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 3 of 4: pH of the ammonia solution
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NHX3+HX2O⇌NHX4X++OHX−:[OHX−]=Kb c=1.8×10−5×8.72×10−3=3.96×10−4;pH=10.59\ce{NH3 + H2O <=> NH4+ + OH-}:\quad [\ce{OH-}] = \sqrt{K_b\,c} = \sqrt{1.8 \times 10^{-5} \times 8.72 \times 10^{-3}} = 3.96 \times 10^{-4};\quad \mathrm{pH} = 10.59
Analysis

Weak-base equilibrium gives [OHX−]=3.96×10−4[\ce{OH-}] = 3.96 \times 10^{-4} mol dm−3^{-3}, so pOH = 3.41 and pH = 10.59 for the container solution.