Q1: 5 mL of dilute nitric acid was added to test tube (1) which contained lead o

Question

Q1: 5 mL of dilute nitric acid was added to test tube (1) which contained lead oxalate and test tube (2) that contained lead sulphate. An appreciable amount of the lead oxalate dissolved but not the lead sulphate, why? (answer is associated with the fact that oxalic acid is a weak acid and sulphuric acid is a strong acid).

Q2: It took 11 drops of 30% calcium chloride added to a solution of 10mL of 10% sodium sulphate to form a precipitate. In another trial, it took 5 drops of 30% calcium chloride added to a solution containing 10ml of 10% sodium sulphate and 8 drops of 50% sulphuric acid. Explain why the number of drops of calcium chloride differs in the two experiments.

Explanation / Answer

Q 1. H2SO4 is a strong acid whereas its anion SO4-2 is a weak conjugate base. PbSO4 reacts with HNO3 to produce Pb (NO3)2 and H2SO4. Since H2SO4 is a strong acid the reaction is favourable. Lead nitrate is soluble in water making a clear solution. For the reason PbSO4 is soluble in dilute solution of HNO3.

But oxalic acid is a weak acid and its anion ; oxalate (C2O4-2) iis a strong conjugate base. For the reason PbC2O4 dodoes nnot react with HNO3 to generate Pb (NO3)2. And lead oxalate being insoluble in water is not dissolved in HNO3 solution.

Q 2. In second case solubility of sodium sulfate decreases in presence of common ion sulfate ( coming from sulfuric acid). So; in 2nd case less amount of sodium sulfate dissolves in solution and less amout of SO4-2 is generated that reacts with CaCl2. As less sulfate ion is produced in 2nd case; less CaCl2 is required.

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