NEET Practice Questions (MCQs) with Answers & Solutions

(c) [H⁺] in monobasic acid = molarity x degree of ionisation

                                        =0.1x 2/100 = 2x10^-3

       ionisation constant of water

          K_w  = [H⁺][OH^-]

           [OH^-] = K_w/[H⁺] = (1x10^-14)/(2x10^-3) = 5x10^-12

(a) For weak base, [OH^-] = $\sqrt{{\mathit{K}}_{\mathit{b}}\mathit{c}}$;

                    .^..        [H⁺] = K_w/$\sqrt{{\mathit{K}}_{\mathit{b}}\mathit{c}}$ ;

             .^..  pK_w -1/2 pK_b + 1/2 logc

 (d)                      2 AB₃ $\rightleftharpoons$ A₂(g) + 3B₂(g)

t=0                       8           0             0

At equilibrium     (8-a)         a/2         3a/2

             Thus, K_c = [A₂][B₂]³/[AB₃]² ; Also, a/2 = 2   .^.. a=4

              .^.. [AB₃] = 4/1 ; [A₂] = 2/1 ; [B₂] = 6/1

                      Thus, K_c = (2 x 6³)/4² = 27 mol²L^-2

(d) N of NH₃ possesses lone pair of electron available for donation.

(d) The corrrect order of pH of isomolar solution of sodium oxide(pH₁), sodium sulphide(pH₂), sodium selenide (pH₃) and sodium telluride (pH₄) is ${\mathrm{pH}}_1 > {\mathrm{pH}}_2 > {\mathrm{pH}}_{3 }> {\mathrm{pH}}_4$ because in aqueous solution, they are hydrolysed  as follows.

${\mathrm{Na}}_2\mathrm{O} + 2{\mathrm{H}}_2\mathrm{O} \stackrel{ }{\to }2\mathrm{NaOH} + {\mathrm{H}}_2\mathrm{O}$

                               base

${\mathrm{Na}}_2\mathrm{S} + 2{\mathrm{H}}_2\mathrm{O} \stackrel{ }{\to } 2\mathrm{NaOH} + {\mathrm{H}}_2\mathrm{S}$

                             strong base      weak acid

${\mathrm{Na}}_2\mathrm{Se} + 2{\mathrm{H}}_2\mathrm{O} \stackrel{ }{\to } 2\mathrm{NaOH} + {\mathrm{H}}_2\mathrm{Se}$

                             strong base      weak acid

${\mathrm{Na}}_2\mathrm{Te} + 2{\mathrm{H}}_2\mathrm{O} \stackrel{ }{\to } 2\mathrm{NaOH} + {\mathrm{H}}_2\mathrm{Te}$

                             strong base      weak acid

order of acidic strength ${\mathrm{H}}_2\mathrm{Te} > {\mathrm{H}}_2\mathrm{Se} > {\mathrm{H}}_2\mathrm{S} > {\mathrm{H}}_2\mathrm{O}$ 

Hence their aqueous solutions have the following order of basic character due to neutralisation of NaOH with H₂O, H₂S, H₂Se > H₂Te. 

${\mathrm{Na}}_2\mathrm{O} > {\mathrm{Na}}_2\mathrm{S} > {\mathrm{Na}}_2\mathrm{Se} > {\mathrm{Na}}_2\mathrm{Te}$ 

($\because$pH of basic solution is higher than acidic or least basic solution)

(d) Consider a hypothetical change, 

                   A + B $\rightleftharpoons$C + D

                         Keq =[C][D]/[A][B]

For the above reaction if the concentration of reactants are doubled then the rate of forward reaction increases for a short time but after sometime equilibrium will be established. So, concentration has no effect on the equilibrium constant. It remains unchanged after increasing the concentration of reactants.

(a) pOH = pK_b + log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}$

           = -log K_b + log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}$

           = -log 1.8 x 10^-5 + log (0.20/0.30)

           = 5-0.25+(-0.176)

           =4.75-0.176=4.57

      .^..     pH =14-4.57

(c) Electron rich species are called Lewis base. Among the given, BF₃ is an electron deficient species, so have a capacity of electron accepting instead of donating. That's why it is least likely to act as a Lewis base. It is a Lewis acid.

(d) Unit of K_c = $(\mathrm{Unit} \mathrm{of} \mathrm{concentration}{)}^{∆\mathrm{n}}.∆=+1.$

(b) [H₃O]⁺ = [H⁺] = 10^-10

     pH + pOH = 14 

             pH = -log[H⁺]

              pH = -log[10^-10]

              pH =10

          pOH +10=14

          pOH=14-10=4