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Please Help! Back titration!
Find the mass of ammonium sulfate in lawn fertiliser.
The experiment in a nutshell:
A sample of 1.301g lawn fertiliser is dissolved in de-ionised water and diluted to 250mL. A 20mL aliquot is taken and an excess of 0.1M NaOH is added. The remaining NaOH is titrated with 0.09933M HCl.
The average titre was 16.335mL.
I get 0.3098g as my mass for ammonium sulfate. But for it to be concordant with my gravimetric analysis, that value should be between 0.600g-0.700g!
Can someone please tell me what's gone wrong or what to do to ge the right answer? Thanks!
My working:
Original n(NaOH) = 0.1 \times 0.02 = 0.002 mol
HCl + NaOH -> NaCl + H_2O
n(HCl) = 0.09933 \times 0.016355 = 0.00162454 mol
n(NaOH)_{unreacted with ammonium sulfate} = 0.00162454 mol
n(NaOH)_{reacted with ammonium sulfate} = 0.002 - 0.00162454 = 0.000375458mol
The ionic reaction between ammonium sulfate and NaOH is as follows:
NH_4^{+} + OH^{-} -> NH_3 + H_2O
n(NH_4^{+}) = n(OH^{-}) = 0.000375458 mol
[NH_4^{+}]_{20.00mL aliquot} = \frac{0.000375458}{0.02} = 0.01877 M
[NH_4^{+}]_{20.00mL aliquot} = [NH_4^{+}]_{250.00mL volumetric flask} = 0.01877 M
n(NH_4^{+})_{250mL volumetric flask} = 0.01877 \times 0.25 = 0.00469 mol
n(NH_4^{+}) = 2 \times n((NH_4)_2SO_4)
so n((NH_4)_2SO_4) = \frac{0.00469}{2} = 0.002345 mol
m((NH_4)_2SO_4) = 0.002345 \times 132.1 = 0.3098g
Find the mass of ammonium sulfate in lawn fertiliser.
The experiment in a nutshell:
A sample of 1.301g lawn fertiliser is dissolved in de-ionised water and diluted to 250mL. A 20mL aliquot is taken and an excess of 0.1M NaOH is added. The remaining NaOH is titrated with 0.09933M HCl.
The average titre was 16.335mL.
I get 0.3098g as my mass for ammonium sulfate. But for it to be concordant with my gravimetric analysis, that value should be between 0.600g-0.700g!
Can someone please tell me what's gone wrong or what to do to ge the right answer? Thanks!
My working:
Original n(NaOH) = 0.1 \times 0.02 = 0.002 mol
HCl + NaOH -> NaCl + H_2O
n(HCl) = 0.09933 \times 0.016355 = 0.00162454 mol
n(NaOH)_{unreacted with ammonium sulfate} = 0.00162454 mol
n(NaOH)_{reacted with ammonium sulfate} = 0.002 - 0.00162454 = 0.000375458mol
The ionic reaction between ammonium sulfate and NaOH is as follows:
NH_4^{+} + OH^{-} -> NH_3 + H_2O
n(NH_4^{+}) = n(OH^{-}) = 0.000375458 mol
[NH_4^{+}]_{20.00mL aliquot} = \frac{0.000375458}{0.02} = 0.01877 M
[NH_4^{+}]_{20.00mL aliquot} = [NH_4^{+}]_{250.00mL volumetric flask} = 0.01877 M
n(NH_4^{+})_{250mL volumetric flask} = 0.01877 \times 0.25 = 0.00469 mol
n(NH_4^{+}) = 2 \times n((NH_4)_2SO_4)
so n((NH_4)_2SO_4) = \frac{0.00469}{2} = 0.002345 mol
m((NH_4)_2SO_4) = 0.002345 \times 132.1 = 0.3098g
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