Chemistry Labs

Problem 2

Anhydrous ammonia is a clean, energy-dense fuel. In a fixed-volume container, gaseous NHX3\ce{NH3} burns according to 4 NHX3(g)+3 OX2(g)→2 NX2(g)+6 HX2O(l)\ce{4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(l)}; initial and final states are at 298 K and after combustion of 14.40 g of OX2\ce{O2} some NHX3\ce{NH3} remains (ΔfH∘(NHX3(g))=−46.11\Delta_fH^{\circ}(\ce{NH3(g)}) = -46.11 kJ mol−1^{-1}, ΔfH∘(HX2O(l))=−285.83\Delta_fH^{\circ}(\ce{H2O(l)}) = -285.83 kJ mol−1^{-1}). (a) Calculate the heat released. (b) To determine dissolved NHX3\ce{NH3}, a 10.00 cm3^3 sample of the resulting aqueous solution was added to 15.0 cm3^3 of HX2SOX4\ce{H2SO4} (c=0.0100c = 0.0100 mol dm−3^{-3}) and back-titrated with NaOH (c=0.0200c = 0.0200 mol dm−3^{-3}), equivalence at 10.64 cm3^3 (Kb(NHX3)=1.8×10−5K_b(\ce{NH3}) = 1.8 \times 10^{-5}; Ka(HSOX4X−)=1.1×10−2K_a(\ce{HSO4-}) = 1.1 \times 10^{-2}). Calculate the pH of the solution in the container. (c) At the equivalence point NHX4X+\ce{NH4+} and SOX4X2−\ce{SO4^{2-}} are present; write the relevant equilibria and predict whether the equivalence-point pH is above, below, or equal to 7.
Step 2 of 4: Ammonia by back titration
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n(NHX3)=2 [n(HX2SOX4)tot−12n(NaOH)]=2 [0.150−12(0.2128)]=0.0872 mmol;c=8.72×10−3 mol dm−3n(\ce{NH3}) = 2\,[n(\ce{H2SO4})_{tot} - \tfrac{1}{2}n(\ce{NaOH})] = 2\,[0.150 - \tfrac{1}{2}(0.2128)] = 0.0872\ \text{mmol};\quad c = 8.72 \times 10^{-3}\ \text{mol dm}^{-3}
Analysis

Excess acid: n(HX2SOX4)=0.150n(\ce{H2SO4}) = 0.150 mmol; NaOH used =0.01064×0.0200=0.2128= 0.01064 \times 0.0200 = 0.2128 mmol neutralises 12(0.2128)=0.1064\tfrac{1}{2}(0.2128) = 0.1064 mmol HX2SOX4\ce{H2SO4}, leaving 0.0436 mmol that reacted with NHX3\ce{NH3}, i.e. 0.0872 mmol NHX3\ce{NH3} in 10.00 cm3^3.