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 4 of 4: Equivalence-point acidity
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Ka(NHX4X+)=KwKb=5.6×10−10>Kb(SOX4X2−)=KwKa(HSOX4X−)=9.1×10−13⇒pH<7K_a(\ce{NH4+}) = \dfrac{K_w}{K_b} = 5.6 \times 10^{-10} > K_b(\ce{SO4^{2-}}) = \dfrac{K_w}{K_a(\ce{HSO4-})} = 9.1 \times 10^{-13} \Rightarrow \mathrm{pH} < 7
Analysis

At equivalence the solution contains NHX4X+\ce{NH4+} (weak acid, Ka=Kw/Kb=5.6×10−10K_a = K_w/K_b = 5.6 \times 10^{-10}) and SOX4X2−\ce{SO4^{2-}} (very weak base, Kb=Kw/Ka(HSOX4X−)=9.1×10−13K_b = K_w/K_a(\ce{HSO4-}) = 9.1 \times 10^{-13}). The acid dominates, so pH < 7.