Question 14.E.E: An ammonia gas-sensing electrode gave the following calibrat...
An ammonia gas-sensing electrode gave the following calibration points when all solutions contained 1 M NaOH.
A dry food sample weighing 312.4 mg was digested by the Kjeldahl procedure (Section 10-8) to convert all nitrogen into NH4^{+}. The digestion solution was diluted to 1.00 L, and 20.0 mL were transferred to a 100-mL volumetric flask. The 20.0-mL aliquot was treated with 10.0 mL of 10.0 M NaOH plus enough NaI to complex the Hg catalyst from the digestion and diluted to 100.0 mL. When measured with the ammonia electrode, this solution gave a reading of 339.3 mV.
Calculate the wt% nitrogen in the food sample.
| NH_{3} (M) | E (mv) | NH_{3} (M) | E (mv) |
| 1.00 × 10^{-5} | 268.0 | 5.00 × 10^{-4} | 368.0 |
| 5.00 × 10^{-5} | 310.0 | 1.00 × 10^{-3} | 386.4 |
| 1.00 × 10^{-4} | 326.8 | 5.00 × 10^{-3} | 427.6 |
A graph of E(mV) versus \log[NH_{3}(M)] gives a straight line whose equation is E = 563.4 + 59.05 × \log [NH_{3}]. For E = 339.3 mV, [NH_{3}] = 1.60 × 10^{−4} M. The sample analyzed contains (100 mL)(1.60 × 10^{−4} M) = 0.016 0 mmol of nitrogen. But this sample represents just 2.00% (20.0 mL/1.00 L) of the food sample. Therefore, the food contains (0.016 mmol N)/0.020 0 = 0.800 mmol N = 11.2 mg of N = 3.59 wt% N.