| 100 gmal 105 1.6 mo-3 QUESTIONS 1. A0.821 gram sample of pure NH,F was treated with 25.0 mL of 1.00 M NAOH and heated to drive off the NH,. How many mols of NaOH are needed to react with the NH,F? Molar Mass of NH4F = 37.037 grams a. 0.82) %3D 0.02216 ニ molar wass 37.037 0:0 222 mols| b. Нow many milliliters of 0.500 M HCI are needed for thc back titration? l,00mol 25.0 mL > 0.025 mol 1000 m/L し 0.00281 mol 0.500 mol /L 0.025 - 0.0222 = 0,00562L 0.00562L x 1000 mL/m 2. How is the calculated molarity of the NaOH affected (higher, lower, or unaffected) if the KHP used for standardization is contaminated with dirt (i.e., is not 100% pure)? Explain. If the KHP is contaminated, the actual mass of kHP useol Will be less then the meacured oot mass , So ty moles of 5.62mL %3D KHP will be higher resulting in a NAOH Moles.'Se the molaiity of NaOH will be higher. higher nomber of 3. Assume that 0.160 g of an unknown ammonium salt, NH X, is treated with 23.00 mL of 0.200 M NaOH, and 30.00 mỉ of 0.0500 M H,C,O, are necded for the back titration. Calculate the molar mass of the ammonium salt. 23 x 0.2 = 4.6= Torial Aa OH Token %3D 30 ml x 0.05M 2 = 3.0 4.6 -3=1,6 NG OH reacteal with ammouium salF 0.160g/larmass = l,l6 x 103M) Molar Mass =| 100 gmal 105 1.6 mo-3