A mass of ).497-g sulfamic acid, NH2SO3H, dissolved in 50.0 mL of water is neutralized by 28.4 mL of NaOH at the phenolphthalein endpoint. WHat is the molarity of the NaOH solution? the Formula weight of NH2SO3H is 97.1 g/mol: NH2SO3H + NaOH----->NH2SO3 -NA+ + H20
2007-01-17 12:25:22 · 3 answers · asked by help ^_^ 1 in Science & Mathematics ➔ Chemistry
n(NH2SO3H) = 0.497 / 97.1 = 0.00512 mol.
n(NaOH) = n(NH2SO3H) = 0.00512 mol
So [NaOH] = 0.00512 / 0.0284 = 0.180 mol / L.
2007-01-17 12:31:52 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
isnt NH2SO3H a diprotic acid?
2007-01-18 10:35:00 · answer #2 · answered by itsbryan 1 · 0⤊ 0⤋
(mass/Molar mass)*1000/voln.(in ml)=molarity or molarity*voln/1000=no. of moles=mass/molar mass But because of neutralization, we can equate easily so, (0.959/x)=0.176*30.95/1000 x=(0.959*1000)/(0.176*30.95) x=176 gms.
2016-05-24 01:46:57 · answer #3 · answered by ? 4 · 0⤊ 0⤋