Two litre solution of a buffer mixture containing 1.0 M – NaH2PO4 and 1.0 M – Na2HPO4 is placed in t — Electrochemistry Chemistry Question
Question
Two litre solution of a buffer mixture containing 1.0 M – NaH2PO4 and 1.0 M – Na2HPO4 is placed in two compartments (one litre in each) of an electrolytic cell. The platinum electrodes are inserted in each compartment and 1.25 A current is passed for 965 min. Assuming electrolysis of only water at each compartment, what will be pH in each compartment after passage of above charge? (pKa for H2PO4^– = 2.15, log 7 = 0.85)
💡 Solution & Explanation
Initial moles in each 1L compartment: 1.0 mol H2PO4^– and 1.0 mol HPO4^2–. Charge passed = 1.25 A × 965 × 60 s = 72375 C = 0.75 F. At the anode, 0.75 mol H^+ is produced, converting HPO4^2– to H2PO4^–. Moles at anode: HPO4^2– = 1.0 - 0.75 = 0.25; H2PO4^– = 1.0 + 0.75 = 1.75. pH_anode = pKa + log(salt/acid) = 2.15 + log(0.25/1.75) = 2.15 - log(7) = 2.15 - 0.85 = 1.30. At the cathode, 0.75 mol OH^– is produced, converting H2PO4^– to HPO4^2–. Moles at cathode: H2PO4^– = 1.0 - 0.75 = 0.25; HPO4^2– = 1.0 + 0.75 = 1.75. pH_cathode = 2.15 + log(1.75/0.25) = 2.15 + 0.85 = 3.00. Therefore, correct answer is B,C.