VITEEE 2015 :: Chemistry :: General questions

• Chemistry - General questions

A saturated solution of H₂ S in 0.1 \M HCl at 25⁰ C  contains S^2- ion concentration of 10^-23 mol L^-1. The solubility product of some sulphides are CuS= 10^-44, FeS =10^-14 , MnS=10^-15 , CdS=10^-25 . If 0.01 M solution of these salts in 1 M  HCl are saturated with H₂S which of these will be precipitated?

Answer:

Explanation:

$[S^{2-}]=10^{-23} mol L^{-1}$

$[M^{2+}]=10^{-2} M$

 Ionic product  ,  $K_{1p}=[M^{2+}][S^{2-}] =10^{-25}$

  $\therefore$ ionic product is greater than  K_sp of CuS and CdS.

 Therefore , all except MnS and FeS are prescipitated

 When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide  are liberated in equal volumes according to the equation,

$xCu+yHNO_{3}\rightarrow Cu(NO_{3})_{2}+NO+NO_{2}+H_{2}O$

 the coefficient of x and y are respectively

Answer:

Explanation:

 Balanced equation

$3Cu+8HNO_{3}\rightarrow 3Cu(NO_{3})_{2}+2NO+4H_{2}O$....(i)

 $Cu+4HNO_{3}\rightarrow Cu(NO_{3})_{2}+2NO_{2}+2H_{2}O$

 Here NO and NO₂ are evolved in equal volumes, hence on adding Eqs. (i) and (ii)

$4Cu+12HNO_{3}\rightarrow 4Cu(NO_{3})_{2}+2NO+2NO_{2}+6H_{2}O$

or  

$2Cu+6HNO_{3}\rightarrow 2Cu(NO_{3})_{2}+NO+NO_{2}+3H_{2}O$

 Hence, coefficients x and y of Cu  and HNO₃ are 2 and 6 respectively

• Chemistry - General questions