Qs: A 0.5033g sample of an unknown salt was dissolved in waterand passed through
ID: 685945 • Letter: Q
Question
Qs: A 0.5033g sample of an unknown salt was dissolved in waterand passed through a cation exchange resin. The total elute fromthe colum was diluted to 250ml and a 50ml aliquot was titrated with11.35mL of 0.10359 M sodium hydroxide. Calculate the percentage ofammonium ion in the sample. Answer:21.07% I know that answer for this Qs is 21.07% but Ireally don't understand how it works.. help me please ?? Qs: A 0.5033g sample of an unknown salt was dissolved in waterand passed through a cation exchange resin. The total elute fromthe colum was diluted to 250ml and a 50ml aliquot was titrated with11.35mL of 0.10359 M sodium hydroxide. Calculate the percentage ofammonium ion in the sample. Answer:21.07% I know that answer for this Qs is 21.07% but Ireally don't understand how it works.. help me please ??Explanation / Answer
NH4+ + OH- => NH3 + H2O 0.01135 L * ( 0.10359 mol OH-/ L) * ( 1 mol NH4+ / 1mol OH-) * (18.03829 g /mol NH4) = 0.0212084563 grams NH4+ (per 50 mL) 250 mL * ( 0.0212084563 g NH4+/ 50 mL) = 0.106042282 g NH4+ 0.106042282 g NH4+ / 0.5033 g = 0.210693984 ~ 21.07 %