Ncert Solutions Chemistry Class 11th

(a) In case of a non-polar (covalent) solvent, solvation enthalpy is small and hence, not sufficient to overcome lattice enthalpy of the salt. Consequently, the salt does not dissolve in non-polar solvent. As a general rule, for a salt to be able to dissolve in a particular solvent its solvation enthalpy must be greater than its lattice enthalpy so that the latter may be overcome by former. Each salt has its characteristic solubility which depends on temperature.

(d) Higher order ionization constants (K_a2, K_a3) are smaller than the lower order ionization constant (K_a1) of a polyprotic acid. The reason for this is that it is more difficult to remove a positively charged proton from a negative ion due to electrostatic forces. This can be seen in the case of removing a proton from the uncharged H₂CO₃ as compared from a negatively charged HCO^3–.Similarly, it is more difficult to remove a proton from a doubly charged HPO₄^2– anion as compared to H₂PO₄^–.

(c) Ammonium hydroxide is a weak base (Kb = 1.77 * 10^–5) and not a weak acid. Therefore, it remains almost unionised in solution. This results in increased of H⁺ ion concentration in solution making the solution acidic. Thus, the pH ofNH₄Cl solution in water is less than 7.

(a) The cations (e.g., Na⁺, K⁺, Ca²⁺, Ba²⁺, etc.) of strong bases and anions (e.g., Cl^–, Br^–, NO^3–, ClO^4– etc.) of strong acids simply get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral i.e., their pH is 7. However, the other category of salts does undergo hydrolysis.

(b) Both are correct statements but R is not the reason for A. The acid-base pair that differs only by one proton is called a conjugate acid-base pair. Therefore, OH^– is called the conjugate base of an acid H₂O and NH⁴⁺ is called conjugate acid of the base NH₃. If Brönsted acid is a strong acid then its conjugate base is a weak base and vice-versa. It may be noted that conjugate acid has one extra proton and each conjugate base has one less proton.

5.8. Calculation of partial pressure of H₂ in 1L vessel:

P₁= 0.8 bar, P₂=? V₁= 0.5 L, V₂ = 1.0 L
As temperature remains constant, P₁V₁ = P₂V₂
=> (0.8 bar) (0.5 L) = P₂  (1.0 L)

=>P₂ = 0.40 bar, i.e., P_H2 = 0.40 bar
Calculation of partial pressure of O₂ in 1 L vessel
P₁V₁ = P₂V₂
(0.7 bar) (2.0 L) = P₂  (1L)

=>P₂ = 1.4 bar

=>Po₂= 1.4 bar
Total pressure =P_H2 + P_Q2 = 0.4 bar + 1.4 bar = 1.8 bar

CaSO₄ (s)↔Ca²⁺ (aq)+SO²⁻₄ (aq)

Ksp= [Ca²⁺] [ SO²⁻₄]

Let the solubility of CaSO4 be s.

[Ca²⁺] = [ SO²⁻₄] = s

Then, Ksp=s²

9.1*10−6=s²

s =3.02*10⁻³mol/L

Molecular mass of CaSO₄=136g/mol

Solubility of CaSO₄ in gram/L= 3.02*10⁻³*136=0.41g/L

This means that we need 1L of water to dissolve 0.41g of CaSO₄.

Therefore, to dissolve 1g of CaSO₄ we require = 10.41L= 2.44Lof water.

Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2.

∴ [FeSO₄]= [Na₂S]=x / 2

Then, [Fe2⁺]= [FeSO₄]=x/ 2

Also, [S²⁻]= [Na₂S]=x/2

FeS (x)↔Fe²⁺ (aq)+S²⁻ (aq)

K_sp= [Fe²⁺] [S²⁻]

= >6.3*10⁻¹⁸= (x/2) (x/2)

x²/4=6.3*10⁻¹⁸

⇒x= 5.02*10⁻⁹

If the concentrations of both solutions are equal to or less than 5.02*10⁻⁹M, then there will be no precipitation of iron sulphide.