Integer Answer Type Questions: Equilibrium | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q. 1. 0.1 M NaOH is titrated with 0.1 M HA till the end point; K_a for HA is 5.6 × 10^–6 and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point.

Ans. 9

Solution. HA + NaOH → NaA + H 2O

At the end point, the solution contains only NaA whose concentration is 0.1/2 = 0.05 M

Since the salt NaA is formed by strong alkali (NaOH) and weak acid HA (indicated by its low K_a value), its pH can be evaluated by the following relation.

Q. 2. The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10^–4. The pH of a 0.01 M solution of its sodium salt is

Ans. 8

Solution. pH of sodium salt of weak acid

Q. 3. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is 

KCN, K₂SO₄, (NH₄)₂C₂O₄, NaCl, Zn(NO₃)₂, FeCl₃, K₂CO₃, NH₄NO₃ and LiCN

Ans. 3

Solution. KCN, K₂CO₃ and LiCN are the salts of weak acid and strong base. So, their aqueous solutions turns red litmus paper blue.

Q. 4. The total number of diprotic acids among the following is: 

H₃PO₄, H₂SO₄, H₃PO₃, H₂CO₃, H₂S₂O₇, H₃BO₃, H₃PO₂, H₂CrO₄ and H₂SO₃.

Ans. 6

Solution. Diprotic acids are H₂SO₄, H₃PO₃, H₂CO₃, H₂S₂O₇, H₂CrO₄ and H₂SO₃.

Q. 5. In 1 L saturated solution of AgCl [Ksp(AgCl) = 1.6 x 10^–10], 0.1 mol of CuCl [K_sp(CuCl) = 1.0 x 10^–6] is added. The resultant concentration of Ag⁺ in the solution is 1.6 x 10^–x.

The value of “x” is

Ans. 7

Solution. Let the solubility of AgCl is x mol litre^-1 and that of CuCl is y mol litre^-1

∴ K_sp of  AgCl =  [Ag⁺] [Cl^–]

1.6 × 10^–10 = x (x + y) ....(i)

Similarly,  K_sp of CuCl = [Cu⁺][Cl^–]

1.6 × 10^–6 = y(x + y) … (ii)

On solving, (i) and (ii)

[Ag⁺] = 1.6 × 10^–7 ∴ x = 7