Hi, I did the Volumetric determination of Nitrogen content in Lawn fertilser back titration prac yesterday, someone care to help me out with a few of these questions?
already know:
Average titre of HCl = 5.5mL (concordant titre)
c(HCl) = 0.1406
n(HCl) = 0.0007733 mol
v(NaOH) = 0.02L
c(NaOH) = 0.097M
n(NaOH) = 0.00194 mol
Equations:
NH4+ + OH- ---> NH3 + H20 (Ammonium compounding in the fertiliser)
HCl + NaOH ---> NaCl + H20
I'm now stuck when I get to here.
2d. Find the amount, in mol of Sodium Hydroxide that did not react with the fertiliser solution in each conical flask (is this just 0.00194 - 0.0007733)?
3. Then I need to calculate the amount of mol of Sodium Hydroxide that did react with the fertiliser solution (seems pretty easy once I know if my 2d is correct)
4. Find the amount of NH4+ ions present in each 20mL aliqout
5. Calculate the amount, in mol of NH4+ ions that were originally present in the 250mL volumetric flask
6. I should be about to find out the mass of nitrogen and the percentage by mass once I know this.
Will supply more info if needed, cheers guys!