The percent nitrogen (N) in a protein sample can be determined by reacting the protein with sulfuric acid to produce ammonium sulfate. The ammonium sulfate is reacted with concentrated sodium hydroxide to form ammonia gas. The ammonia gas is collected by bubbling it through a known volume of standardized hydrochloric acid solution (all of the ammonia reacts in this step). The unreacted hydrochloric acid is then back-titrated with standardized sodium hydroxide solution. The reactions involved are shown below: Protein-N + H2SO4 (aq) -> (NH4)2SO4 (aq) 2 NH4+ (aq) + 2 NaOH (aq) -> 2 NH3 (g) + 2 H2O NH3 (g) + HCl (excess) -> NH4Cl (aq) + HCl (unreacted) HCl (unreacted) + NaOH (aq) -> NaCl (aq) + H2O If a 1.85 g sample of protein required 25.00 mL of 0.1535 M HCl and 21.55 mL of 0.0986 M NaOH, determine the mass percent nitrogen in the protein sample.
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First, we need to find the moles of unreacted HCl. We can do this using the volume and molarity of NaOH used in the back-titration: moles of unreacted HCl = volume of NaOH × molarity of NaOH moles of unreacted HCl = 21.55 mL × 0.0986 mol/L = 0.002125 mol Show more…
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A 0.608-g sample of fertilizer contained nitrogen as ammonium sulfate, $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$. It was analyzed for nitrogen by heating with sodium hydroxide. $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(s)+2 \mathrm{NaOH}(a q) \longrightarrow$ $$ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NH}_{3}(g) $$ The ammonia was collected in $46.3 \mathrm{~mL}$ of $0.213 \mathrm{M} \mathrm{HCl}$ (hydrochloric acid), with which it reacted. $$ \mathrm{NH}_{3}(g)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ This solution was titrated for excess hydrochloric acid with $44.3 \mathrm{~mL}$ of $0.128 \mathrm{M} \mathrm{NaOH}$. $$ \mathrm{NaOH}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What is the percentage of nitrogen in the fertilizer?
The amount of nitrogen in an organic substance can be determined by an analytical method called the Kjeldahl method, in which all the nitrogen in the organic substance is converted to ammonia. The ammonia, which is a weak base, can be neutralized with hydrochloric acid, as described by the equation: NH3(aq) + HCl(aq) ⟶ NH4Cl(aq) If 58.0 mL of 0.150 M HCl(aq) is needed to neutralize all the NH3(g) from a 2.25 g sample of organic material, calculate the mass percentage of nitrogen in the sample. mass percentage:
Dinesh S.
In some laboratory analyses, the preferred technique is to dissolve a sample in an excess of acid or base and then "back-titrate" the excess with a standard base or acid. This technique is used to assess the purity of a sample of $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} .$ Suppose you dissolve a $0.475-\mathrm{g}$ sample of impure $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ in aqueous KOH. $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{KOH}(\mathrm{aq}) \rightarrow$ $$ 2 \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The $\mathrm{NH}_{3}$ liberated in the reaction is distilled from the solution into a flask containing $50.0 \mathrm{mL}$ of $0.100 \mathrm{M}$ HCl. The ammonia reacts with the acid to produce $\mathrm{NH}_{4} \mathrm{Cl},$ but not all of the HCl is used in this reaction. The amount of excess acid is determined by titrating the solution with standardized NaOH. This titration consumes $11.1 \mathrm{mL}$ of $0.121 \mathrm{M} \mathrm{NaOH} .$ What is the weight percent of $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ in the 0.475 -g sample?
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