TN State Board Samacheer Kalvi Online Test for 11th Std - Chemistry - English - Ionic Equilibrium

Question 1 Of 10

Using Gibb’s free energy change, ∆G0 = 57.34 KJ mol-1, for the reaction, X2Y(s) ( ightleftharpoons) 2X+ + Y2-(aq), calculate the solubility product of X2Y in water at 300K (R = 8.3 J K-1 Mol-1)

10-10 10-12 10-14 can not be calculated from the given data

Question 2 Of 10

Dissociation constant of NH4OH is 1.8 x 10-5 the hydrolysis constant of NH4Cl would be …………….

1.8 x 10-19 5.55 x 10-10 5.55 x 10-5 1.80 x 10-5

Question 3 Of 10

The pH of an aqueous solution is Zero. The solution is ……………..

slightly acidic strongly acidic neutral basic

Question 4 Of 10

(a) (iv)

(i) (ii) (iii) (iii) 75 mL (frac { M }{ 5 }) HCI + 25 mL (frac { M }{ 5 }) NaOH

Question 5 Of 10

Conjugate base for bronsted acids H2O and HF are ………………

OH– and H2FH+, respectively H3O+ and F–, respectively OH– and F–, respectively H3O+ and H2F+, respectively

Question 6 Of 10

The solubility of AgCl (s) with solubility product 1.6 x 10-10 in 0. 1 M NaCl solution would be ………….

1.26 x 10-5 M 1.6 x 10-9 M 1.6 x 10-11 M Zero

Question 7 Of 10

Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H+ ion concentration in the mixture?

37 x 10-2 10-6 0.111 none of these

Question 8 Of 10

The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5NH) in a 0.10M aqueous pyridine solution (Kb for C5H5N = 1.7 x 10-9) iS ……………..

0.006% 0.013% 0.77% 1.6%

Question 9 Of 10

What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?

2.0 3 7.0 12.65

Question 10 Of 10

If the solubility product of lead iodide is 3.2 x 10-8, its solubility will be …………..

2 x 10-3M 4 x 10-4 M 1.6 x 10-5 M 1.8 x 10-5 M

Tamil Nadu State Board ( Samacheer Kalvi ) Online Test based on Book Back Questions - 11th Standard Chemistry - English Medium - Ionic Equilibrium

Samacheer Kalvi 11th Chemistry - Ionic Equilibrium - Book Back Questions in English Medium