A 0.608 g sample of fertilizer contained nitrogen as ammonium sulfate, ( NH 4 ) 2 SO 4 . It was analyzed for nitrogen by heating with sodium hydroxide. ( NH 4 ) 2 SO 4 ( s ) + 3 NaOH ( aq

Question

A 0.608 g sample of fertilizer contained nitrogen as ammonium sulfate, $(NH_{4})_{2} SO_{4}$ . It was analyzed for nitrogen by heating with sodium hydroxide.

$(NH_{4})_{2} SO_{4} (s) + 3 NaOH (aq) \to Na_{2} SO_{4} (aq) + 2 H_{2} O (l) + 2 NH_{3} (g)$

The ammonia was collected in 36.3 mL of 0.213 M HCl.

This solution was titrated for excess HCl with 44.3 mL of 0.128 M NaOH. What is the percentage of nitrogen in the fertilizer?

TestHub · Accepted Answer

Equivalent of acid for NH 3 ​ = equivalent of NH 3 ​ from fertilizer first we determine unreacted Hcl ( V mL ) , V mL of 0.213 M HCl = 44.3 m L of 1.28 M NaOH V = 0.213 44.3 × 0.128 ​ = 26.62 m L used H Cl = 46.30 − 26.62 = 19.68 m L amount of nitrogen in fertilizer = 1000 NE V ​ equivalent to 19.68 mL of 0.213 M HCl = 1000 0.213 × 14 × 19.68 ​ g = 0.0587 g This amount is present in 0.608 g sample of fertilizers, % of nitrogen = 0.687 0.0587 ​ × 100 = 9.65%

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