NEET Practice Questions (MCQs) with Answers & Solutions

Equilibrium constant for the reaction $ SO_{ 2 (g) } + { 1 \over 2 } O_{ 2 (g) } \rightleftharpoons SO_{3(g)} $ and for reaction $ 2 SO_{ 2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} $ $ K_C = ( { 1 \over 4.9 \times 10 ^ {-2} } ) ^ 2 = 416 .49 $

$ P^H = 3 \therefore [ H^+_3 ] = 1 \times 10 ^ {-3} $ $ Ka = { [ H_3 ^ + O ] ^2 \over C } = { ( 1 \times 10 ^ {-3} ) ^2 \over 0.1 } = 1 \times 10 ^ {-5} $

According to lowry-bronsted acid base theory in (ii) reaction $ H_2 PO_4 ^- donates H^+ ion to H_2 O$ it acts as an acid.

$ K_p = Kc ( RT) ^ { \triangle n (g) } \triangle n (g) = 1 $ $ = Kc \times 0.0831 \times 457 it means Kp \gt Kc $

Water is an amphiprotic solvent as it can accept protons as well as give protons.

Ammonium ion $ (NH _4 ^ +) is a conjugate acid of NH_3 $ $ NH_3 + H_2O \rightleftharpoons NH_4 ^ + + OH ^ - $ Bronsted base Conjugate acid

$ HSO_4 ^ - $ can act as a bronsted acid as well as bronsted base. $ HSO_4 ^ - + H_2 O \rightleftharpoons SO_4 ^ {2-} + H_3 ^ + O $ Acid $ HSO_4 ^ - + H_2 O \rightleftharpoons H_2 SO_4 + OH^ - $ Base

$ P ^ H = 4.7 \therefore [ H_3 ^ + O ] = 1.995 \times 10 ^ {-5} pK_w = 14 \therefore K_w = 1 \times 10 ^ {-14} $ $ = - log [ H_3 ^ + O ] Now K_w = [ H_3 ^ + O ] [ OH^- ] $ $ \therefore [OH^-] = { 1 \times 10 ^ { -14} \over 1.995 \times 10 ^ {-5} } = 5 \times 10 ^ {-10} M $

Conjugate base of $ H_2 PO_4 ^ - is HPO_4 ^ {2-} $ $ H_2 PO_4 ^ - + H_2 O \rightleftharpoons HPO_4 ^ {2-} + H_3 O ^ + $ Acid

Conjugate base of $ OH^- $ $ OH^- + H_2 O \rightleftharpoons O ^ {2-} + H_3 O ^ + $