Anhydrous neutral salt A(51.3% Ba and 24.0% O) is a good reducing agent(E degree
ID: 531328 • Letter: A
Question
Anhydrous neutral salt A(51.3% Ba and 24.0% O) is a good reducing agent(E degree = -0.39 V) for the preparation of pure metals. Ba(OH_2) and H_2O are placed in a bulb to prepare A. The flask is purged with nitrogen, changed with 0.496 g of a white substance B, and refluxed until B is dissolved completely. The reaction is accompanied with formation of a gas C and 30 mL of 0.2 M solution of salt A. CO_2 is bubbled through solution to remove the excess of Ba(OH)_2. Salt A can be isolated in 98% yield (1.570 g). Gas C is trapped with H_2O. It leads to C oxidation. The product of oxidation forms 49.50 mL of 8 middot 10^-2 M solution of acid D(K_a1 = 7.6 middot 10^-3; K_a2 = 6.2 middot 10^-8; K_a3 = 4.2 middot 10^-13). In addition, during the synthesis of salt A one can observe rare flashes and red film of B modification is formed. Thermolysis of salt A produces H_2O, C, and salt E(61.2% Ba), which is insoluble in CH_3COOH. Salt A dissolves in acid D giving rise to the acid F(K_a = 5.9 middot 10^-2). Identify compounds A - F, if the quantities of C and B are equal. Write chemical equations of the corresponding reactions. Propose structural formulae for B, C, D, F and for the anion of the salt E. Give a reasonable explanation for the difference in basicities of D and F. Indicate the metals which theoretically can be obtained by treatment of the anion of the salt A at standard conditions from the following acidic solutions(potentials of reduction to free metals to free metals are presented in the table): Calculate the values of E degree for these reactions and write ionic equations. The reaction between A and solution of CuSO_4 gives rise to G (98.46% of Cu). Write the chemical equation. Derive mathematical equation for the calculation of [H^+] in the solution of A in D. Estimate pH of this solution using your equation. Provide chemical equation explaining flashes and red film formation observed during salt A synthesis.Explanation / Answer
Solution:
#4
A is BaO
CuSO4+BaO ---> BaSO4+CuO
Hence oxidation yields 98.46% recovery of copper in the oxide form.
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