Ionic EquilibriumHard
Question
A solution is prepared by dissolving 1.5 g of a monoacidic base into 1.5 kg of water at 300 K, which showed a depression in freezing point by 0.165oC. When 0.496 g of the same base titrated, after dissolution, requires 40 mL of semimolar H2SO4 solution. If Kf of water is 1.86 K kg mol-1, then select the correct sattements (s) out of the following (assuming molarity = molality)
Options
A.The pH of the solution of weak base is 12.9.
B.The ionisation constant of the base is 8 × 10-3.
C.he osmotic pressure of the aqueous solution of base is 21.67 atm
D.The base is 10% ionized in aqueous solution.
Solution
Calculation of molecular weight of BOH:
Eq of base = Eq of H2SO4 used
× 1 = 40 × 10-3 × 
× 2(n factor)
∴ Mw of BOH = 12.4 gmol-1
Now: ᐃfT = iKfm
0.165 = i × 1.86 ×
= i × 1.86 × 0.8 (So m = 0.8)
∴ i =
= 1.1
For weak base: 1 + α = 1.1, ⇒ α = 0.1 = 10% (Answer d)
Thus [OH-] = Cα = 0.8 × 0.1 = 8 × 10-2,
∴ pOH = 1.1, pH = 12.9 (Answer a)
So, Kb = Cα2 = 0.8 × (0.1)2 = 8 × 10-3, (Answer b)
Also; π = iMRT
= 1.1 × 0.8 × 0.0821 × 300
= 21.67 atm (Answer c)
Eq of base = Eq of H2SO4 used
× 1 = 40 × 10-3 × × 2(n factor)∴ Mw of BOH = 12.4 gmol-1
Now: ᐃfT = iKfm
0.165 = i × 1.86 ×
= i × 1.86 × 0.8 (So m = 0.8)∴ i =
= 1.1 For weak base: 1 + α = 1.1, ⇒ α = 0.1 = 10% (Answer d)
Thus [OH-] = Cα = 0.8 × 0.1 = 8 × 10-2,
∴ pOH = 1.1, pH = 12.9 (Answer a)
So, Kb = Cα2 = 0.8 × (0.1)2 = 8 × 10-3, (Answer b)
Also; π = iMRT
= 1.1 × 0.8 × 0.0821 × 300
= 21.67 atm (Answer c)
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