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11. Which of these lists of molecules is arranged in order of increasing acid strength? H2S <H20 <H2Se 6. HS <H2Se <H20 H2O <
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Answer #1

The first four questions are solved here. Kindly post the the other questions separately.

11. The correct option is c)

In moving down the oxygen group, size of central atom increases from O to S to Se. In moving down a group, electronegativity effects are less important as compared to moving along a period.

Therefore the bond length between H-and the atom goes on becoming weaker. Hence, it is easier to release H+ ion as one move from O to S to Se.

12. PbI2 dissociates as

PbI2(s)\rightleftharpoons Pb+2 (aq) + 2 I-(aq)

Ksp = [Pb+2 ] [ 2 I- ]

Ksp is rate constant for this reaction. It is only temperature dependent.

considering each option: a) lowering the pH means that acidity of the reaction is increased. It will not have any effect on its Ksp.

                                         b) addition of Pb(NO3)2 will increase the concentration of Pb+2 ions. However the reaction will shift towards the left according to Le Chatelier's principle, and the Ksp would be constant.

                                         c) Similarly addition of KI will increase the concentration of I- ions. Again, the reaction will shift towards the left according to Le Chatelier's principle, and the Ksp would be constant.

                                         d) None of the above would change Ksp. Therefore this is the correct option.

13. Option a) is the correct one.

In a weak base-strong acid titration, the acid and base will react to form an acidic solution.

It is because, the conjugate acid of the base formed after titration will undergo hydrolysis in water and produce H+ ions. This makes the pH of the solution slightly acidic.

Other options are incorrect because:

b) Here the final pH will be slightly basic, because of hydrolysis of conjugate base of the weak acid.

c) The final pH will be neutral pH (at 7) if Ka = Kb

d) The final pH will be neutral in this case also.

14. HNO2\rightleftharpoons NO2- (aq) + H+(aq)

If the 0.14 M solution is 5.7 % ionized, means [H+] = 0.14 x 0.057 M = 0.00798 M

Therefore, pH = -log [H+] = -log (0.00798 ) = 2.1. Hence option c) is correct

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