Phosphoric(V) acid, H₃PO₄, is used in both inorganic and organic reactions.
(a) H₃PO₄ is made in a two-step process from phosphorus.
step 1 Phosphorus reacts with an excess of oxygen to form a white solid.
step 2 The white solid then reacts with water to form H₃PO₄.
(i) Write an equation for each step.
(ii) H₃PO₄ is a weak Brønsted–Lowry acid. Define weak Brønsted–Lowry acid.
(b) H₃PO₄ is also formed in the process shown in reaction 1.
reaction 1   \(4H_3PO_3 \to 3H_3PO_4 + PH_3\)
Table 3.1 shows some relevant thermodynamic data.
(i) Define enthalpy change of formation.
(ii) Use the data in Table 3.1 to calculate the enthalpy change, \(\Delta H_r\), of reaction 1.
(iii) Explain why reaction 1 is a disproportionation reaction. Explain your reasoning with reference to relevant oxidation numbers.
(c) Fig. 3.1 shows a reaction scheme that involves H₃PO₄ in several reactions.
(i) Identify A, which reacts with propene in the presence of H₃PO₄ in reaction 2.
(ii) Draw the structure of B.
(iii) Name the type of reaction that occurs in reaction 3.
(iv) Reaction 3 is monitored using infrared spectroscopy. It is not possible to use the O—H absorption frequency to monitor the reaction. Use Table 3.2 to identify a suitable bond whose absorption frequency can be used to monitor the progress of reaction 3. State the change you would see in the infrared spectrum during reaction 3.
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(d) H₃PO₄ also reacts with alcohols to form organophosphates. Organophosphates are compounds similar to esters. They have the general structure shown in Fig. 3.2.
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(i) Complete the equation to suggest the products of the reaction of H₃PO₄ with methanol, CH₃OH.
(ii) Compound T is a simple organophosphate. The mass spectrum of T shows a molecular ion peak at m/e = 182. This peak has a relative intensity of 12.7. The relative intensity of the M+1 peak is 0.84. Deduce the number of carbon atoms in T. Hence suggest the molecular formula of T. Assume that phosphorus and oxygen exist as single isotopes. Show your working.
▶️ Answer/Explanation
(a)(i)
Step 1: \(4P + 5O_2 \to P_4O_{10}\)
Step 2: \(P_4O_{10} + 6H_2O \to 4H_3PO_4\)
Explanation: Phosphorus reacts with oxygen to form phosphorus pentoxide (\(P_4O_{10}\)), which then hydrolyzes with water to produce phosphoric acid (\(H_3PO_4\)).
(a)(ii) A weak Brønsted–Lowry acid is a proton (H⁺) donor that partially dissociates in solution.
Explanation: \(H_3PO_4\) does not fully ionize in water, releasing only a fraction of its protons.
(b)(i) Enthalpy change of formation is the energy change when one mole of a compound is formed from its constituent elements in their standard states.
(b)(ii) \(\Delta H_r = +9 + 3(-1281) – 4(-972) = +54 \text{ kJ mol}^{-1}\)
Explanation: The enthalpy change is calculated using the given stoichiometry and standard enthalpies of formation.
(b)(iii) This is a disproportionation reaction because phosphorus in \(H_3PO_3\) is both oxidized (to \(H_3PO_4\)) and reduced (to \(PH_3\)).
Explanation: The oxidation state of P changes from +3 to +5 (oxidation) and +3 to -3 (reduction).
(c)(i) A is \(H_2O\) (water/steam).
(c)(ii) Structure of B:
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(c)(iii) The reaction type is condensation.
(c)(iv) Bond: C=O
Change in infrared spectrum: The absorption shifts from 1670–1740 cm⁻¹ to 1710–1750 cm⁻¹.
(d)(i) \(H_3PO_4 + 3CH_3OH \to (CH_3O)_3PO + 3H_2O\)
(d)(ii) Number of carbon atoms = \(\frac{0.84}{12.7} \times \frac{100}{1.1} = 6\)
Molecular formula: \(C_6H_{15}O_4P\)
Explanation: The M+1 peak intensity ratio is used to determine the number of carbon atoms, leading to the molecular formula.
