2007 NJC Prelims Paper 3 FAQs

A Note for Q1b
2 mol of OH- reacts with 1 mol of A to give a white ppt B. The thermal decomposition of B yields a white ppt C. Thus M is likely to be a Group II element.

NOTE: If M = Zn, you will also get similar observations (except that the mass of B formed is NOT 0.0215 g).

Explanation for Q3b(i)
The pH of the solution is 2.5, so [H+] = 10^(-2.5) mol/dm3.
The initial concentration of CrO42- is 1.00 x 10^(-3) mol/dm3.
2 mol of CrO42- react with 2 mol of H+ to form 1 mol of Cr2O72-.
Thus, [CrO42-] : [Cr2O72-] = (1.00 x 10^(-3) - 2y) : y = 200 : 1
[Cr2O72-] = y = 4.98 x 10^(-6) mol/dm3
[CrO42-] = 1.00 x 10^(-3) - 2[4.98 x 10^(-6)] = 9.90 x 10^(-4) mol/dm3
[H+] = 10^(-2.5) - 2[4.98 x 10^(-6)] = 3.15 x 10^(-3) mol/dm3