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For a weak monobasic acid, pKa = 4 then van’t Hoff factor for 0.01 M acid is
  • a)
    1.01
  • b)
    1.02
  • c)
    1.10
  • d)
    1.20
Correct answer is option 'C'. Can you explain this answer?
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For a weak monobasic acid, pKa = 4 then vant Hoff factor for 0.01 M ac...
Explanation:
The formula for vant Hoff factor (i) is given as:
i = (1 + α - β)
where α is the degree of dissociation and β is the degree of association.

For a weak monobasic acid, the dissociation equation is as follows:
HA ⇌ H+ + A-

α = [H+] / [HA]
Since the concentration of acid is given as 0.01 M, the concentration of [HA] is also 0.01 M.
Let x be the concentration of [H+] at equilibrium.
α = x / 0.01
pKa = -logKa
Ka = 10^-pKa
Ka = [H+][A-] / [HA]
Substituting the values and simplifying, we get:
Ka = x^2 / (0.01 - x)
Since the acid is weak, we can assume that x is much smaller than 0.01. Hence, we can simplify the equation as follows:
Ka ≈ x^2 / 0.01
pKa = -logKa
4 = -log(x^2 / 0.01)
x = 10^-2
α = x / 0.01
α = 0.01 / 0.01
α = 1

For a weak monobasic acid, the degree of association (β) is negligible. Hence, we can assume β = 0.

i = (1 + α - β)
i = (1 + 1 - 0)
i = 2
The vant Hoff factor for the given acid is 2.

However, the question asks for the vant Hoff factor for a 0.01 M solution of the acid. Since the question implies that the acid does not fully dissociate in solution, we need to consider the effect of association on the vant Hoff factor. The actual value of the vant Hoff factor will be less than 2.

The formula for the apparent molar mass (Mapp) is given as:
Mapp = (i / α) x M
where M is the true molar mass of the solute.

For a weak monobasic acid, we can assume that the true molar mass (M) is approximately equal to the molar mass of the acid molecule. Let us assume the molar mass of the acid to be 100 g/mol.

Mapp = (2 / 1) x 100
Mapp = 200 g/mol

The apparent molar mass of the acid is 200 g/mol.

The formula for the degree of association (β) is given as:
β = (1 - α) / i
β = (1 - 1) / 2
β = 0.5

The degree of association for the given acid is 0.5.

The formula for the vant Hoff factor (i') is given as:
i' = (1 + α - β)
i' = (1 + 1 - 0.5)
i' = 1.5

The vant Hoff factor for the given acid in a 0.01 M solution is 1.5.

However, the question asks for the answer in terms of the vant Hoff factor, not the apparent vant Hoff factor. Hence, we need to divide the apparent vant Hoff factor by the degree of association.

i = i' / (1 - β
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For a weak monobasic acid, pKa = 4 then vant Hoff factor for 0.01 M acid isa)1.01b)1.02c)1.10d)1.20Correct answer is option 'C'. Can you explain this answer?
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