CBSE 11th Standard Chemistry Subject Equilibrium Ncert Exemplar 3 Mark Questions 2021
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CBSE 11th Standard Chemistry Subject Equilibrium Ncert Exemplar 3 Mark Questions 2021
11th Standard CBSE
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Reg.No. :
Chemistry
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A reaction between ammonia and boron trifluoride is given below : NH3+BF3⟶H3N:BF3 . Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of B and N in the reactants?
(a) -
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3M. what is its pH ?
(a) -
How can you predict the following stages of a reaction by comparing the value \({ K }_{ c }\) and \({ Q }_{ c }\) ?
Net reaction proceeds in the forward direction.(a) -
The ionisation constant of HF,HCOOH and HCN at 298 are 6.8 x 10-4 and 4.8 x 10 -9 respectively.Calculate the ionisation constant of the corresponding conjucate bases.
To find Kb of a conjugate base , use the formula Ka.Kb = Kw = 1 x 10-14 Conjugate bases of HF,HCOOH and HCN are F- HCOO- and CN- respectively.(a)
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CBSE 11th Standard Chemistry Subject Equilibrium Ncert Exemplar 3 Mark Questions 2021 Answer Keys
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Although BF3 does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with NH3 by accepting the lone pair of electrons from NH3 and complete its octet. the reaction can be represented by
BF3+:NH3⟶BF3\(\leftarrow\)NH3
Lewis electronic theory of acids and bases can explain it. Boron in BF3 is sp2 hybridised where N in NH3 is sp3 hybridised. -
pH = -log[H+] = -log [3.8\(\times \)10-3]
= - [log[3.8] + log[10-3]
= - [(0.58) + (-3.0)] = - [-2.42] = 2.42
Therefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic. -
Prediction of the following stages of a reaction by comparing the value \({ K }_{ c }\) and \({ Q }_{ c }\) are If \({ Q }_{ c }\)<\({ K }_{ c }\), the reaction will proceed in the direction of the products (forward reaction)
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If Ka is the ionisation constant of weak acid (HA) and Kb is the ionisation constant of its conjugate base(A-) then Ka Kb = Kw
\({ K }_{ b }({ F }^{ - })=\frac { { K }_{ w } }{ { K }_{ a }(HF) } =\frac { 1\times { 10 }^{ -14 } }{ 6.8\times { 10 }^{ -4 } } =1.47\times { 10 }^{ -11 }\)
\({ K }_{ b }(HCOO^{ - })=\frac { { K }_{ w } }{ { K }_{ a }(HCOOH) } =\frac { 1\times { 10 }^{ -14 } }{ 1.8\times { 10 }^{ -4 } } =5.56\times { 10 }^{ -11 }\)
\({ K }_{ b }(Cn^{ - })=\frac { { K }_{ w } }{ { K }_{ a }(HCn) } =\frac { 1\times { 10 }^{ -14 } }{ 4.8\times { 10 }^{ -4 } } =2.08\times { 10 }^{ -6 }\)
