Question

100g of salt AB cocnsisting of weak base and strong acid is dissolved in2.5L solution, ph oh which was 5 at 298K. if salt AB has BCC structure with radius of cation=1.6A and radius of anion=1.86A, then find degree of hydrolysism molecular mass, density and edge length given that Kb of AOH=4 x 10^-5

Answer 1

• Degree of hydrolysism (α): 2.5×10⁻⁵
• Molecular Mass: 100 g/mol
• Edge Length: 4.0 Å
• Density: 2.6 g/cm³

Answer 2

Solution:

1. Hydrolysis equilibrium:

The salt AB (made from a weak base A and a strong acid B) dissolves to give A⁺ ions which undergo hydrolysis:

A⁺ + H₂O ⇌ AOH + H⁺

For such cations, the acid constant is:

Kₐ = Kw/K_b = (10⁻¹⁴)/(4×10⁻⁵) = 2.5×10⁻¹⁰.

If the salt concentration is c (in mol/L) and α is the degree of hydrolysis then at equilibrium:

[A⁺] = c(1–α), [AOH] = [H⁺] = cα.

Assuming α is very small so that (1–α) ≈ 1, the equilibrium gives:

Kₐ ≈ (cα)²/(c) = cα².

2. Finding c and α:

100 g of salt is dissolved in 2.5 L. Let the molecular mass be M (in g/mol), so

moles of salt = 100/M, c = (100/M)/2.5 = 40/M mol/L.

The pH is 5 so that [H⁺] = 10⁻⁵ mol/L. But since [H⁺] = cα,

α = (10⁻⁵)/(40/M) = M/(4×10⁶).

3. Using the equilibrium constant:

We have from above:

cα² = 2.5×10⁻¹⁰

Substitute c = 40/M and α = M/(4×10⁶):

(40/M)×(M/(4×10⁶))² = 2.5×10⁻¹⁰

Simplify:

(40/M)×(M²/(16×10¹²)) = 2.5×10⁻¹⁰

=> (40 M)/(16×10¹²) = 2.5×10⁻¹⁰

=> (5 M)/(2×10¹²) = 2.5×10⁻¹⁰

Multiply both sides by 2×10¹²:

5M = 2.5×10⁻¹⁰×2×10¹² = 500

Thus, M = 500/5 = 100 g/mol.

And then, α = 100/(4×10⁶) = 2.5×10⁻⁵.

4. Crystal structure (BCC, similar to CsCl):

In a CsCl-type (BCC) salt, the cation is at the center and the anion at the corners. They touch along the body diagonal.

The body diagonal = a√3 is equal to twice the sum of the ionic radii:

a√3 = 2(r₊ + r₋) ⇒ a = [2(r₊ + r₋)]/√3.

With r₊ = 1.6 Å and r₋ = 1.86 Å,

a = 2(1.6 + 1.86)/1.732 ≈ 2×3.46/1.732 ≈ 6.92/1.732 ≈ 4.0 Å.

5. Density determination:

For a BCC CsCl type structure, there is 1 formula unit per unit cell.

Mass of one formula unit = M/Avogadro’s number = 100 g/mol ÷ 6.022×10²³ ≈ 1.66×10⁻²² g.

Volume of unit cell = a³ = (4×10⁻⁸ cm)³ = 64×10⁻²⁴ cm³ = 6.4×10⁻²³ cm³.

Thus, density ρ = (mass)/(volume) ≈ (1.66×10⁻²² g)/(6.4×10⁻²³ cm³) ≈ 2.6 g/cm³.