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Question: A quantity of 40 g of a salt (AB) of strong acid and weak base is dissolved in water to form 0.5 L a...

A quantity of 40 g of a salt (AB) of strong acid and weak base is dissolved in water to form 0.5 L aqueous solution. At 298 K, the pH of the solution is found to be 5.0. If AB forms CsCl type crystal and the radius of A+^+ and B−^- ions are 100 pm and 160 pm respectively, then the incorrect statement is (Given Kb_b of AOH = 4 × 10−5^{-5}; 3\sqrt{3} = 1.732)

A

The degree of hydrolysis of salt (AB) is 2.5 x 10−5^{-5}

B

Unit cell edge length of AB crystal is approximately 300 pm

C

Molar mass of salt (AB) is 50 g/mol

D

Density of solid (AB) is 12.3 g/mL

Answer

Molar mass of salt (AB) is 50 g/mol

Explanation

Solution

The problem involves two main parts: the hydrolysis of a salt of a strong acid and weak base, and the crystal structure of the salt. We need to evaluate each statement to find the incorrect one.

Part 1: Hydrolysis of salt (AB)

The salt AB is formed from a strong acid and a weak base (AOH). Therefore, the cation A+^+ undergoes hydrolysis:

A+^+ + H2_2O ⇌\rightleftharpoons AOH + H+^+

Given:

  • Kb_b of AOH = 4 × 10−5^{-5}
  • pH of the solution = 5.0
  • Volume of solution = 0.5 L
  • Mass of salt = 40 g
  • Temperature = 298 K (so Kw_w = 10−14^{-14})

  1. Calculate the hydrolysis constant (Kh_h):
    For a salt of strong acid and weak base, Kh_h = Kw_w / Kb_b.
    Kh_h = 10−14^{-14} / (4 × 10−5^{-5}) = 0.25 × 10−9^{-9} = 2.5 × 10−10^{-10}.

  2. Determine [H+^+] from pH:
    pH = 5.0, so [H+^+] = 10−pH^{-pH} = 10−5^{-5} M.

  3. Relate [H+^+] to concentration and degree of hydrolysis:
    Let C be the initial concentration of the salt AB and α be its degree of hydrolysis.
    At equilibrium:
    [A+^+] = C(1-α)
    [AOH] = Cα
    [H+^+] = Cα

    From [H+^+] = Cα, we have 10−5^{-5} = Cα.
    The hydrolysis constant Kh_h = [AOH][H+^+] / [A+^+] = (Cα)(Cα) / C(1-α) = Cα2^2 / (1-α).
    Since α is generally small for weak hydrolysis, we can approximate (1-α) ≈ 1.
    So, Kh_h ≈ Cα2^2.

    Substitute Cα = 10−5^{-5} into Kh_h ≈ Cα2^2:
    Kh_h = Cα * α = 10−5^{-5} * α
    α = Kh_h / 10−5^{-5} = (2.5 × 10−10^{-10}) / 10−5^{-5} = 2.5 × 10−5^{-5}.

    Statement A: The degree of hydrolysis of salt (AB) is 2.5 x 10−5^{-5}.
    This statement is correct.

  4. Calculate the concentration C:
    From Cα = 10−5^{-5} and α = 2.5 × 10−5^{-5}:
    C = 10−5^{-5} / α = 10−5^{-5} / (2.5 × 10−5^{-5}) = 1 / 2.5 = 0.4 M.

  5. Calculate the molar mass (M) of salt (AB):
    Concentration C = (moles of salt) / (Volume of solution)
    Moles of salt = (mass of salt) / Molar mass (M) = 40 g / M
    C = (40 / M) / 0.5 L = 80 / M
    We found C = 0.4 M.
    0.4 = 80 / M
    M = 80 / 0.4 = 200 g/mol.

    Statement C: Molar mass of salt (AB) is 50 g/mol.
    This statement is incorrect.

Part 2: Crystal structure of AB

Given:

  • AB forms CsCl type crystal (Body-Centered Cubic, BCC-like structure).
  • Radius of A+^+ (r+^+) = 100 pm
  • Radius of B−^- (r−^-) = 160 pm
  • 3\sqrt{3} = 1.732

  1. Calculate the unit cell edge length (a):
    In a CsCl-type structure, ions touch along the body diagonal. The body diagonal length is a√3.
    a√3 = 2(r+^+ + r−^-)
    a = 2(r+^+ + r−^-) / √3
    a = 2(100 pm + 160 pm) / 1.732
    a = 2(260 pm) / 1.732 = 520 pm / 1.732 ≈ 300.23 pm.

    Statement B: Unit cell edge length of AB crystal is approximately 300 pm.
    This statement is correct.

  2. Calculate the density (ρ) of solid (AB):
    Density ρ = (Z × M) / (NA_A × a3^3)
    Where:

    • Z = number of formula units per unit cell. For CsCl type, Z = 1.
    • M = Molar mass of AB = 200 g/mol (calculated above).
    • NA_A = Avogadro's number = 6.022 × 1023^{23} mol−1^{-1}.
    • a = unit cell edge length = 300.23 pm. Convert to cm: 1 pm = 10−10^{-10} cm.
      a = 300.23 × 10−10^{-10} cm = 3.0023 × 10−8^{-8} cm.
      a3^3 = (3.0023 × 10−8^{-8} cm)3^3 ≈ 2.706 × 10−23^{-23} cm3^3.

    ρ = (1 × 200 g/mol) / (6.022 × 1023^{23} mol−1^{-1} × 2.706 × 10−23^{-23} cm3^3)
    ρ = 200 / (6.022 × 2.706) g/cm3^3
    ρ = 200 / 16.298 g/cm3^3
    ρ ≈ 12.27 g/cm3^3.
    Since 1 g/cm3^3 = 1 g/mL, ρ ≈ 12.27 g/mL.

    Statement D: Density of solid (AB) is 12.3 g/mL.
    This statement is correct (approximately).

Based on the calculations, statements A, B, and D are correct, while statement C is incorrect. The question asks for the incorrect statement.