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Back titration!

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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 [tex]NaOH[/tex] is added. The remaining [tex]NaOH[/tex] is titrated with 0.09933M [tex]HCl[/tex].

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) = [tex]0.1 \times 0.02[/tex] = 0.002 mol

[tex]HCl + NaOH -> NaCl + H_2O[/tex]

[tex]n(HCl) = 0.09933 \times 0.016355[/tex] = 0.00162454 mol

[tex]n(NaOH)_{unreacted with ammonium sulfate}[/tex] = 0.00162454 mol

[tex]n(NaOH)_{reacted with ammonium sulfate}[/tex] = 0.002 - 0.00162454 = 0.000375458mol

The ionic reaction between ammonium sulfate and NaOH is as follows:

[tex]NH_4^{+} + OH^{-} -> NH_3 + H_2O[/tex]

[tex]n(NH_4^{+}) = n(OH^{-}) = 0.000375458 mol[/tex]

[tex][NH_4^{+}]_{20.00mL aliquot} = \frac{0.000375458}{0.02} = 0.01877 M[/tex]

[tex][NH_4^{+}]_{20.00mL aliquot} = [NH_4^{+}]_{250.00mL volumetric flask} = 0.01877 M[/tex]

[tex]n(NH_4^{+})_{250mL volumetric flask} = 0.01877 \times 0.25 = 0.00469 mol[/tex]

[tex]n(NH_4^{+}) = 2 \times n((NH_4)_2SO_4)[/tex]

so [tex]n((NH_4)_2SO_4) = \frac{0.00469}{2} = 0.002345[/tex] mol

[tex]m((NH_4)_2SO_4) = 0.002345 \times 132.1 = 0.3098g[/tex]
 
Last edited:

chemisttree

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Re: Please Help! Back titration!

How much 0.1 M NaOH did you add?
 

symbolipoint

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Re: Please Help! Back titration!

I have the same question as post #2. You apparantly also confused the volume of aloquat with the unstated volume of NaOH.
 

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